Answer:
2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.
12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution
Explanation:
First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:
Ca: 40 g/moleF: 19 g/moleSo the molar mass of CaF₂ is:
CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole
Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?
[tex]moles=\frac{0.0016 grams*1 mole}{78 grams}[/tex]
moles=2.05*10⁻⁵
2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.
Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?
[tex]mass of CaF_{2}=\frac{1000 mL*1g}{1mL}[/tex]
mass of CaF₂= 1000 g
Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?
[tex]moles=\frac{1000 grams*1 mole}{78 grams}[/tex]
moles=12.82
12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution
A metal M react with sulphur ro form MS. If 3.6g of M reacts with 0.09mol of sulphur to form MS . what is the name of M.
Answer:
Calcium
Explanation:
We must first put down the reaction equation as this will serve as a guide in solving the question.
M(s) + S(s) -----> MS(s)
According to the reaction equation; 1 mole of metal reacts with 1 mole of sulphur. Hence 0.09 moles of sulphur reacts with 0.09 moles of metal.
Now recall that;
Number of moles (n) = mass(m)/ molar mass(M)
Since
mass of metal reacted= 3.6g
Number of moles of metal= 0.09 moles
Then;
Making molar mass of metal the subject of the formula;
Molar mass of metal = mass of metal / number of moles of metal
Molar mass of metal = 3.6g /0.09 moles
Molar mass of metal= 40 gmol-1
The metal having a molar mass of 40gmol-1 is calcium, therefore the metal is calcium.
Determine the expected diffraction angle for the first-order diffraction from the (111) set of planes for FCC nickel (Ni) when monochromatic radiation of wavelength 0.1937 nm is used. The lattice parameter for Ni is 0.3524 nm
Answer:
56°
Explanation:
First calculate [tex]a:[/tex]
[tex]a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}[/tex]
The interplanar spacing can be calculated from:
[tex]d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}[/tex]
The diffraction angle is determined from:
[tex]\sin \theta=\frac{n \lambda}{2 d_{111}}=\frac{1(0.1927)}{2(0.2035)}=0.476[/tex]
Solve for [tex]\theta[/tex]
[tex]\theta=\sin ^{-1}(0.476)=28^{\circ}[/tex]
The diffraction angle is:
[tex]2 \theta=2\left(28^{\circ}\right)=56^{\circ}[/tex]
A 38.8 gram piece of metal absorbs 181J as it temperature increases from 25.0 degree celsius to 36.0 degree celsius. What is the specific heat of the metal?
__________________________J/g degree C
Answer: 0.424 J/g°C
Explanation:
For this problem, we would have to manipulate the equaiton for heat, q=mCT. Specific heat is the C in the equation. Since we are looking for specific heat, we manipulate the equation so that it says C=.
[tex]C=\frac{q}{m(deltaT)}[/tex]
*I didn't know how to type in delta so I just wrote the word delta, but pretend you see a Δ.
Now that we have our equation, we can plug in our values and solve.
[tex]C=\frac{181J}{(38.8g)(36-25°C)}[/tex]
*Please ignore the capital A in the equation. It pops up every time I type in the ° sign.
[tex]C=0.424J/g°C[/tex]
A 1.8 g mass of fructose is added to 0.100 kg of water and it is
found that the freezing point has decreased by 0.186 °C. Given
that the Kf value of water is 1.86 °C kg/mol, what is the molar
mass of fructose (van't Hoff factor, i = 1)?
g/mol
Round your answer to the nearest whole number. Do not include units in your answer
The molar mass of fructose will be "180 g/mol". To understand the calculation check below.
Molar mass and MolalityAccording to the question,
Fructose mass = 1.8 g
Water's mass = 0.100 kg
Molal freezing point depression constant, [tex]K_f[/tex] = 1.86°C/m
Freezing point change, Δ[tex]T_f[/tex] = 0.186°C
Freezing point constant, [tex]K_f[/tex] = 1.86°C/m
We know the relation,
→ Δ[tex]T_f[/tex] = i × [tex]K_f[/tex] × m
or,
= i × [tex]K_f[/tex] × [tex]\frac{Fructose \ mass}{Fructose \ molar \ mass\times Solvent's \ mass}[/tex]
By substituting the values, we get
0.186 = 1 × 1.86 × [tex]\frac{1.8}{Fructose \ molar \ mass\times 0.100}[/tex]
By applying cross-multiplication,
Molar mass = 180 g/mol
Thus the above approach is right.
