The chemical reaction for the formation of syngas is: CH4 + H2O -> CO + 3 H2 What is the rate for the formation of hydrogen, if the rate of the formation of carbon monoxide is 0.35 M/s ? g

Answers

Answer 1

Answer:

Answer : The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-(1)/(a)(d[A])/(dt)$

$\text{Rate of disappearance of B}=-(1)/(b)(d[B])/(dt)$

$\text{Rate of formation of C}=+(1)/(c)(d[C])/(dt)$

$\text{Rate of formation of D}=+(1)/(d)(d[D])/(dt)$

$Rate=-(1)/(a)(d[A])/(dt)=-(1)/(b)(d[B])/(dt)=+(1)/(c)(d[C])/(dt)=+(1)/(d)(d[D])/(dt)$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$CH_4+H_2O\rightarrow CO+3H_2$

The expression for rate of reaction :

$\text{Rate of disappearance of }CH_4=-(d[CH_4])/(dt)$

$\text{Rate of disappearance of }H_2O=-(d[H_2O])/(dt)$

$\text{Rate of formation of }CO=+(d[CO])/(dt)$

$\text{Rate of formation of }H_2=+(1)/(3)(d[H_2])/(dt)$

The rate of reaction expression is:

$\text{Rate of reaction}=-(d[CH_4])/(dt)=-(d[H_2O])/(dt)=+(d[CO])/(dt)=+(1)/(3)(d[H_2])/(dt)$

As we are given that:

$+(d[CO])/(dt)=0.35M/s$

Now we to determine the rate for the formation of hydrogen.

$+(1)/(3)(d[H_2])/(dt)=+(d[CO])/(dt)$

$+(1)/(3)(d[H_2])/(dt)=0.35M/s$

$(d[H_2])/(dt)=3* 0.35M/s$

$(d[H_2])/(dt)=1.05M/s$

Thus, the rate for the formation of hydrogen is, 1.05 M/s

Related Questions

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Which of the following electron configurations represents an excited state of the indicated atom? Group of answer choices a.Na: 1s2 2s2 2p6 3s2 3p2 3s1
b.Ne: 1s2 2s2 2p6
c.N: 1s2 2s2 2p3
d.P: 1s2 2s2 2p6 3s2 3p2 4s1
e.He: 1s2

Answers

Answer: The electronic configuration of the atom that represents excited state is P: $1s^22s^22p^63s^23p^24s^1$

Explanation:

There are 2 states classified under energy levels:

Ground state: This is the lower energy state which is termed as stable state.
Excited state: This is the upper energy state and is termed as the unstable state. All the electrons which are present in this state always come back to the ground state.

For the given options:

For a: Sodium

The atomic number of sodium element is 11. The ground state electronic configuration of this element is $1s^22s^22p^63s^1$

For b: Neon

The atomic number of neon element is 10. The ground state electronic configuration of this element is $1s^22s^22p^6$

For c: Nitrogen

The atomic number of nitrogen element is 7. The ground state electronic configuration of this element is $1s^22s^22p^3$

For d: Phosphorus

The atomic number of phosphorus element is 15. The ground state electronic configuration of this element is $1s^22s^22p^63s^23p^3$

One electron from the valence shell jumps into outer shell and the excited state electronic configuration becomes $1s^22s^22p^63s^23p^24s^1$

For e: Helium

The atomic number of helium element is 2. The ground state electronic configuration of this element is $1s^2$

Hence, the electronic configuration of the atom that represents excited state is P: $1s^22s^22p^63s^23p^24s^1$

Final answer:

The sodium atom with the electron configuration 1s2 2s2 2p6 3s2 3p2 3s1 is in an excited state because other sodium atom stages are not completely filled.

Explanation:

An atom is in an excited state when one or more electrons have moved to a higher energy level. Normal electron configurations have the electrons in the lowest possible energy states (or orbitals). In this case, the answer choice is (a) sodium (Na) with the electron configuration 1s2 2s2 2p6 3s2 3p2 3s1. Sodium normally has the 3s state fully occupied, so the presence of an electron in a higher energy state (3p) and the vacancy in the lower energy state (3s) indicates an excited state.

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During an experiment, a student adds 2.90 g CaO to 400.0 mL of 1.500 M HCl . The student observes a temperature increase of 6.00 °C . Assuming that the solution's final volume is 400.0 mL , the density is 1.00 g/mL , and the heat capacity is 4.184 J/g⋅°C , calculate the heat of the reaction, ΔHrxn .

