Question

What volume of 0.27 M H2SO4 can be prepared by diluting 119 mL of 6.0 M H2SO4?

What is ΔH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide?

CaCO3(s) → CaO(s) + CO2(g)

Substance	ΔH°f(kJ/mol)
CaCO3(s)	–1206.9
CaO(s)	–635.6
CO2(g)	–393.5

–2236.0 kJ/mol

–1449.0 kJ/mol

–177.8 kJ/mol

177.8 kJ/mol

2236.0 kJ/mol

How much heat is released if 35.0 g of ethanol (C2H5OH) burns in excess oxygen?

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH°rxn = –1367 kJ/mol

1797 kJ

1367 kJ

9.61 × 10^–4 kJ

4.78 × 10⁴ kJ

1040 kJ

How much heat is required to raise the temperature of 12.0 g of water from 15.4°C to 93.0°C? The specific heat of water is 4.184 J/g·°C.

223 J

773 J

503 J

4.67 ×10³ J

3.90 ×10³ J

Answer 1

#

Use the formula C₁V₁ =C₂V₂

Where, C₁ = initial concentration = 6.00M

V₁ = initial volume = ?

C₂ = final concentration = 0.27 M

V₂ = final volume = 119 ml

V₁ = C₂V₂ /C₁

Substitute the value in above equation

V₁ = 0.27 119 / 6 = 5.355 ml

5.355 ml of solution required.

#

_fH⁰ CaCO_3(s) = -1206.9 KJ/mol

_fH⁰ CaO_(s) = - 635.6 KJ/mol

_fH⁰ CO_2(g) = -393.5 KJ/mol

H⁰ = _fH⁰ (Product) - _fH⁰ (Reactant)

H⁰ = [(_fH⁰ CO₂) + (_fH⁰ CaO)] - [(_fH⁰ CaCO₃)]

= [( (-393.5)) + ((-635.6))] - [(-1206.9) ]

= [-1029.1] + [1206.9]

H⁰ = 177.8 KJ/mol

H⁰ for reaction = 177.8 KJ/mol

#

C₂H₅OH_(l)  + 3O_2(g)    2CO_2(g)  + 3H₂O_(l)   H⁰ for reaction = -1367 KJ/mol

molar mass of C₂H₅OH = 46 gm / mole

According to above reaction 1 mole that mean 46 gm of ethyl alcohole combution reaction heat of reaction = -1367 KJ / mol then for 35 gm have heat of reaction = -1367 35 / 46.06844 = -1040 KJ

-1040 KJ

#

Temperature change of water 15.4⁰C to 93⁰C

T = 93 - 15.4 = 77.6⁰C

specific of water = 4.184 J/ g K

mass of water = 12 gm

q = mass of water   specific heat of H₂O_(l)      T

= 12   4.184   77.6

q = 3896 J = 3.9 10³ J

= 3.9 10³​ J