Find out more information about molar mass here:
https://brainly.com/question/24727597
A 1.8 g mass of fructose is added to 0.100 kg of water and it is found that the freezing point has decreased by 0.186 °C. The molar mass of fructose is 180 g/mol.
Given:
ΔTf = 0.186 °C
Kf = 1.86 °C kg/mol
m = mass of fructose/mass of water
mass of fructose = 1.8 g
mass of water = 0.100 kg = 100 g
Use the equation:
ΔTf = Kf mi
Where:
ΔTf is the change in freezing point (in °C)
Kf is the cryoscopic constant of water (in °C kg/mol)
m is the molality of the solute (in mol/kg)
i is the van't Hoff factor
m = (mass of fructose) / (mass of water)
m = 1.8 g / 100 g
m = 0.018 mol/kg
Rearrange the equation to solve for the molar mass (M):
ΔTf = Kfmi
Substitute the values in the above equation:
0.186 = 1.86 × 0.018 × 1 / M
By applying cross multiplication,
Molar mass (M) = 180 g/mol
To learn more about the molar mass, follow the link:
https://brainly.com/question/12127540
#SPJ6
Draw a mechanism for the reaction of methylamine with 2-methylpropanoic acid. Draw any necessary curved arrows. Show the products of the reaction. Include any nonzero formal charges and all lone pairs of electrons. Indicate which side of the reaction is favored at equilibrium.
Answer:
See figure 1
Explanation:
On this case we have a base (methylamine) and an acid (2-methyl propanoic acid). When we have an acid and a base an acid-base reaction will take place, on this specific case we will produce an ammonium carboxylate salt.
Now the question is: ¿These compounds can react by a nucleophile acyl substitution reaction? in other words ¿These compounds can produce an amide?
Due to the nature of the compounds (base and acid), the nucleophile (methylamine) doesn't have the ability to attack the carbon of the carbonyl group due to his basicity. The methylamine will react with the acid-producing a positive charge on the nitrogen and with this charge, the methylamine loses all his nucleophilicity.
I hope it helps!
How do I calculate the moles consumed in a vinegar titration?
Answer: Use the titration formula
Explanation:
When 50.0 g of nitrogen react with excess hydrogen to form ammonia gas, 164.5 kJ of heat are liberated (released) at standard state conditions. Calculate the standard enthalpy of formation (in kJ/mol) for ammonia gas.
Answer:
THE STANDARD ENTHALPY OF FORMATION OF AMMONIA GAS IS 293.75kJ OF HEAT.
Explanation:
To solve this question, you must first write out the equation for the reaction.
Equation:
N2 (g) + 3H2(g) <-------> 2NH3(g)
So therefore, when 50 g of N2 reacts, 164.5 kJ of Heat was liberated.
First equate the number of moles of Nitrogen and ammonia gas
1 mole of N2 produces 2 moles of ammonia
Calculate the molar mass of each variables:
Molar mass of N2 = 14*2 = 28 g/mol
Molar mass of ammonia = ( 14 + 1*3) = 17 g/mol
So, 1 mole of N2 = 2 moles of NH3
28 g/mol of N2 = 17 * 2 g/mol of NH3
If 50 g of nitrogen was used to react with excess hydrogen, the mass of ammonia formed is;
28 g of N2 = 34 g/mol of NH3
50 g of N2 = ( 50 * 34 / 28 ) g of NH3
= 1700 / 28
= 60 .71 g of ammonia.
At standard conditions, 34 g of ammonia will liberate 164.5 kJ of heat. What amonut would be generated by 60.71 g of ammonia?
34 g of ammonia = 164.5 kJ of heat
60.71 g of ammonia = ( 60.71 * 164.5 / 34) kJ of heat
= 9987.5 / 34
= 293.75 kJ of heat.
In other words, the standard enthalpy of formulation for ammonia gas is 293.75 kJ of heat.
Steam reforming of methane (CH4 ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a tank with of methane gas and of water vapor, and when the mixture has come to equilibrium measures the amount of carbon monoxide gas to be 18 mol . Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.
The question is incomplete, here is the complete question:
Steam reforming of methane [tex](CH_4)[/tex] produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125 L tank with 20 mol of methane gas and 10 mol of water vapor at 38 degrees celsius. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of hydrogen gas to be 18 mol . Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.
