Explanation:
a) E = P × t
E = 750 × 1 s = 750 J
b) E = P × t
E = 750 × 3s = 2250 J
c) E = P × t
1500 = 750 × t
t = 1500/750
t = 2 s
Since some amino acids, such as Glu and His, can be viewed as weak acids or bases, the ratio of their protonated form to their deprotonated form is dependent on pH. Furthermore, you can use this knowledge to predict whether a given amino acid will exist predominantly in its protonated or deprotonated form at a given pH. To do this, it is easier to think about the pK, of the amino acid functional group. Recall from general chemistry that the pK, is equal to -log(K). Using the K, value for the Glu side chain, which is 5.62 x 10-5, calculate the pK, for the Glu side chain. pKa =
Answer:
The pKa of Glu is [tex]pK_a = 4.25[/tex]
Explanation:
From the question we are told that
The [tex]K_a[/tex] value of Glu is [tex]K_a = 5.62*10^{-5}[/tex]
The [tex]pKa[/tex] is mathematically represented as
[tex]pK_a = -log (K_a )[/tex]
substituting values
[tex]pK_a = -log (5.62 *10^{-5} )[/tex]
[tex]pK_a = 4.25[/tex]
According to the Clausius theorem, the cyclic integral of
for a reversible cycle is zero.
Answer:
Yes it is true
Explanation:
This is because the Clausius theorem states that The cyclic integral always has two defined results. The results include it being less than or equal to zero under certain conditions.
When the system consists of only reversible processes, the cyclic integral is equal to zero. If it consists of and irreversible processes, the integral is usually less than zero.
The following initial rate data are for the reaction of mercury(II) chloride with oxalate ion: 2 HgCl2 + C2O42- 2 Cl- + Hg2Cl2 + 2 CO2
Experiment [HgCl2] [C2O42-]o, M Initial Rate
1 0.124 0.115 1.61E-5
2 0.248 0.015 3.23E-5
3 0.124 0.229 6.40E-5
4 0.248 0.229 1.28E-4
1) Complete the rate law for this reaction in the box below. Use the form k[A]m[B]n , where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n.
2) From these data, the rate constant is ____ M-2s-1?
Answer:
Explanation:
2 HgCl₂ + C₂O₄²⁻ = 2 Cl⁻ + Hg₂Cl₂ + 2CO₂
1 )
Rate of reaction [tex]= k[HgCl_2]^m[C_2O_4^{-2}]^n[/tex]
[HgCl₂] [C₂O₄²⁻ ] Rate
1 . .124 .115 1.61 x 10⁻⁵
2 . .248 .115 3.23 x 10⁻⁵
3 . .124 .229 6.4 x 10⁻⁵
4 . .248 .229 1.28 x 10⁻⁴
comparing 1 and 3 , when concentration of HgCl₂ remains constant and concentration of C₂O₄²⁻ becomes twice , rate becomes 4 times so rate is proportional to square of concentration of C₂O₄²⁻ .
Hence n = 2
comparing 1 and 2 , when concentration of HgCl₂ becomes twice and concentration of C₂O₄²⁻ remains constant , rate becomes 2 times so rate is proportional to simply concentration of C₂O₄²⁻ .
Hence m = 1
Putting the data of 1 in the rate equation found
1.61 x 10⁻⁵ = k x .124 x .115²
k = 11.3 x 10⁻⁴ M⁻² s⁻¹
. In which reaction is nitric acid acting as an oxidising agent?
A. CuO + 2HNO3 → Cu(NO3)2 + H2O
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
C. Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2
D. NaOH + HNO3 → NaNO3 + H2O. In which reaction is nitric acid acting as an oxidising agent?
A. CuO + 2HNO3 → Cu(NO3)2 + H2O
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
C. Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2
D. NaOH + HNO3 → NaNO3 + H2O
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
A substance is acting as oxidising agent when is reducing its chemical state in the reaction.
In HNO3, Nitrogen has an oxidation state of +5 (Each oxygen is -2, 3 oxygens, -6, 1 hydrogen, +1). Thus, if the oxidation state of the nitrogen in the products is < +5, the nitrogen is reducing its oxidation state acting as oxidising agent.
