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mr Goodwill [35]
2 years ago
11

What is the empirical formula of a compound that is 41.4% Strontium, 13.24%

Chemistry
1 answer:
Maurinko [17]2 years ago
5 0

Empirical Formula StepsN2O6Sr Strontium Nitrate

Explanation:

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In a generic chemical reaction involving reactants A and B and products C and D, aA + bB → cC + dD, the standard enthalpy ΔH∘rxn
velikii [3]

Answer:

A. ΔH∘rxn =  21.9 KJ/mol

B. ΔH∘rxn = 103 KJ/mol

C. C2H5OH + 3O2 → 2CO2 + 3H2O

Explanation:

A.

The standard reaction equation is given as:

aA + bB → cC + dD

Its standard enthalpy is given as:

ΔH∘rxn = cΔH∘f(C) + dΔH∘f(D) − aΔH∘f(A) − bΔH∘f(B)

Reaction given to us is:

H2O(l) + CCl4(l) → COCl2(g) + 2HCl(g)

So, its standard enthalpy will be:

ΔH∘rxn = (1)ΔH∘f(CoCl2 (g)) + (2)ΔH∘f(HCl(g)) − (1)ΔH∘f(H2O(l)) − (1)ΔH∘f(CCl4(l))

using the values from table:

ΔH∘rxn = - 218.8 KJ/mol + (2)(- 92.3 KJ/mol) - (- 285.8 KJ/mol) - (- 139.5 KJ/mol)

<u>ΔH∘rxn =  21.9 KJ/mol</u>

<u></u>

B.

Reaction given to us is:

2A + B ⇌ 2C + 2D

So, its standard enthalpy will be:

ΔH∘rxn = (2)ΔH∘f(C) + (2)ΔH∘f(D) − (2)ΔH∘f(A) − (1)ΔH∘f(B)

using the values from table:

ΔH∘rxn = (2)181 KJ/mol + (2)(- 523 KJ/mol) - (2)(- 225 KJ/mol) - (- 337 KJ/mol)

<u>ΔH∘rxn = 103 KJ/mol</u>

<u></u>

C.

Balanced equation for combustion of ethanol is:

<u>C2H5OH + 3O2 → 2CO2 + 3H2O</u>

3 0
3 years ago
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