Answer: The equilibrium constant for the reaction is [tex]3.99\times 10^{-2}[/tex]
Explanation:
We are given:
Initial moles of methane gas = 20 moles
Initial moles of water vapor = 10 moles
Equilibrium moles of carbon monoxide = 18 moles
Volume of the tank = 125 L
The chemical equation for the reaction of methane and water vapor follows:
[tex]CH_4(g)+H_2O(g)\rightleftharpoons CO(g)+3H_2(g)[/tex]
Initial: 20 10
At eqllm: 20-x 10-x x 3x
Evaluating the value of 'x':
[tex]\Rightarrow 3x=18\\x=6[/tex]
So, equilibrium moles of methane gas = (20 - x) = [20 - 6] = 14 mol
Equilibrium moles of water vapor = (10 - x) = [10 - 6] = 4 mol
Equilibrium moles of carbon monoxide gas = x = 6 mol
The expression of equilibrium constant for the above reaction follows:
[tex]K_{eq}=\frac{[H_2]^3[CO]}{[CH_4][H_2O]}[/tex]
We are given:
[tex][H_2]=\frac{18}{125}=0.144M[/tex]
[tex][CO]=\frac{6}{125}=0.048M[/tex]
[tex][CH_4]=\frac{14}{125}=0.112M[/tex]
[tex][H_2O]=\frac{4}{125}=0.032M[/tex]
Putting values in above expression, we get:
[tex]K_{eq}=\frac{(0.144)^3\times 0.048}{0.112\times 0.032}\\\\K_{eq}=3.99\times 10^{-2}[/tex]
Hence, the equilibrium constant for the reaction is [tex]3.99\times 10^{-2}[/tex]
A gas at 350 K and 12 atm has a molar volume 12 per cent larger than that calculated from the perfect gas law. Calculate (i) the compression factor under these conditions and (ii) the molar volume of gas. Which are dominating in the sample, the attractive or repulsive forces
Answer:
(i)The compression factor is 1.12
(ii) The molar volume of the gas is 2.68 L/mol
Since the compression factor is greater than 1, the attractive forces are dominating.
Explanation: Please see the attachments below
Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Show the formal charges of all atoms in the correct structure. To change the symbol of an atom, double-click on the atom and enter the letter of the new atom for:SeO2 and CO2?3
Answer:
See explanation below
Explanation:
In this case, let's see both molecules per separate:
In the case of SeO₂ the central atom would be the Se. The Se has oxidation states of 2+, and 4+. In this molecule it's working with the 4+, while oxygen is working with the 2- state. Now, how do we know that Se is working with that state?, simply, let's do an equation for it. We know that this molecule has a formal charge of 0, so:
Se = x
O = -2
x + (-2)*2 = 0
x - 4 = 0
x = +4.
Therefore, Selenium is working with +4 state, the only way to bond this molecule is with a covalent bond, and in the case of the oxygen will be with double bond. See picture below.
In the case of CO₂ happens something similar. Carbon is working with +4 state, so in order to stabilize the charges, it has to be bonded with double bonds with both oxygens. The picture below shows.
Calculate ΔHrxnΔHrxn for the following reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l)CH4(g)+2O2(g)→CO2(g)+2H2O(l) Use the following reactions and given ΔHΔH values. CH4(g)+O2(g)→CH2O(g)+H2O(g)CH4(g)+O2(g)→CH2O(g)+H2O(g), ΔH=−ΔH=−284 kJkJ CH2O(g)+O2(g)→CO2(g)+H2O(g)CH2O(g)+O2(g)→CO2(g)+H2O(g), ΔH=−ΔH=−527 kJkJ H2O(l)→H2O(g)H2O(l)→H2O(g), ΔH=ΔH= 44.0 kJ
Answer: the enthalpy of reaction is, -155 kJ
Explanation:-
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The final reaction is:
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)[/tex] [tex]\Delta H_{rxn}=?[/tex]
The intermediate balanced chemical reaction will be,
(1) [tex]CH_4(g)+O_2(g)\rightarrow CH_2O(g)+H_2O(g)[/tex] [tex]\Delta H_1=-284kJ[/tex]
(2) [tex]CH_2O(g)+O_2(g)\rightarrow CO_2(g)+H_2O(g)[/tex] [tex]\Delta H_2=-527kJ[/tex]
(3) [tex]H_2O(l)\rightarrow H_2O(g)[/tex] [tex]\Delta H_3=44.0kJ[/tex]
Now multiplying (3) by 2 and adding all the equations, we get :
[tex]\Delta H_{rxn}=\Delta H_1+\Delta H_2+2\times \Delta H_3[/tex]
[tex]\Delta H_{rxn}=(-284)+(-527)+2\times (44)[/tex]
[tex]\Delta H_{rxn}=-155kJ[/tex]
Therefore, the enthalpy of reaction is, -155 kJ
Which is the best method for collection of co2
Answer:
The best way is to burn fuel with oxygen instead of airConsider four beakers labeled A, B, C, and D, each containing an aqueous solution and a solid piece of metal. Identity the beakers in which a chemical reaction will occur and those in which no reaction will occur. Drag each item to the correct bin.
