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Andrej [43]
3 years ago
15

A 7.100 L sample of gas is cooled from 73.0 °C to a temperature at which its volume is 5.700 L. What is this new temperature? As

sume no change in pressure of the gas.
Chemistry
1 answer:
arsen [322]3 years ago
8 0
The  mew  temperature  of  a  sample of   gas  is  calculated  using Charles  law  formula

that  is  v1/T1  = V2/T2
V1=  7.100 L
T1=73 + 273 = 346 K
V2 = 5.700 L
T2=?

by  making the  T2  the   subject  of the  formula
T2 =V2T1/V1

T2= 5.700 x  346/7.100 =277.8 K
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Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

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