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Mandarinka [93]
2 years ago
14

A large balloon contains 1.00 mol of helium in a volume of 22.4 L at 0.00oC. What pressure will the helium exert on its containe

r? Give your answer in atm and in mmHg.
Chemistry
1 answer:
Yuki888 [10]2 years ago
8 0

Answer:

The pressure exerted by helium on the walls of the container is approximately 760.6 mmHg

Explanation:

The parameters given in the question are;

Amount of moles of helium in balloon = 1.00 mole

Volume of balloon = 22.4 L = ‪0.0224 m³

Temperature of balloon = 0.00°C = 273.15 K

Therefore, we have the pressure given by the universal gas equation as follows;

P·V = n·R·T

So that

P =  \frac{nRT}{V} = \frac{1 \times 8.3145 \times 273.15}{0.0224 } = 101388.646 \ Pa

Converting the resultant pressure in pascals, Pa to millimetre mercury, mmHg we have;

1 mmHg = ‪133.3 Pa

∴ 1 Pa = 1/133.33 mmHg and

101388.646 Pa = 101388.646× 1/133.3 mmHg = ‪760.604996 mmHg

Therefore the pressure exerted by helium on the container ≈ 760.6 mmHg.

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A waste collection tank can hold 18754 kg of methanol, which has a density of 0.788 g/cm3.
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Given:
Mass of methanol, m = 18754 kg
Density of methanol, ρ = 0.788 g/cm³

By definition, the volume of methanol in the collection tank is
Volume  = mass/density
               = \frac{(18754 \, kg)*(10^{3} \,  \frac{g}{kg}) }{(0.788 \,  \frac{g}{cm^{3})} }  \\ \\ =2.38 \times 10^{7} \,  \frac{g}{cm^{3}}

Answer:  2.38 x 10⁷ g/cm³
3 0
2 years ago
A sample of a compound of Cl and O reacts with an excess of H2 to give 0.233g of HCl and 0.403g of H2O. Determine the empirical
Vsevolod [243]

Answer:

Cl₂O₇

Explanation:

For the reaction:

ClₓOₙ + H₂ → HCl + H₂O

Moles of HCl and moles of H₂O are:

HCl: 0.233g HCl ₓ (1mol / 36.46g) = 6.39x10⁻³ mol HCl

H₂O: 0.403g H₂O ₓ (1mol / 18.02g) = 2.236x10⁻² mol H₂O

As you can see, moles of HCl are equivalent to moles of Cl in the compound and moles of H₂O are equivalent to moles of O in the compound, that means:

6.39x10⁻³ mol Cl

2.236x10⁻² mol O

Empirical formula is the simplest ratio of atoms presents in a molecule. If Cl is <em>1</em>, Oxygen will be:

2.236x10⁻² mol / 6.39x10⁻³ = <em>3.5</em>

As empirical formula must be given in natural numbers, the empirical formula is:

<em>Cl₂O₇</em>

<em></em>

4 0
2 years ago
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An electrically neutral atom of gallium has 31 electrons and 39 neutrons. What is the mass number for an atom of gallium?. A. 31
hammer [34]
If the atom is neutral (meaning, not charged) the number of electron is equal to the number of protons. The mass number of an atom is the sum of the number of proton and the number of neutrons. From the given above, the mass number of gallium is 31 + 39. The answer is letter D. 70.
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Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC
Salsk061 [2.6K]

<u>Answer:</u> The \Delta G for the reaction is 54.6 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}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm

Putting values in above equation, we get:

K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92

  • 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 = 22.92

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol

Hence, the \Delta G for the reaction is 54.6 kJ/mol

7 0
2 years ago
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