1. Mn(s) Ca(NO3)2(aq)
2. KOH(aq) Fe(s)
3. Pt(NO3)2(aq) Cu(s)
4. Cr(s) H2SO4(aq)
A. Reaction
B. Non-reaction
Consider the following data for five hypothetical element: Q, W, X, Y, and Z. Rank the elements from most reactive to least reactive.
Combination Observation of reaction
Q + W+ Reaction occurs
X + Z+ No reaction
W + Z+ Reaction occurs
Q+ + Y Reaction occurs
Place the element symbol from most to least reactive.
1. Q
2. W
3. X
4. Y
5. Z
A. Most reactive
B. Least reactive
Use the following reactions to arrange the elements A, B, C, and D in order of their rodox reactivity from most reactive to least reactive. Rank from most reactive to least reactive.
1. A + D+ righ tarrow A+ + D
2. C+ + D righ tarrow C + D+
3. B+ + D righ tarrow B + D+
4. B + C+ righ tarrow B+ + C
A. Most reactive
B. Least reactive
Answer:
A Reaction
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
B Non Reaction
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
Y > Q > W > Z > X
Explanation:
The first question is whether a reaction will occur base on the chemical equation below.
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
Firstly, some element are more reactive than others , base on this criteria element can be arranged base on it reactivity .
1. Mn(s) + Ca(NO₃)₂(aq)
This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.
2. KOH(aq) + Fe(s)
The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.
3. Pt(NO₃)₂(aq) + Cu(s)
Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.
4. Cr(s) + H₂SO₄(aq)
Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.
Base on the reaction
Q + W+ Reaction occurs
Since the reaction occurred element Q is more reactive as it displace element w from it compound.
X + Z+ No reaction
No reaction occurred because element x is less reactive than z therefore, it cannot displace z from it compound.
W + Z+ Reaction occurs
Element w is more reactive than z as it displaces z form it compound.
Q+ + Y Reaction occurs
Element Y is more reactive than element Q as it displaces Q from it compound.
Therefore, the order of reactivity from the most reactive to the least reactive will be Y > Q > W > Z > X
A. The beakers in which a chemical reaction will occur:
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
B. The beakers in which there is no reaction:
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
C. The elements from most reactive to least reactive is:
Y > Q > W > Z > X
Solving for each part:A.
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
Elements can be arranged on the basis of reactivity:
1. Mn(s) + Ca(NO₃)₂(aq)
This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.
2. KOH(aq) + Fe(s)
The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.
3. Pt(NO₃)₂(aq) + Cu(s)
Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.
4. Cr(s) + H₂SO₄(aq)
Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.
According to reactions:
Q + W→ Reaction occurs
Since the reaction occurred element Q is more reactive as it displace element w from it compound.
X + Z →No reaction
No reaction occurred because element X is less reactive than Z therefore, it cannot displace z from it compound.
W + Z→ Reaction occurs
Element W is more reactive than Z as it displaces Z form it compound.
Q + Y →Reaction occurs
Element Y is more reactive than element Q as it displaces Q from it compound.
Thus, the order of reactivity from the most reactive to the least reactive will be: Y > Q > W > Z > X
Find more information about Reactivity series here:
brainly.com/question/17469010
Which type of semiconductor is created by doping with atoms that contain more valence electrons than the semiconductor material ?
Answer:
N-Type Semiconductor
Explanation:
When I was a boy, Uncle Wilbur let me watch as he analyzed the iron content of runoff from his banana ranch. A 25.0-mL sample was acidified with nitric acid and treated with excess KSCN to form a red complex. (KSCN itself is colorless.) The solution was then diluted to 100.0 mL and put in a variable-path length cell. For comparison, a 10.0-mL reference sample of 6.74 times 10^-4 M Fe^3+ was treated with HNO_3 and KSCN and diluted to 50.0 mL, The reference was placed in a cell with a 1.00-cm light path. The runoff sample exhibited the same absorbance as the reference when the path length of the runoff cell was 2.41 cm. What was the concentration of iron in Uncle Wilbur's runoff?
