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SSSSS [86.1K]
2 years ago
6

A balloon with a volume of 2.0 l at 25°c is placed in a hot room at 35°c. the pressure on the balloon is constant at 1.0 atm. fo

rmula to use: v1 t1 = v2 t2 how does the volume of the balloon change after moving it to the hot room?
Chemistry
2 answers:
shusha [124]2 years ago
5 0

The volume increases and the final volume is 2.1 L

<h3>Why volume increases?</h3>

The volume of a gas is directly related to the heat and pressure. If temperature increases, volume increases.

If pressure increases, volume decreases

Based on Boyle's law,

The absolute temperature of a gas is directly proportional to the volume of the gas.

As the balloon change increases its temperature, the volume increases

The Boyle's equation is:

V₁T₂ = V₂T₁

Where,

  • V is volume
  • T is absolute temperature of 1, initial state and 2, final state of the gas.

Replacing :

V₁ = 2.0L

T₂ = 35°C + 273.15 = 308.15K

V₂ = ?

T₁ = 25°C + 273.15 = 298.15 K

2.0L x 308.15 K = V₂ x 298.15 K

2.1 L = V₂

Final volume is 2.1L

Hence, The volume increases and the final volume is 2.1 L

Learn more about gas laws here ;

brainly.com/question/12669509

#SPJ1

Archy [21]2 years ago
4 0

Answer:

increases, 2.1

Explanation:

I got it right.

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3 years ago
CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

7 0
3 years ago
Which are balanced and which are unbalanced?
erica [24]

A. is balanced

B. is not balanced

C. is not balanced

3 0
3 years ago
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Answer:

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Explanation:

4 0
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Why are numbers placed before the elements and molecules in a<br>chemical equation?​
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Answer:

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Explanation:

they tell you how many atoms there are of its type.

ex. 4 C²H³. there are 4 of the c² and h³ (couldn't find a way to use subscript so used the power signs to show.)

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