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3 years ago

What volume does 2800mg of nitrogen gas occupy at 98Kpa and -10 C?

Chemistry

1 answer:

Evgen [1.6K]3 years ago

6 0

Answer:

2.23L

Explanation:

Using the general gas law as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas constant (0.0821 Latm/molK)

T = temperature (K)

According to the information provided in this question,

Mass of nitrogen = 2800mg

Since 1g = 1000mg

2800mg = 2800/1000

= 2.8g

Using mole = mass ÷ molar mass

Molar mass of nitrogen gas (N2) = 28g/mol

mole = 2.8/28

mole of N2 gas = 0.1mol

Pressure = 98kPa

1 kilopascal (kPa) = 1000pascal (pa)

98kPa = 98 × 1000

= 98000pascal

Since 1 Pascal = 9.869 × 10^-6 atmosphere (atm)

98000pascal = 98000 × 9.869 × 10^-6

= 0.967atm

Temperature = -10°C

Kelvin = °C + 273

Kelvin = -10 + 273

K = 263K

Hence, using PV = nRT

0.967 × V = 0.1 × 0.0821 × 263

0.967V = 2.159

V = 2.159/0.967

V = 2.23

Volume of Nitrogen gas = 2.23 L

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I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

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Answer: The balanced chemical equation is $2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

Explanation:

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For the given unbalanced chemical equation:

$KClO_3(s)\rightarrow KCl(s)+O_2(g)$

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