Answer:
C = 2.24x10⁻⁴ M
Explanation:
The concentration of iron in Uncle Wilbur's runoff can be calculated using Beer-Lambert law:
[tex] A = \epsilon*C*l [/tex] (1)
Where:
A: is the absorbance of the compound
ε: is the molar absorptivity of the compound
C: is the concentration of the compound
l: is the optical path length
Since the runoff sample exhibited the same absorbance as the reference sample, we can find the concentration using equation (1):
[tex] \epsilon*C_{1}*l_{1} = \epsilon*C_{2}*l_{2} [/tex] (2)
Where:
Subscripst 1 and 2 refer to Uncle Wilbur's runoff and to reference sample, respectively.
l₁ = 2.41 cm
l₂ = 1.00 cm
We can find C₂ as follows:
[tex] C_{2} = \frac{C_{2i}*V_{i}}{V_{f}} [/tex] (3)
Where:
[tex]C_{2i}[/tex]: is the initial concentration of the reference sample = 6.74x10⁻⁴ M
[tex]V_{i}[/tex]: is the initial volume = 10.0 mL
[tex]V_{f}[/tex]: is the final volume = 50.0 mL
[tex] C_{2} = \frac{6.74 \cdot 10^{-4} M*10.0 mL}{50.0 mL} = 1.35 \cdot 10^{-4} M [/tex]
Now, we can find C₁ using equation (2):
[tex] C_{1} = \frac{C_{2}*l_{2}}{l_{1}} = \frac{1.35 \cdot 10^{-4} M*1.00 cm}{2.41 cm} = 5.60 \cdot 10^{-5} M [/tex]
Finally, since the runoff solution was diluted to 100.0 mL, the initial concentration can be calculated using equation (3) for [tex]C_{1i}[/tex]:
[tex]C_{1i} = \frac{C_{1}*V_{f}}{V_{i}} = \frac{5.60 \cdot 10^{-5} M*100.0 mL}{25.0 mL} = 2.24 \cdot 10^{-4} M[/tex]
Therefore, the concentration of iron in Uncle Wilbur's runoff is 2.24x10⁻⁴ M.
I hope it helps you!
Consider the reaction below. 2C6H14 + 19O2 Right arrow. 12CO2 + 14H2O How many moles of hexane (C6H14) must burn to form 18.4 mol of carbon dioxide?
Answer:
[tex]3.06mol~C_6H_14[/tex]
Explanation:
We have to start with the reaction:
[tex]2C_6H_1_4 + 19O_2->12CO_2 + 14H_2O[/tex]
We have 2 carbons, 6 hydrogens and 38 oxygens on both sides.
Now the molar ratio between [tex]C_6H_14[/tex] and [tex]CO_2[/tex] is 2:12 or [tex]2~mol~C_6H_14=~12~mol~CO_2[/tex]. With this in mind we can calculate the moles of [tex]C_6H_14[/tex], so:
[tex]18.4~mol~CO_2\frac{2~mol~C_6H_14}{12~mol~CO_2}=3.06mol~C_6H_14[/tex]
I hope it helps!
Answer:
B) 3.07 mol
Explanation:
Analyses of enzymes found in the blood are used as indicators that tissue damage (heart, liver, muscle) has occurred and resulted in the leakage of cellular enzymes into the bloodstream. One of the common enzymes measured for this purpose is the one that interconverts lactate and pyruvate. Identify the name of this enzyme.
Answer:
Lactate dehydrogenase (LDH)
Explanation:
LDH is tetrameric enzyme found in the muscles (M-type) and the heart (H-type) of living cells, responsible for the conversion of pyruvate to lactate and back, to promote generation of nicotinamide adenine dinucleotide (NAD).
What is the concentration of a 34% solution converted to parts per million (ppm).?
34,000,000 ppm
340.000 ppm
34,000 ppm
3,400 ppm
Answer:
340.000 ppm
Explanation:
Parts per million (ppm) is a unit of measure for concentration that measures the number of units of substance per million units of the set. It is a concept homologous to the percentage, only in this case it is not parts per percent but per million. The ppm calculation method is different for solids, liquids and gase. There are many formulas, however in this case, you can use this fact:
[tex]10000ppm=1\%[/tex]
Using Cross-multiplication:
[tex]\frac{10000ppm}{x} =\frac{1\%}{34\%}[/tex]
Solving for [tex]x[/tex] :
[tex]x=34*10000=340000ppm[/tex]
Therefore, the concentration of a 34% solution converted to parts per million is:
[tex]340000ppm[/tex]
A sample tube consisted of atomic hydrogen in their ground state. A student illuminated the atoms with monochromatic light, that is, light of a single wavelength. If only two separate emission lines in the visible region are observed, what is the wavelength (or wavelengths) of the incident radiation?