Answers

Answer:

ΔHrxn = 193107.69 J/mol

Explanation:

ΔHrxn = mcΔT

m = mass

c = heat capacity

ΔT = temperature variation

density = m/V

m = density x V

m = 1.00 g/mL x 400.0 mL

m = 400.0 g

ΔHrxn = mcΔT

ΔHrxn = 400 g x 4.184 J/g°C x 6.00 °C

ΔHrxn = 10041.6 J

CaO + 2HCl → CaCl₂ + H₂O

CaO = 56.0774 g/mol

2.90 g CaO = 0.052 mol

400.0 mL of 1.500 mol/L HCl = 0.6 mol HCl

ΔHrxn = 10041.6 J is for 0.052 mol of CaO

ΔHrxn = 193107.69 J is for 1 mol of CaO

Identify the true statements about introns.a- they code for polypeptide proteinsb- they have a branch site located 20 to 50 nucleotides upstream of the 3' splice sitec- they end with the nucleotides AG at the 3' endd- they begin with the nucleotides GU at the 5' ende- they tend to be common in bacterial genes

Answers

Answer:

The answer is "Option b, c, and d".

Explanation:

In such a gene, Autosomes are also the sequence for code and transposable elements, not the series of encoding. Through the expression of genes, such fragments of its introns are split through protein complexes throughout the translation process. There has been no kenaf fiber in the genomes of prokaryotic cells.

At standard temperature and pressure (0 ∘C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L. What volume would the same amount of gas occupy at the same pressure and 55 ∘C ?

Answers

Taking into account the Charles's law, the same amount of gas at the same pressure and 55 ∘C would occupy a volume of 26.91 L.

Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, which is maintained at a constant pressure.

This law states that the volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

$(V)/(T)=k$

Studying an initial state 1 and a final state 2, it is satisfied:

$(V1)/(T1)=(V2)/(T2)$

In this case, you know:

V1= 22.4 L
T1= 0 C= 273 K
V2= ?
T2= 55 C= 328 K

Replacing:

$(22.4 L)/(273 K)=(V2)/(328 K)$

Solving:

$V2=328 Kx(22.4 L)/(273 K)$

V2= 26.91 L

Finally, the same amount of gas at the same pressure and 55 ∘C would occupy a volume of 26.91 L.

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Answer:26.9L

Explanation: this is Charles' law which states that the volume of a gas is directly proportional to the absolute temperature at contant pressure. The expression is V1/T1 = V2/T2

Making V2 the subject of the formula we have

V2 = V1 xT2/T1

= 22.4 x 328/273

= 26.9L

A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with distilled water. What is the final molarity of the dilute solution?

Answers

A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with distilled water. the final molarity of the dilute solution is 0.102 M.

From the question given above, the following data were obtained:

Volume of stock solution (V1) = 5 mL

Molarity of stock solution (M₁) = 5.103 M

Volume of diluted solution (V₂) = 250 mL

Molarity of diluted solution (M₂) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

5.103 × 5 = M2 × 250

25.515 = M2 × 250

Divide both side by 250

M2 = 25.515 / 250

M2 = 0.102 M

Thus, the molarity of the diluted solution is 0.102 M.

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Answer:

0.102 M.

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V1) = 5 mL

Molarity of stock solution (M1) = 5.103 M

Volume of diluted solution (V2) = 250 mL

Molarity of diluted solution (V2) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M1V1 = M2V2

5.103 × 5 = M2 × 250

25.515 = M2 × 250

Divide both side by 250

M2 = 25.515 / 250

M2 = 0.102 M

Thus, the molarity of the diluted solution is 0.102 M.

(e) Which sample contains more atoms: 2.0 g of Fe2O3 or 2.0 g of CaSO4? Support your answer with calculations.

Answers

CaSO4. It has 6 atoms while Fe2O3 has 5 atoms

Explanation:

Fe2O3 contains 2 atoms of Iron(Fe) and 3 atoms of Oxygen(O) = 2 + 3 = 5 atoms

CaSO4 contains 1 atom of Calcium(Ca), 1 atom of Sulphur(S) and 4 atoms of Oxygen(O) = 1 + 1 + 4 = 6 atoms

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How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid? How many mL of 0.100 M NaOH are needed to neutralize 50.00 mL of a 0.150 M solution of CH3CO2H, a monoprotic acid? a. 37.50 mL b. 50.00 mL c. 75.00 mL d. 100.00 mL e. 25.00 mL

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Please ignore my answers because they are wrong.