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2 years ago

Determine the total volume of a balloon if it is inflated with 2.50g helium at 180C and 0.87atm.

Chemistry

1 answer:

puteri [66]2 years ago

5 0

Answer:

V = 267.2 L

Explanation:

Given data:

Volume of balloon = ?

Mass of helium = 2.50 g

Temperature of gas = 180°C

Pressure = 0.87 atm

Solution:

Number of moles of helium:

Number of moles = mass/molar mass

Number of moles = 2.50 g/ 4 g/mol

Number of moles = 0.625 g

Volume of helium:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

0.87 atm× V = 0.625 mol × 0.0821 atm.L/mol.K×453 K

V = 232.45 atm.L /0.87 atm

V = 267.2 L

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Click the DeltaH is an Extensive Property button within the activity, and analyze the relationship between the two reactions tha

DaniilM [7]

Answer:

∴ΔH₂ = - 12,258 KJ

Explanation:

Enthalpy:

Enthalpy is a property of a thermodynamic system. Enthalpy of a system is equal to the sum of internal energy of the system and presser times volume of the system.

The heat absorbes or releases in a closed system is the change of enthalpy of the system.

Given reactions are:

Reaction 1: C₃H₈(g)+5O₂(g)→ 3CO₂(g)+4H₂O, ΔH₁= - 2043 KJ

Reaction 2: 6C₃H₈(g)+30 O₂(g)→ 18 CO₂(g)+24 H₂O, ΔH₂=?

Take a look at reaction 1 and reaction 2, the only difference is that 1 molecule of C₃H₈ is combusted in reaction 1 and 6 molecules of C₃H₈ is combusted in reaction 2.

We can think the reaction 2 as occurring 6 different container and each containers contains 1 molecule of C₃H₈. The enthalpy is an extensive property. Total enthapy of the 6 containers is = 6×(-2043 KJ)

= - 12,258 KJ

∴ΔH₂ = - 12,258 KJ

6 0

3 years ago

"An aqueous CaCl2 solution has a vapor pressure of 83.1mmHg at 50 ∘C. The vapor pressure of pure water at this temperature is 92

Lynna [10]

Answer : The the concentration of $CaCl_2$ in mass percent is, 41.18 %

Solution : Given,

Molar mass of water = 18 g/mole

Molar mass of $CaCl_2$ = 110.98 g/mole

First we have to calculate the mole fraction of solute.

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

$\fac{p^o-p_s}{p^o}=X_B$

where,

$p^o$ = vapor pressure of the pure component (water) = 92.6 mmHg

$p_s$ = vapor pressure of the solution = 83.1 mmHg

$X_B$ = mole fraction of solute, $(CaCl_2)$

Now put all the given values in this formula, we get the mole fraction of solute.

$\fac{92.6-83.1}{92.6}=X_B$

$X_B=0.102$

Now we have to calculate the mole fraction of solvent (water).

As we know that,

$X_A+X_B=1\\\\X_A=1-X_B\\\\X_A=1-0.102\\\\X_A=0.898$

The number of moles of solute and solvent will be, 0.102 and 0.898 moles respectively.

Now we have to calculate the mass of solute, $(CaCl_2)$ and solvent, $(H_2O)$ .

$\text{Mass of }CaCl_2=\text{Moles of }CaCl_2\times \text{Molar mass of }CaCl_2$

$\text{Mass of }CaCl_2=(0.102mole)\times (110.98g/mole)=11.32g$

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(0.898mole)\times (18g/mole)=16.164g$

Mass of solution = Mass of solute + Mass of solvent

Mass of solution = 11.32 + 16.164 = 27.484 g

Now we have to calculate the mass percent of $CaCl_2$

$Mass\%=\frac{\text{Mass of}CaCl_2}{\text{Mass of solution}}\times 100=\frac{11.32g}{27.484g}\times 100=41.18\%$

Therefore, the the concentration of $CaCl_2$ in mass percent is, 41.18 %

4 0

3 years ago

Which of the following compounds will not produce O2 gas when it decomposes? (2 points)

Umnica [9.8K]

LiOH is the right answer.

6 0

3 years ago

The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma

Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

$\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})} \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}$