Answer:
The wavelength of the monochromatic light is 486.2 nm.
Explanation:
The illumination of the hydrogen atom by the monochromatic light causes an absorption of energy by its electrons which causes an excitation. After a period, the particle de-excites (decays) losing the absorbed energy and falls back to its initial state releasing the energy in the form of a photon. This photon can be observed as a colored light of the Balmer series.
From Rydberg's expression,
1/λ=−R([tex]\frac{1}{n_{2} ^{2} }[/tex] − [tex]\frac{1}{n_{1} ^{2} }[/tex])
The transition of the electron is from n = 2 to 4, so that;
1/λ = R ([tex]\frac{1}{2^{2} }[/tex] - [tex]\frac{1}{4^{2} }[/tex])
= 1.097 x [tex]10^{7}[/tex] ([tex]\frac{1}{2^{2} }[/tex] - [tex]\frac{1}{4^{2} }[/tex])
1/λ = 2056875
So that,
λ = [tex]\frac{1}{2056875}[/tex]
= 4.8617 x [tex]10^{-7}[/tex] m
The wavelength of the monochromatic light is 486.2 nm.
Question 5 Tungsten is a solid phase of tungsten still unknown to science. The only difference between it and ordinary tungsten is that Tungsten forms a crystal with an fcc unit cell and a lattice constant . Calculate the density of Tungsten .
Complete Question
The complete question is shown on the first uploaded image
Answer:
The density is [tex]\rho = 21.1 \ g/cm^3[/tex]
Explanation:
From the question we are told that
The lattice constant is [tex]a = 0.387 nm = 0.387 *10^{-9} \ m[/tex]
Generally the volume of the unit cell is [tex]V = a^3[/tex]
=> [tex]V = [0.387 *10^{-9}]^2[/tex]
[tex]V = 5.796 *10^{-29} \ m^3[/tex]
Converting to [tex]cm^3[/tex] We have [tex]5.796 *10^{-29} * 1000000 = 5.796 *10^{-23} cm^3[/tex]
The molar mass of Tungsten is constant with a value [tex]Z = 184 g/mol[/tex]
One mole of Tungsten contains [tex]6.022*10^{23}[/tex] unit cells
Where [tex]6.022*10^{23}[/tex] is a constant for the number of atom in one mole of a substance(Tungsten) which is known as Avogadro's constant
Now for FCC distance the number of atom per unit cell is n = 4
Mass of Tungsten (M) = [tex]= \frac{Z * n }{1 \ mole \ of \ Tungsten}[/tex]
=> Mass of Tungsten (M) = [tex]= \frac{184 * 4 }{6.023*10^{23}}[/tex]
=> Mass of Tungsten (M) = [tex]= 1.222*10^{-21} \ g[/tex]
Now
The density of Tungsten is
[tex]\rho = \frac{M}{V}[/tex]
substituting values
[tex]\rho = \frac{1.222*10^{-21}}{5.796*10^{-23}}[/tex]
[tex]\rho = 21.1 \ g/cm^3[/tex]
Classify each amino acid according to whether its side chain is predominantly protonated or deprotonated at a pH of 7.40. The pKa values of the Asp, His, and Lys side chains are 3.65, 6.00, and 10.53, respectively.
Answer and Explanation:
7.40 becomes lower or shorter than the pKa of functional groups or chains of all the 3 amino acids respectively Lys, His, as well as Asp, although since pH provided throughout the statement. However, all of the covalent chains or bonds of the three compounds or acids (amino) will serve a purpose in the pro-toned pattern.
The pH during which everyone's binding sites will mostly keep track of customer for someone's comparison is:
Lys:
[tex]pH > 10.53[/tex]
His:
[tex]pH > 6.00[/tex]
Asp:
[tex]pH > 3.65[/tex]
2NO2(g)= 2NO(g) + O2(g)
Nitrogen dioxide is dissociated to the extent of 56.6 percent and 494°C and 99kPa pressure. At what pressure will the dissociation be 80 percent at 494°C?