In the reactions:
A. CuO + 2HNO3 → Cu(NO3)2 + H2O . In Cu(NO₃)₂, Nitrogen has an oxidation state of +5. Thus, nitric acid is not acting as oxidising agent.
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 . Also, oxidation state in Cu(NO₃)₂ does not have any change but there is another product, NO₂, where nitrogen has an oxidation state of +4. That means nitric acid is acting as oxidising agent.
C. Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2 . Here, in NaNO₃, oxidation state of nitrogen is +5 (Na: +1, 3 O: -6). Thus, nitrogen is not changing is oxidation state and nitric acid is not acting as oxidising agent.
D. NaOH + HNO3 → NaNO3 + H2O. Here, also, the product is NaNO₃. That means nitric acid is not acting as oxidising agent.
How many moles of HCl can be produced from 0.226 g of SOCl2? SOCl2 + H2O ----> SO2 + 2HCl
Explanation would be helpful c:
Answer:
3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂
Explanation:
The balanced reaction is:
SOCl₂ + H₂O ----> SO₂ + 2 HCl
By stoichiometry of the reaction they react and produce:
SOCl₂: 1 moleH₂O: 1 moleSO₂: 1 moleHCl: 2 moleBeing:
S: 32 g/moleO: 16 g/moleCl: 35.45 g/moleH: 1 g/molethe molar mass of the compounds participating in the reaction is:
SOCl₂: 32 g/mole + 16 g/mole + 2*35.45 g/mole= 118.9 g/moleH₂O: 2*1 g/mole + 16 g/mole= 18 g/moleSO₂: 32 g/mole + 2*16 g/mole= 64 g/moleHCl: 1 g/mole + 35.45 g/mole= 36.45 g/moleThen, by stoichiometry of the reaction, the following amounts of mass react and are produced:
SOCl₂: 1 mole* 118.9 g/mole= 118.9 gH₂O: 1 mole* 18 g/mole= 18 gSO₂: 1 mole* 64 g/mole= 64 gHCl: 2 mole* 36.45 g/mole= 72.9 gThen the following rule of three can be applied: if by stoichiometry of the reaction 118.9 grams of SOCl₂ produce 2 moles of HCl, 0.226 grams of SOCl₂ how many moles of HCl do they produce?
[tex]moles of HCl=\frac{0.226 grams of SOCl_{2}*2 mole of HCl }{118.9 grams of SOCl_{2} }[/tex]
moles of HCl= 3.80*10⁻³
3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂
How many grams of CO2 are used when 6.0 g of O2 are produced? Express your answer with the appropriate units.
Answer:
that is why co2 is in the power of 2ik
How many moles are in 50 grams of cobalt?
help asapp
hey there!
atomic mass cobalt = 58.9332 amu
therefore:
1 mol Co ------------ 58.9332 g
moles ------------ 50 g
moles = 50 x 1 / 58.9332
moles = 50 / 58.9332
= 0.8484 moles of Co
Hope this helps!
0.8484 moles are in 50 grams of cobalt.
What are moles?A mole is defined as 6.02214076 × [tex]10^{23}[/tex] of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Atomic mass cobalt = 58.9332 amu
Therefore:
1 mol Co ------------ 58.9332 g
moles ------------ 50 g
moles = 50 x 1 ÷ 58.9332
moles = 50 ÷ 58.9332
= 0.8484 moles of Co
Learn more about moles here:
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what is a molecule containing only carbon and hydrogen called?
Answer:
d
Explanation:
it's called the hydrocarbon
Answer:
A hydrocarbon
Explanation:
The correct name for CaH2 is
Answer:
Calcium hydride
Explanation:
What functional groups are in ch2=chch2oh?
Explanation:
H
|
H-C=C-C-OH
| | |
H H H
in this organic compound the functional groups are OH and alkene
prop-2-enol
since it has a double bond, alkene is the functional group. and also it has OH group so hydroxyl group also the other functional group
Helium is a very important element for both industrial and research applications. In its gas form it can be used for welding, and since it has a very low melting point (only 0.95 K under 2.5 MPa) it can be used in liquid form to cool superconducting magnets, such as those found in particle physics experiments. Say we have a cylinder of n = 125 moles of Helium gas at room temperature (T = 20° C). The cylinder has a radius of r = 17 cm and a height h = 1.64 m.What pressure is Helium gas under?
