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kolezko [41]
2 years ago
6

Enter your answer in the provided box. Liquid methanol (CH3OH) can be used as an alternative fuel in pickup and SUV engines. An

industrial method for preparing it uses the catalytic hydrogenation of carbon monoxide: CO(g) + 2H2(g) CH3OH(l) How much heat (in kJ) is released when 15.0 L of CO at 85°C and 112 kPa reacts with 17.0 L of H2 at 75°C and 744 torr?
Chemistry
1 answer:
andreyandreev [35.5K]2 years ago
3 0

Answer: E=∆H*n= -40.6kj

Explanation:

V(CO) =15L=0.015M³

P=11200Pa

T=85C=358.15K

PV=nRT

n=(112000×0.015)/(8.314×358.15)

n(Co)= 0.564mol

V(Co)= 18.5L = 0.0185m³

P=744torr=98191.84Pa

T= 75C = 388.15k

PV=nRT

n= (99191.84×0.0185)/(8.314×348.15)

n(H2) = 0.634mol

n(CH30H) =1/2n(H2)=1/2×0.634mol

=0.317mol

∆H =∆Hf{CH3OH}-∆Hf(Co)

∆H=-238.6-(-110.5)

∆H = 128.1kj

E=∆H×n=-40.6kj.

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<h3><u>Condensation of gases into liquids by kinetic molecular theory:</u></h3>

The "kinetic molecular theory" explains the states of matter based on the matter composed of very tiny little particles that are constantly in motion. The theory also explains the observable properties and behaviors of solids, liquids, and gases.

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2 years ago
The molar solubility of pbi2 is 1.5 103 m.
Vsevolod [243]

Answer: -

Concentration of PbI₂ = 1.5 x 10⁻³ M

PbI₂ dissociates in water as

PbI₂ ⇄ Pb²⁺ + 2 I⁻

So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.

Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =

= 1.5 x 10⁻³ x 2 M

= 3 x 10⁻³ M

PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.

[Pb²⁺] = 1.5 x 10⁻³ M

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Given 2Na + Cl2=2NaCl, what is the excess reactant? What is the limiting reactant?
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Answer: The limiting reactant is Na

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A fluid occupying has a mass of 4mg. Calculate its density and specific volume in SI, EE, and BG units.
kondaur [170]

The question is incomplete, complete question is:

A fluid occupying 3.2 m^3 of volume has a mass of 4 Mg. Calculate its density and specific volume in SI, EE, and BG units.

Explanation:

1) Mass of liquid = m = 4 Mg = 4 × 1,000 kg = 4,000 kg

(1 Mg = 1000 kg)

Volume of the fluid = V = 3.2 m^3

Density of the fluid = D

D=\frac{m}{V}=\frac{4,000 kg}{3.2 m^3}=1,250 kg/m^3

Specific volume is the reciprocal of the density :

V_{specific}=\frac{1}{Density}

Specific volume of the fluid = S_v

S_v=\frac{1}{D}=\frac{1}{1,250 kg/m^3}=0.0008 m^3/kg

2)

Density of the fluid in English Engineering units  = D (lb/ft^3)

1 kg = 2.20462 lb

1 m = 3.280 ft

D=\frac[1,250\times 2.20462 lb}{(3.280 ft)^3=78.95 lb/ft^3

Specific volume of the fluid :

=\frac{1}{78.95 lb/ft^3}=0.0127 ft^3/lb

3)

Density of the fluid in British Gravitational System units  = D (slug/ft^3)

1 kg = 0.06852 slug

1 m = 3.280 ft

D=\frac[1,250\times 0.0685218 slug}{(3.280 ft)^3=2.43 slug/ft^3

Specific volume of the fluid :

=\frac{1}{2.43 slug/ft^3}=0.412 ft^3/slug

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