Answer:
AT THE END OF 80% DISSOLUTION, THE PRESSURE OF NO2 HAS CHANGED FROM 99kPa TO 139.97kPa
Explanation:
P1 = 99 kPa
P2 = unknown
From the reaction,
2 mole of NO2 will produce 2 mole of NO
We can also say that 1 mole of NO2 will produce 1 mole of NO
At 56.6 % of NO2, 0.566 mole of NO2 will be consumed
At STP, 1 mole of a substance will occupy 22.4 dm3 volume
0.566 mole will occupy ( 22.4 * 0.566 / 1) dm3 volume
= 39.58 dm3 volume
V1 = 39.56 dm3
At the new percent of 80%, 0.80 mole of NO2 will be consumed
Since, 1 mole = 22.4 dm3
0.80 mole = (22.4 / 0.80) dm3
= 28 dm3
V2 = 28 dm3
Using the equation of Boyle's law which shows the relationship between pressure and volume of a given mass of gas at constant temperature, we have:
P1 V1 = P2 V2
Re-arranging to make P2 the subject of formula:
P2 = P1V1 / V2
P2 = 99 kPa * 39.56 / 28
P2 = 3916.44 kPa / 28
P2 = 139.87 kPa
So at 80 % dissociation of NO2, the pressure has changed from 99 kPa to 139.97 kPa.
Consider the experimental data for multiple experiments with different pressures and volumes of gases.
Experiment PV Value (Pa-mL)
1 1.10 x 106
2 1.14 x 106
3 1.18 x 106
4 1.16 x 106
5 1.19 x 106
Boyle's law that pressure and volume are inversely related because___________.
Answer: Boyle's law states that pressure and volume are inversely related because the product of pressure and volume is approximately constant.
Explanation:
In each of the given values product of P and V, that is, PV remains constant. As according to Boyle's law, at constant temperature the pressure will be inversely proportional to volume of the gas.
Mathematically, [tex]P \propto \frac{1}{V}[/tex]
or, PV = k
where, k = Proportionality constant
P = pressure of gas
V = volume of gas
Product of values will only remain constant when pressure is inversely proportional to volume.
Therefore, we can conclude that Boyle's law states that pressure and volume are inversely related because the product of pressure and volume is approximately constant.
Boyle's law mentioned that pressure and volume should be inversely related since the product of pressure and volume is approximately constant.
Boyle's lawAccording to this law, at a constant temperature, the pressure should be inversely proportional to the volume of the gas.
PV = k
Here
k = Proportionality constant
P = Pressure
V = volume of gas
So based on this we can conclude that Boyle's law mentioned that pressure and volume should be inversely related since the product of pressure and volume is approximately constant.
Learn more about volume here: https://brainly.com/question/21960832
A 1.34 mole sample of LiCl dissolves in water, The volume of the final
solution is 0.86L. Find the Molarity of the Solution.
Answer:
[tex]1.56 \,\,mol/L[/tex]
Explanation:
Molarity of the solution measures number of moles of a solute per litre of solution.
Molarity = volume of solution in litres/number of moles of solute dissolved in solution
Volume of solution in litres = 0.86 L
Also, 1.34 mole sample of LiCl dissolves in water
So,
Molarity of the Solution = [tex]\frac{1.34}{0.86}=1.56 \,\,mol/L[/tex]
How many moles of NaOH will be required to produce 375.4 g of any Na2S04
Answer:
5.28 moles of NaOH.
Explanation:
Step 1:
Determination of the number of mole in
375.4 g of any Na2S04.
Molar mass of Na2SO4 = (2x23) + 32 + (16x4) = 142g
Mass of Na2SO4 = 375.4g
Number of mole of Na2SO4 =..?
Mole = Mass /Molar Mass
Number of mole of Na2SO4 = 375.4/142 = 2.64 moles
Step 2:
The balanced equation for the reaction.
H2SO4 + 2NaOH —> Na2SO4 + 2H2O
Step 3:
Determination of the number of mole of NaOH needed for the reaction.
From the balanced equation above,
2 moles of NaOH reacted to produce 1 mole of Na2SO4.
Therefore, Xmol of NaOH will react to produce 2.64 moles of Na2SO4 i.e
Xmol of NaOH = 2 x 2.64
Xmol of NaOH = 5.28 moles
Therefore, 5.28 moles of NaOH is needed for the reaction.
Write condensed and bond-line structural formulas for all of the constitutional isomers Practice problem 4.1 with the molecular formula C7H16. (There are a total of nine constitutional isomers.)
Answer:
See figure 1.
