Answer:
1.00 M
Explanation:
Step 1: Write the balanced equation
HBr + LiOH ⇒ LiBr + H₂O
Step 2: Calculate the reacting moles of lithium hydroxide
40.00 mL of 0.500 M solution react. The reacting moles of LiOH are:
[tex]40.00 \times 10^{-3} L \times \frac{0.500mol}{L} = 0.0200 mol[/tex]
Step 3: Calculate the reacting moles of hydrobromic acid
The molar ratio of HBr to LiOH is 1:1. The reacting moles of hydrobromic acid are 1/1 × 0.0200 mol = 0.0200 mol.
Step 4: Calculate the molarity of hydrobromic acid
0.0200 moles of HBr are in 20.00 mL of the solution. The molarity of the hydrobromic acid solution is:
[tex]M= \frac{0.0200 mol}{20.00 \times 10^{-3} L } =1.00 M[/tex]
Draw the structure of the bromohydrin formed when (Z)-3-hexene reacts with Br2/H2O. Use the wedge/hash bond tools to indicate stereochemistry where it exists. If the reaction produces a racemic mixture, draw both stereoisomers. Separate multiple products using the sign from the drop-down menu.
Answer:
(3R,4R)-4-bromohexan-3-ol
Explanation:
In this case, we have reaction called halohydrin formation. This is a markovnikov reaction with anti configuration. Therefore the halogen in this case "Br" and the "OH" must have different configurations. Additionally, in this molecule both carbons have the same substitution, so the "OH" can go in any carbon.
Finally, in the product we will have chiral carbons, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)
I hope it helps!
How do I calculate the moles consumed in a vinegar titration?
Answer: Use the titration formula
Explanation:
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]
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.
Citric acid is one component of some soft drinks. Suppose that 8 L of solution are made from 0.24 g of citric acid, C6H8O7. What is the molarity of citric acid in the solution?
Answer:
The correct answer is 0.15 × 10⁻³ M.
Explanation:
C₆H₈O₇ is the molecular formula of citric acid. The mass of the one mole of the substance that comprise Avogadro's no of molecules is termed as the molar mass of the substance.
The molar mass of the citric acid (C₆H₈O₇) will be,
6 × atomic mass of carbon + 8 × atomic mass of hydrogen + 7 × atomic mass of oxygen = 6×12 + 8×1 + 7×16 = 192. Thus, the molar mass of citric acid is 192 g/mol.
The value of the solution given in the question is 8 L.
The mass of citric acid given is 0.24 g or 240 mg, which can also be written as 240/1000 g or 240 × 10⁻³ g.
The number of moles can be calculated by using the formula = mass / molar mass. Thus, by putting the values we get,
= 240 × 10⁻³ g / 192 g/mol
= 1.25 × 10⁻³ mol
The number of moles of the solute present in the 1 liter of the solution is termed as the molarity of the solution. The formula of molarity or M is,
= no of moles of solute / volume of solution in L
Now putting the values we get,
= 1.25 × 10⁻³ mol / 8 L
= 0.156 × 10⁻³ M
Hence, the molarity of citric acid in the given solution is 0.15 × 10⁻³ M
Compare a 3s orbital to a 3p orbital. Which of the following are true? I. The orbitals have the same shape. II. The orbitals have different shapes. III. The orbitals have the same energy. IV. The orbitals have different energies.
Answer:
II. The orbitals have different shapes.
IV. The orbitals have different energies.
Explanation:
1. Shapes
Below are pictures of a 3s and a 3p orbital.
The 3s orbital has a spherical shape, and the 3p orbital has a dumbbell shape.
2. Energies
For every element except H, the order of energies is
3s < 3p
For a hydrogen atom,
3s = 3p
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.
in what type of reaction or process does heat flow into the system
Answer: The correct answer is: " endothermic . "
______________________________________
Note: Heat flows into [heat may be absorbed within] an "endothermic" reaction or system
To the contrary, heat flows out [heat may exit from or may be released from] an "exothermic" reaction or process.
Hint: Think of the "prefixes" of: "endothermic" and "exothermic" :
_____________________________________
1) endo- = "within" (as in "endothermic" —heat tends to be absorbed/"within"/"released within"/released within"/into" ;
2) exo- = " outwards"/"exit" (as in "exothermic") —heat tends to '"exit"/leave/escape from/"be released out of/form".
_____________________________________
Hope this is helpful to you!
Best wishes to you in your academic pursuits
—and within the "Brainly" community"!
_____________________________________
Match the term to its description.
Gravitational energy
Answer 1
Mechanical energy
Answer 2
Nuclear energy
Answer 3
Thermal energy
Answer 4
Internal energy caused by vibrations of atoms and molecules
The potential energy of position
The energy of an object due to its movement
Energy stored in the nucleus of an atom
Answer:
Gravitational Energy: The energy of an object due to its movement
Mechanical Energy: The potential energy of position
Nuclear Energy: internal energy caused by vibrations of atoms and molecules
Thermal Energy: Energy stored in the nucleus of an atom
Answer:
radiant
nuclear
thermal
potential
Explanation:
Next, a chemical reaction of interest was conducted in the same constant volume calorimeter. The neutralization reaction of HCl(aq) with NaOH(aq) caused the temperature of the calorimeter to rise by 2.46 °C. What is the change in internal energy ΔU of the neutralization reaction in kJ?
Answer:
A constant volume calorimeter (bomb calorimeter) was calibrated by performing in it a reaction in which 5.23 kJ of heat energy was released, causing the calorimeter to rise by 7.33 °C. What is the heat capacity Cy of the calorimeter
Explanation:
the heat capacity of the calorimeter is
[tex]C_v = \frac{q}{\Delta T} \\\\=\frac{5.23}{7.33} \\\\=0.714kJ /^\circ C[/tex]
Now that our heat capacity of the calorimeter is 0.714kJ/°C
we can easily calculate the change in internal energy ΔU of the neutralization reaction
[tex]\Delta U = C_v \ dt[/tex]
or
[tex]\Delta U = C_V \Delta T[/tex]
Δ T = 2.46 °C
[tex]C_v = 0.714 kJ/ ^\circ C[/tex]
[tex]\Delta U = 0.714*2.46\\\\=1.7564kJ[/tex]
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
The volume measuring system for a gasoline pump at the service station is calibrated at 20.0°C. If the temperature of the gasoline drops to 10.0°C, what percentage extra amount of mass of gasoline do you receive when making a purchase? The coefficient of volume expansion for gasoline is 9.5 × 10-4/K.
Answer:
answer = -0.95
Explanation:
express your answer to three significant figures and include appropriate units
Answer:
58.702
Explanation:
Data obtained from the question include:
Mass number of isotope 1 (M1) = 57.93
Abundance of isotope 1 (M1%) = 67.76%
Mass number of isotope 2 (M2) = 59.93
Abundance of isotope 2 (M2%) = 26.16%
Mass number of isotope 3 (M3) = 60.93
Abundance of isotope 3 (M3%) = 1.25%
Mass number of isotope 4 (M4) = 61.93
Abundance of isotope 4 (M4%) = 3.66%
Mass number of isotope 5 (M5) = 63.93
Abundance of isotope 5 (M5%) = 1.16%
Relative atomic mass =...?
The relative atomic mass(RAM) of the element can be obtained by doing the following:
RAM = [(M1×M1%) /100] +[(M2×M2%) /100] + [(M3×M3%) /100] + [(M4×M4%) /100] + [(M5×M5%) /100]
RAM = [(57.93×67.76)/100] + [(59.93×26.16)/100] + [(60.93×1.25) /100] + [(61.93×3.66)/100] + [(63.93×1.16)/100]
RAM = 39.253 + 15.678 + 0.762 + 2.267 + 0.742
RAM = 58.702
Therefore, the relative atomic mass (RAM) of the element is 58.702
The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. What is the final level of the electron? (c = 3.00 x 108m/s, h = 6.63 x 10−34 J • s, RH = 2.179 x 10−18 J)
a. 5
b. 6
c. 8
d. 9
e. 1
Answer:
The level is [tex]n_1 = 5[/tex]
Explanation:
From the question we are told that
The level of the hydrogen atom is [tex]n_2 = 8[/tex]
The wavelength of the photon is [tex]\lambda = 3745 nm = 3745 *10^{-9} \ m[/tex]
Generally the wave number is mathematically represented as
[tex]k = \frac{1}{\lambda }[/tex]
Now this wave number can also be mathematically represented as
[tex]k = R_{\infty} [\frac{1}{n_1^2} + \frac{1}{n_2^2} ][/tex]
This implies that
So
Here [tex]R_{\infty}[/tex] is the Rydberg constant, with a value [tex]1.097 * 10^7[/tex]
and [tex]n_1 \ and \ n_2[/tex] are the principal quantum levels
substituting values
[tex]0.0243= [\frac{1}{n_1^2} - \frac{1}{8^2} ][/tex]
[tex]0.0243= \frac{1}{n_1^2} - 0.015625[/tex]
[tex]0.0243 + 0.015625= \frac{1}{n_1^2}[/tex]
[tex]n_1 = 5[/tex]
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.
Be sure to answer all parts. Give the name and the formula of the compound formed from the following elements: (a) bromine and rubidium Name: (select) (select) Formula: (b) strontium and sulfur Name: (select) (select) Formula: (c) barium and chlorine Name: (select) (select) Formula
Answer:
These are the chemical names and molecular formula of the compounds formed from these elements:
Rubidium + Bromine = Rubidium bromide, RbBrStrontium + Sulphur = Strontium sulfide, SrSBarium and Chlorine = Barium chloride, [tex]BaCl_{2}[/tex]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 is [tex]\lambda =97.3 nm[/tex]
Explanation:
Generally the series whose emission line show on the visible spectrum is the
Balmar series so this two emission line seen on the visible spectrum could either be due to the move of electron from
[tex]n=3 \to \ n=2[/tex]
OR
[tex]n=4 \to n=2[/tex]
This implies that the first excitement is from [tex]n_i=1 \to \ n_f=4[/tex]
So the energy change due to the excitement is mathematically represented as
[tex]\Delta E = R_H [\frac{1}{n_i^2} -\frac{1}{n_f^2} ][/tex]
substituting values
[tex]\Delta E= R_H [\frac{1}{1^2} -\frac{1}{4^2} ][/tex]
[tex]\Delta E= \frac{15}{16} R_H[/tex]
This energy change can also be represented as
[tex]\Delta E = \frac{hc}{\lambda}[/tex]
So [tex]\frac{15}{16} R_H = \frac{hc}{\lambda}[/tex]
=> [tex]\lambda = \frac{16hc}{15 R_H}[/tex]
Where [tex]R_H[/tex] is the Rydberg constant with a value of [tex]R_H = 2.18 * 10^{-18} J.[/tex]
h is the Planck's constant with values [tex]h = 6.626 * 10^{-34} m^2 kg / s[/tex]
c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
So
[tex]\lambda = \frac{16(6.626 *10^{-34})(3*10^{8})}{15 * (2.18*10^{-18})}[/tex]
[tex]\lambda = 9.73 *10^{-8} \ m[/tex]
[tex]\lambda =97.3 nm[/tex]
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.
Need help with these 7 problems
Answer:
1. 2.3
2. 4.5
3. 10.5
4. 5.6
5. 3.5
6. 3.9
7. 1.6
Explanation:
pH = - log [H+]
1. Lemon juice
[H+] = 0.005
pH = - log [H+]
pH = - log 0.005
pH = 2.3
2. Beer
[H+] = 3.16x10^-5
pH = - log [H+]
pH = - log 3.16x10^-5
pH = 4.5
3. Milk of Magnesia
[H+] = 3.16x10^-11
pH = - log [H+]
pH = - log 3.16x10^-11
pH = 10.5
4. Rain water
[H+] = 2.51x10^-6
pH = - log [H+]
pH = - log 2.51x10^-6
pH = 5.6
5. Soda
[H+] = 3.16x10^-4
pH = - log [H+]
pH = - log 3.16x10^-4
pH = 3.5
6. Tomatoes
[H+] = 1.23x10^-4M
pH = - log [H+]
pH = - log 1.23x10^-4
pH = 3.9
7. Bleach
[OH-] = 2.3x10^-2M
pOH = - log [OH-]
pOH = - log 2.3x10^-2
pOH = 1.6
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.
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!
50 mL of CH3CH2Br (bromoethane) and 50 mL of water are poured into a separatory funnel. Bromoethane is a water‑insoluble compound with a density of 1.460gmL . The funnel is stoppered and the mixture is shaken vigorously. After standing, two layers separate. Which substance is in which layer? Explain. Bromoethane is more dense than water. Water is added second. The top layer is bromoethane and the bottom layer is water. The top layer is water and the bottom layer is bromoethane. Bromoethane has a higher molar mass than does water. b) Into the same funnel is poured carefully 50 mL of hexane (density = 0.660 g/mL) so that the other two layers are not disturbed. The hexane forms a third layer. The funnel is stoppered, and the mixture is shaken vigorously. After standing, two layers separate. Which compound(s) are in which layer? The top layer contains: The bottom layer contains:
Answer:
See explanation below
Explanation:
In this case, we can explain this in a very basic way.
We know that heavier objects will go always at the bottom when we are carrying two objects, one heavier than the other right?
In the same manner works the density of two liquids. In this case, we have a mix of water and bromoethane. Bromoethane is an organic compound and it's less polar than water which is extremely polar. When we mix these two liquids, we can see that both of them are insoluble, so no matter how much we shake the funnel, the liquids will not mix to form a solution.
Instead of that, both of them will be in the funnel, and they'll be gradually separating into two layers. The bromoethane has a higher density than water, this means that in the bottom layer we will have the bromoethane and in the top layer we will have the water.
In case you are wondering what happens if we added water first and then, the bromoethane?, it will happen the same, it does not matter the order you add the liquid, because density here is a very important factor, so when the water is added no matter which position, it will go to the top layer after the bromoethane is added.
Now when the hexane is added, it will form now three layers, and again, density plays an important factor. The higher density will go to the bottom, and the lowest to the top.
In this case, the order of layer will be:
Top layer: hexane (d = 0.66 g/mL)
Middle layer: water (d = 1 g/mL)
Bottom layer: bromoethane (d = 1.46 g/mL)
Hope this helps
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:
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
For the general reaction aA + bB cC + DD, has the general rate law, rate k[A] [B].". What is the
correct algebraic equation for determining the units of the rate constant, k, when concentration is in M
and time is in seconds.
O
M
M's
k=
k=
MX+y.s
M*+y
O
Ma+b
O
M:s
k=
k
s
MX+y
O
M
o
S
k=
k=
Ma+b.s
M
O
M*+y
k=
S
Answer:
K=CHANGE IN CONCENTRATION/TIME TAKEN
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).
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.
Which type of semiconductor is created by doping with atoms that contain more valence electrons than the semiconductor material ?
Answer:
N-Type Semiconductor
Explanation:
Using the relationship given in part 1, calculate Ecell if [Cu2+] = 1.453 M and [Cr3+] = 0.00176 M, assuming that the temperature remains at 25°C. Give your answer to two places after the decimal.
Answer:
1.05 V
Explanation:
Now recall Nernst equation;
Ecell= E°cell - 0.0592/n log [Red]/[Ox]
Now we look at the values of E°cathode and E°anode
E°cathode= 0.34 V
E°anode = -0.74 V
E°cell= 0.34-(-0.74)
E°cell= 1.08 V
[Red] = 1.453 M
[Ox] = 0.00176 M
n= 6
Ecell= E°cell - 0.0592/n log [Red]/[Ox]
Ecell= 1.08 - 0.0592/6 × log[1.453]/[0.00176]
Ecell= 1.08 - 0.0592/6 × 2.917
Ecell= 1.08 - 0.02878
Ecell= 1.05 V