Answer:
P = 20.1697 atm
Explanation:
In this case we need to use the ideal gas equation which is:
PV = nRT (1)
Where:
P: Pressure (atm)
V: Volume (L)
n: moles
R: universal gas constant (=0.082 L atm / K mol)
T: Temperature
From here, we can solve for pressure:
P = nRT/V (2)
According to the given data, we have the temperature (T = 20 °C, transformed in Kelvin is 293 K), the moles (n = 125 moles), and we just need the volume. But the volume can be calculated using the data of the cylinder dimensions.
The volume for any cylinder would be:
V = πr²h (3)
Replacing the data here, we can solve for the volume:
V = π * (17)² * 164
V = 148,898.93 cm³
This volume converted in Liters would be:
V = 148,898.93 mL * 1 L / 1000 mL
V = 148.899 L
Now we can solve for pressure:
P = 125 * 0.082 * 293 / 148.899
P = 20.1697 atmIn a photoelectric experiment a student uses a light source whose frequency is greater than that needed to eject electrons from a certain metal. However, after continuously shining the light on the same area of the metal for a long period of time the student notices that the maximum kinetic energy of ejected electrons begins to decrease, even though the frequency of the light is held constant. How would you account for this behaviour?
Answer:
The metal surface becomes more positive as electrons are lost from it.
Explanation:
Let us note that photoelectric effect refers to a phenomenon in which electrons are ejected from a clean metal surface irradiated with light of appropriate frequency. This photon must possess a frequency above the threshold frequency of the metal and its energy must exceed the work function of the metal. When these conditions are met, electrons are emitted from a clean metal surface, having a constant kinetic energy as long as the frequency of the incident photon remains constant.
However, as photoelectric effect progresses and electrons are lost from the metal surface, the metal surface becomes more positive. The more positive the surface, the greater the attraction of the positive surface for the emitted electrons. This reduces the kinetic energy of the emitted photons even though the frequency of incident photons is held constant.
Under which set of conditions is ΔGrxn for the reaction A(g)→B(g) most likely to be negative? PA=10.0atm ; PB=0.050atm PA=10.0atm; PB=10.0atm PA=0.050atm; PB=0.050atm PA=0.050atm; PB=10.0atm
Answer:
ΔGrxn for the reaction A(g)→B(g) will most likely to be negative under the PA=10.0atm ; PB=0.050atm condition
The FIRST option is correct
Explanation:
From Thermodynamics,
Using the Gibbs Free Energy (G) formula to determine the reaction A(g)→B(g) most likely to be negative.
Gibbs Free Energy (G) which is the energy associated with a chemical reaction that can be used to do work.
CHECK THE ATTACHMENT FOR THE DETAITALED EXPLANATION
1. Potassium (K) has an atomic mass of 39.0983 amu and only two naturally-occurring isotopes. The K-41 isotope (40.9618 amu) has a natural abundance of 6.7302%. What is the mass (in amu) of the other isotope
Answer:
38.96383282 amu
Explanation:
39.0983 = (40.9618 [tex]\times[/tex] 0.067302) + ( ? [tex]\times[/tex] (1-0.067302)
39.0983 = 2.756811064 + ( ? [tex]\times[/tex] 0.932698)
subtract 2.756811064 from both sides
36.34148894 = ( ? [tex]\times[/tex] 0.932698)
divide both sides by 0.932698
? = 38.96383282 amu
Answer:
38.96383282 amu
Explanation:
39.0983 = (40.9618 0.067302) + ( ? (1-0.067302)
39.0983 = 2.756811064 + ( ? 0.932698)
subtract 2.756811064 from both sides
36.34148894 = ( ? 0.932698)
divide both sides by 0.932698
? = 38.96383282 amu
QUICK!!!
The specific heat of mercury is 0.138 J/g Celsius. If 452 g of mercury at 85.0 Celsius are placed in 145 g of water at 23.0 Celsius, what will be the final temperature for both the mercury and the water?
Answer: Thus the final temperature for both the mercury and the water is [tex]28.8^0C[/tex]
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]heat_{released}=heat_{absorbed}[/tex]
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)][/tex]
Q = heat absorbed or released
[tex]m_1[/tex] = mass of mercury= 452 g
[tex]m_2[/tex]= mass of water = 145 g
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_1[/tex] = temperature of mercury = [tex]85.0^0C[/tex]
[tex]T_2[/tex] = temperature of water = [tex]23.0^0C[/tex]
[tex]c_1[/tex] = specific heat of mercury = [tex]0.138J/g^0C[/tex]
[tex]c_2[/tex] = specific heat of water= [tex]4.184J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]-[452\times 0.138\times (T_f-85.0)^0C]=145\times 4.184\times (T_f-23.0)^0C[/tex]
[tex]T_f=28.8^0C[/tex]
Thus the final temperature for both the mercury and the water is [tex]28.8^0C[/tex]
What product(s) are formed during the complete combustion reaction that occurs when methane (CH4) and molecular oxygen (O2) react? CO2 and H4 C and H2O CO2 and H2O C(OH)4
Answer: CO2 and H2O
Explanation: I already took the test it's right
chemical reactions occur in predictable ways
Answer:
Yes that is why you can even see how some chemicals might react but that is if you are not considering temperature, pressure, or if catalyst are involved.
4. If the DNA nitrogen bases were TACCGGAT, how would the other half of
the attached DNA strand read?
Answer:
ATGGCCTA
Explanation:
For this we have to keep in mind that we have a specific relationship between the nitrogen bases:
-) When we have a T (thymine) we will have a bond with A (adenine) and viceversa.
-) When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa.
Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:
TACCGGAT
ATGGCCTA
I hope it helps!
A booth renter is responsible for conducting all necessary business blank
Answer:
building and managing their own clients, purchasing supplies and keeping records
hope this helps you
Consider the following reaction. 2C2H2(g)+5O2(g)⟶4CO2(g)+2H2O(g) 2C2H2(g)+5O2(g)⟶4CO2(g)+2H2O(g) Compound ΔH∘f (kJ/mol)ΔHf∘ (kJ/mol) C2H2(g)C2H2(g) 227.4227.4 O2(g)O2(g) 00 CO2(g)CO2(g) −393.5−393.5 H2O(g)H2O(g) −241.8−241.8 What is the ΔH∘ΔH∘ of the reaction?
Answer:
ΔH° = -1255.8 kJ
Explanation:
Step 1: Data given
ΔHf∘ C2H2 = 227.4 kJ/mol
ΔHf∘ O2 = 0 kJ/mol
ΔHf∘ CO2 = -393.5 kJ/mol
ΔHf∘ H2O = -241.8 kJ/mol
Step 2: The balanced equation
2C2H2(g)+5O2(g) ⟶ 4CO2(g)+2H2O(g)
C2H2(g)+5/2O2(g) ⟶ 2CO2(g)+H2O(g)
Step 3: Calculate ΔH° of the reaction
ΔH° = (2*ΔHf∘ CO2 + ΔHf∘ H2O) - (ΔHf∘ C2H2)
ΔH° = (2* -393.5 kJ/mol + (-241.8) kJ/mol) - 227.4 kJ/mol
ΔH° = -787 - 241.8 - 227. kJ/mol
ΔH° = -1255.8 kJ
Calculate the molar solubility of CaCO3 in 0.250M Na2CO3
Kps CaCO3 is 4.96x10-9
Answer:
solubility is 1.984x10⁻⁹M
Explanation:
When CaCO₃ is in water, the equilibrium that occurs is:
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Kps = [Ca²⁺] [CO₃²⁻] = 4.96x10⁻⁹
If you have a 0.250M solution of Na₂CO₃, [CO₃²⁻] = 0.250M:
[Ca²⁺] [0.250M] = 4.96x10⁻⁹
Assuming you are adding an amount of CaCO₃:
[X] [0.250 + X] = 4.96x10⁻⁹
Where X is the amoun of CaCO₃ you can add, that means, solubility
X² + 0.250X - 4.96x10⁻⁹ = 0
Solving for X:
X = -0.25M → False answer, there is no negative concentrations.
X = 1.984x10⁻⁹M.
That means, solubility is 1.984x10⁻⁹M
Answer:
[tex]1.984x10^{-8}M[/tex]
Explanation:
Hello,
In this case, the equilibrium reaction for the solubility of calcium carbonate is:
[tex]CaCO_3(s) \rightleftharpoons Ca^{2+}(aq)+CO_3^{-2}(aq)[/tex]
In such a way, since 0.250 M sodium carbonate solution is the solvent, we assume an initial concentration of carbonate anion to be also 0.250 M since sodium carbonate is completely dissolved, for that reason the equilibrium equation turns out:
[tex]Ksp=[Ca^{2+}][CO_3^{2-}]\\\\4.96x10^{-9}=x*(0.250+x)[/tex]
Hence, solving for [tex]x[/tex] we have:
[tex]x=1.984x10^{-8}M[/tex]
Which corresponds to the molar solubility if calcium carbonate as well.
Regards.
A 2.300×10^−2 M solution of glycerol (C3H8O3) in water is at 20.0°C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.0 mL . The density of water at 20.0°C is 0.9982 g/mL.
Required:
a. Calculate the molality of the glycerol solution.
b. Calculate the mole fraction of glycerol in this solution.
c. Calculate the concentration of the glycerol solution in percent by mass.
d. Calculate the concentration of the glycerol solution in parts per million.
Answer:
a) Based on the given information, the volume of the solvent given is 999 ml and the density of water given is 0.9982 gram per ml.
The mass of solvent can be calculated by the formula mass = density * volume
mass = 0.9982 * 999 = 0.997 Kg
The molarity of the solution given is 2.300 * 10^-2 M
Molality of the glycerol solution can be calculated by using the formula,
Molality = molarity/solvent (kg) = 2.300 * 10^-2 / 0.997 = 0.023 m
b) Molarity or the moles of the solute given is 2.300 * 10^-2 moles
The moles of solvent can be determined by using the formula, n = mass of solvent/mol.wt = 997/18 (mol.wt of solvent is 18 g/mol), now putting the values we get,
n = 997/18 = 55.4
The mole fraction of the glycerol will be = 2.300 * 10^-2 M/(2.300 * 10^-2)+55.40
= 4.15 * 10^-4
c) The mass percent of glycerol can be determined by using the formula,
mass of solute/mass of solution * 100% ----- (i)
The mass of solute can be determined by using the formula,
n = mass of solute/mol.wt of solute
The n or the no. of moles is 2.300 * 10^-2 and the molecular weight of glycerol is 92.09 g/mol. Now putting the values we get,
mass = 2.300 * 10^-2 * 92.09 = 2.12 grams
Now putting the values in equation (i) we get,
mass percent = 2.12 / 997 * 100% = 0.21%
d) Based on the above calculation, the mass of solute (glycerol) is 2.12 g or 2.12 * 1000 mg
The volume of water is 999 ml or 999 * 10^-3 L
The concentration of the glycerol solution will be,
Concentration = 2.12 * 10^3 mg/999 * 10^-3 L
= 2.12/999 * 10^6 mg/L
= 2122.1 ppm
Considering the solution of molality, mole fraction, mass percentage and ppm, you obtain that:
a) The molality of the glycerol solution is 0.02306 molal.
b) The mole fraction of glycerol in the solution is 4.15×10⁻⁴.
c) The percent by mass is 0.212%.
d) The concentration of the glycerol solution is 2118.07 ppm.
a. MolalityMolality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.
The Molality of a solution is determined by the expression:
[tex]molality=\frac{number of moles of solute}{kilogramsof solvent}[/tex]
In this case, you have a 2.300×10⁻² M solution of glycerol (C₃H₈O₃) in water. The sample was created by dissolving a sample of C₃H₈O₃ in water and then bringing the volume up to 1.000 L.
So, being the molarity the number of moles of solute that are dissolved in a certain volume, the number of moles of glycerol can be calculated as:
number of moles of glycerol= 2.300×10⁻² M× 1 L
number of moles of glycerol= 2.300×10⁻² moles
On the other side, the volume of water needed was 999 mL and the density of water at 20.0∘C is 0.9982 [tex]\frac{g}{mL}[/tex]. So, the mass of water needed can be calculated as:
999 mL×0.9982[tex]\frac{g}{mole}[/tex] = 997.2 grams of water= 0.9972 kg of water
Then, the molality of the solution is:
[tex]molality=\frac{2.300x10^{-2} moles}{0.9972 kg}[/tex]
molality= 0.02306 molal
Finally, the molality of the glycerol solution is 0.02306 molal.
b. Mole fractionThe molar fraction is a way of measuring the concentration that expresses the proportion in which a sustance is found with respect to the total moles of the solution.
In this case, the number of moles of glycerol= 2.300×10⁻² moles
Having 997.2 grams of water, the moles of solvent can be determined knowing that the molar mass of water is 18[tex]\frac{g}{mole}[/tex]:
[tex]number of moles of water=997.2 gramsx\frac{1 mole}{18 grams}[/tex]
number of moles of water= 55.4 moles of water
Being the number of total moles the sum of the moles of grycerol and the numbers of moles of water, the mole fraction can be calculated as:
[tex]mole fraction=\frac{2.300x10^{-2} moles}{2.300x10^{-2}moles+55.4 moles}[/tex]
mole fraction= 4.15×10⁻⁴
In summary, the mole fraction of glycerol in the solution is 4.15×10⁻⁴.
c. Percent by massThe Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.
The mass percentage of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.
The mass percentage is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:
[tex]percent by mass=\frac{mass of solute}{mass of solution}x100[/tex]
In this case, remmeber that you have a number of moles of glycerol of 2.300×10⁻² moles .
Being the molar mass glycerol 92.09 [tex]\frac{g}{mole}[/tex], the mass of glycerol can be calculated as:
2.300×10⁻² moles×92.09 [tex]\frac{g}{mole}[/tex]= 2.11807 grams of glycerol
Remembering that you have 997.2 grams of water, the percent by mass is calculated as:
[tex]percent by mass=\frac{mass of glycerol}{mass of glycerol + mass of water}x100[/tex]
Solving:
[tex]percent by mass=\frac{2.11807 grams}{2.11807 grams + 997.2 grams}x100[/tex]
[tex]percent by mass=\frac{2.11807 grams}{999.31807 grams}x100[/tex]
percent by mass= 0.212%
Finally, the percent by mass is 0.212%.
d. Parts per millionParts per million (ppm) is a unit of measurement of concentration that measures the number of units of substance in each million units of the whole. In this case, the concentration measurement refers to mg of glycerol per L of solution.
Being the mass of glycerol 2.11807 grams equal to 2118.07 mg (1 g=1000mg), the concentration is:
[tex]concentration=\frac{2118.07 mg}{1 L solution}[/tex]
Solving:
concentration= 2118.07 ppm
In summary, the concentration of the glycerol solution is 2118.07 ppm.
Learn more about:
molality: brainly.com/question/20366625?referrer=searchResults brainly.com/question/4580605?referrer=searchResultsmole fraction: brainly.com/question/14434096?referrer=searchResults brainly.com/question/10095502?referrer=searchResultsmass percentage: brainly.com/question/19168984?referrer=searchResults brainly.com/question/18646836?referrer=searchResults brainly.com/question/24201923?referrer=searchResults brainly.com/question/9779410?referrer=searchResults brainly.com/question/17030163?referrer=searchResultsppm: brainly.com/question/16727593?referrer=searchResults brainly.com/question/13565240?referrer=searchResults Lithium reacts more slowly with cold water than sodium.
State two ways the reaction can be made to go faster.
Answer:
1. Increase in the temperature of the water
2. Increasing the surface area of the lithium
Explanation:
1. Increase in the temperature of the water
The activation energy for the lithium water reaction is +161 kJ/mol while the activation energy for the sodium is +109 kJ/mol, hence for increased reaction rate, the water temperature will be raised to enable more lithium atoms enter into reaction with the water molecules as their energy is increased lowering the activation energy required for the reaction.
2. Increasing the surface area of the lithium
As the lithium floats on the water, due to its low temperature and the heat evolved from the reaction of lithium with the cold water is below the melting point of lithium, the reaction rate can be increased by increasing the surface area of lithium sample by grinding so as to increase the number of lithium water reaction sites.
is the total disappearance of all members of a species.
Answer: EXTINCTION
Explanation:
Answer:
extinction
Explanation:
The tabulated data were collected for this reaction:
CH3Cl(g) + 3Cl2(g) → CCl4(g) + 3HCl(g)
[CH3Cl] (M) [Cl2] (M) Initial Rate (M/s)
0.050 0.050 0.014
0.100 0.050 0.029
0.100 0.100 0.041
0.200 0.200 0.115
(a) Write an expression for the reaction rate law and calculate the value of the rate constant, k.
(b) What is the overall order of the reaction?
Answer:
ai) Rate law, [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]
aii) Rate constant, k = 1.25
b) Overall order of reaction = 1.5
Explanation:
Equation of Reaction:
[tex]CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)[/tex]
If [tex]A + B \rightarrow C + D[/tex], the rate of backward reaction is given by:
[tex]Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex]
k is constant for all the stages
Using the information provided in lines 1 and 2 of the table:
[tex]0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1[/tex]
Using the information provided in lines 3 and 4 of the table and insering the value of a:
[tex]0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\[/tex]
[tex]0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5[/tex]
The rate law is: [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]
The rate constant [tex]k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex] then becomes:
[tex]k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25[/tex]
b) Overall order of reaction = a + b
Overall order of reaction = 1 + 0.5
Overall order of reaction = 1.5
The rate law is for this reaction is [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]
where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.
Let's consider the following reaction.
CH₃Cl(g) + 3 Cl₂(g) → CCl₄(g) + 3 HCl(g)
What is the rate law?The rate law for a chemical reaction is an equation that relates the reaction rate with the concentrations of the reactants.
The rate law for this reaction is:
r = k [CH₃Cl]ᵃ [Cl₂]ᵇ
where,
r is the initial rate.k is the rate constant.a is the reaction order for CH₃Cl.b is the reaction order for Cl₂.If we write the ratio r₂/r₁, we get:
r₂/r₁ = k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ / k [CH₃Cl]₁ᵃ [Cl₂]₁ᵇ
r₂/r₁ = {[CH₃Cl]₂/ [CH₃Cl]₁}ᵃ
(0.029 M/s)/(0.014 M/s) = {0.100M/0.050 M}ᵃ
a ≈ 1
If we write the ratio r₃/r₂, we get:
r₃/r₂ = k [CH₃Cl]₃ᵃ [Cl₂]₃ᵇ / k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ
r₃/r₂ = {[Cl₂]₃/[Cl₂]₂}ᵇ
(0.041 M/s)/(0.029 M/s) = {0.100 M/0.050 M}ᵇ
b ≈ 0.5
So far, the rate law is:
[tex]r = k [CH_3Cl] [Cl_2]^{0.5}[/tex]
Let's use the values of the first experiment to find the value of the rate constant.
[tex]k = \frac{r}{[CH_3Cl][Cl_2]^{0.5} } = \frac{0.014 M/s}{(0.050 M)(0.050 M)^{0.5} } = 1.25 M^{-0.5} s^{-1}[/tex]
The final rate law is:
[tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]
What is the overall order of the reaction?It is the sum of the individual orders of reaction.
a + b = 1 + 0.5 = 1.5
The rate law is for this reaction is [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]
where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.
Learn more about the rate law here: https://brainly.com/question/14945022
Glyceraldehyde is an aldose monosaccharide. The Fischer projection of D-glyceraldehyde is given below. Draw D-glyceraldehyde using wedge and dash bonds around the chirality center and including ALL hydrogen atoms.
Image is not given in the question, so the image for the question is given below.
Answer:
Fischer projection is a two-dimensional representation of a three-dimensional organic molecule by projection.
Wedge and dash bonds are used to represent the three-dimensional structure of a molecule, in which wedges indicates bonds towards the viewer, solid lines indicates bonds in the plane of the image and dashed lines indicates bonds away from the viewer.
Wedge and dash bonds structure of D-glyceraldehyde is attached below.
A 0.04380 g sample of gas occupies 10.0 mL at 290.5 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound?
Answer:
THE MOLECULAR FORMULA FOR THE COMPOUND IS C2 Cl2
Explanation:
mass = 0.04380 g
Volume = 10 mL = 10 / 1000 L = 0.010 L
Pressure = 1.10 atm
R = 0.082 L atm mol^-1 K^-1
Temperature = 290.5 K
Carbon = 25.305 %
Chlorine = 74.695 %
Atomic mass of carbon = 12
Atomic mass of chlorine = 35.5
First calculate the empirical formula by following these steps:
1. Write the percentage composition of the elements involved and divide by its atomic mass
Carbon = 25.305 / 12 = 2.10875
Chlorine = 74.695 /35.5 = 2.1040
2. Divide each by the smaller value
Carbon = 2.10875 / 2.1040 = 1.002
Chlorine = 2.1040 / 2.1040 = 1
3. Round up to the whole number
The empirical formula is C Cl
Next is to calculate the molar mass of the compound using ideal gas equation
PV = mRT/ MM
Mm = mRT / PV
Mm = 0.04380 * 0.082 * 290.5 / 1.10 * 0.010
Mm = 1.0433 / 0.011
Mm = 94.845 g/mol
Mm = 94.85 g/mol
Now that we know the molar mass, we can go on to calculate the molecular formula:
(C Cl) n = Molar mass
( 12 + 35.5) n = 94.85
(47.5)n = 94.85
n = 94.85 / 47.5
n = 1.9968
n ~ 2
The molecular formula can then be written as C2Cl2.
which would you think would be a stronger interaction and why: an interaction between a sodium ion and the prtial negative charge on the oxygen in ethanol, or the interaction between two ethanol molecules
Answer:
The interaction between sodium ion and the partial negative charge on the oxygen is stronger.
Explanation:
The predominant interaction that exists between sodium ion and the partial negative charge on the oxygen is ion-dipole interaction.
The predominant interaction that exists between two ethanol molecules is hydrogen bonding interaction.
The order of strength of the intermolecular interactions in decreasing order:
Ionic bond > ion-dipole interaction > hydrogen bonding > dipole-dipole interaction > ion-induced dipole interaction > induced dipole- dipole interaction > london force
So, ion-dipole interaction is stronger than hydrogen bonding.
Hence, the interaction between sodium ion and the partial negative charge on the oxygen is stronger.
How many moles of oxygen in 3.5 Mols of CH2O
Calculate the free energy change for the reaction:
2 NO(g) + O2(g) → 2 NO2(g)
(A) - 9.3 kcal
(B) + 24.9 kcal
(C) + 9.3 kcal
(D) - 16.6 kcal
(E) + 16.6 kcal
Answer:
(D) - 16.6 kcal
Explanation:
Hello,
In this case, the Gibbs free energy for the given reaction is computed in terms of the Gibbs free energy of formation of each species involved in the chemical reaction:
[tex]\Delta _rG=2\Delta _fG_{NO_2}-2\Delta _fG_{NO}-\Delta _fG_{O_2}[/tex]
Thus, it is found for nitrogen monoxide, oxygen and nitrogen dioxide the following Gibbs free energies of formation: 87.6, 0 and 51.3 kJ/mol respectively, therefore we compute:
[tex]\Delta _rG=2(51.3kJ/mol)-2(87.6kJ/mol)=-72.6kJ*\frac{1kcal}{4.184kJ} \\\\\Delta _rG=-17.35kcal[/tex]
The closest result is (D) - 16.6 kcal, as such difference is noticed when different sources for thermochemical data are used, in this case, the NIST data were used.
Best regards.