Explanation:
In this case have to take into account that all structures must have the formula: [tex]C_7H_16[/tex]. If we remember the general formula for alkanes: [tex]C_nH_2_n_+_2[/tex] if we have "7" carbons (n=7) we will have 16 hydrogens. Therefore all the structures that fit with this formula are alkanes.
A 1424 gram sample of a liquid at an initial temperature of 30.0 degrees C absorbs 1560 J of heat. Given the specific heat of 2.44 J/g degree C, what is the final temperature of the liquid?
____________________ degree C
Answer:
THE FINAL TEMPERATURE OF THE LIQUID SAMPLE IS 30.45 DEGREE CELSIUS
Explanation:
Mass of the liquid sample = 1424 g
Initail temperature = 30 degree C
Heat evolved = 1560 J
Specific heat of the liquid = 2.44 J/g degree C
Final temperature = unknown
Since the heat evolved by a substance is the product of the mass, specific heat capacity and the change in temperature of the sample
Heat = Mass * Specific heat * change in temp.
H = m c (T2-T1)
Re-arranging the formula by making T2 (final temperature) the subject of the equation, we have:
T2= H/ m c + T1
So therefore, introducing the value of the variables and solving for T2, we have:
T2 = 1560 / 1424 * 2.44 + 30
T2 = 1560 / 3474.56 + 30
T2 = 0.4487 + 30
T2 = 30.4487 degree C
The final temperature of the liquid sample is approximately 30.45 degree C
Aqueous solutions of barium chloride and silver nitrate are mixed to form solid silver chloride and aqueous barium nitrate. The complete ionic equation contains which of the following species (when balanced in standard form)?
A. NO (aq)
B. 2Ba (aq)
C. 2Ag (aq)
D. CI(aq)
Answer:
Option C
Explanation:
Consider the ionic equation of this chemical equation. We are given barium chloride and silver nitrate as the reactants, and silver chloride and barium nitrate as the products. We can thus conclude that the ionic equation ( not balanced yet ) should be as follows -
Ba( 2 + ) + Cl ( - ) + Ag ( + ) + NO3 ( - ) ------> AgCl + Ba( 2 + ) + NO3( - )
As you can see these compounds are present in aqueous solutions, and are thus dissociated.
______________________________________________________
Now let us take a look at the number of elements on the reactant and product sides, and balance this chemical equation out -
Ba( 2 + ) + 2Cl ( - ) + 2Ag ( + ) + 2NO3 ( - ) ------> 2AgCl + Ba( 2 + ) + 2NO3( - )
Solution = Option C!
A reaction is at equilibrium when...
A. the amounts of the reactants and products are not longer changing.
B. the reactants have all turned into products.
C. the products have all turned into reactants.
D. the amounts of the reactants and products are equal.
Answer:
Option A. is correct
Explanation:
A reaction is at equilibrium when the amounts of the reactants and products are no longer changing.
Chemical equilibrium means that the rate of formation of products is equal to the rate at which the products re-form reactants.
Products are formed by the forward reaction and by the reverse reaction, the products re-form reactants.
Classify each titration curve as representing a strong acid titrated with a strong base, a strong base titrated with a strong acid, a weak acid titrated with a strong base, a weak base titrated with a strong acid, or a polyprotic acid titrated with a strong base. A. Strong acid/ strong base.B. Strong base/ strong acid.C. Weak acid/ strong base.D. Weak base/ strong acid.E. Polyprotic acid/ strong base.
Answer:
Acid base titration curves shows the pH at equivalence point
Explanation:
Since the images were not shown, I will proceed to give a general description of the following acid-base titration curves:
In a strong acid-strong base titration, the acid and base will react to form a neutral solution. At the equivalence point of the reaction, hydronium (H+) and hydroxide (OH-) ions will react to form water, leading to a pH of 7.
The titration curve reflects the strengths of the corresponding acid and base. If one reagent is a weak acid or base and the other is a strong acid or base, the titration curve is irregular, and the pH shifts less with small additions of titrant near the equivalence point.
Polyprotic acids are able to donate more than one proton per acid molecule, in contrast to monoprotic acids that only donate one proton per molecule. In the titration curve of a polyptotic acid and a strong base, The curve starts at a higher pH than a titration curve of a strong base. There is always a steep climb in pH before the first midpoint. Gradually, the pH increases until it passes the midpoint; Right before the equivalence point there is a very sharp increase in pH.
Answer:
awzsexdrcftvgybhunimoknljbkhvjgchfx
Explanation: