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3 years ago

What volume in liters will be occupied by 1.20 moles of helium gas at a pressure of 3.675 atm and 100*C?

Chemistry

1 answer:

lesya [120]3 years ago

5 0

Answer:

V = 121.82 L

Explanation:

Given data:

Number of moles of gas = 1.20 mol

Pressure of gas = 3.675 atm

Temperature = 100°C

Volume occupied = ?

Solution:

Formula:

General gas equation:

PV = nRT

R = general gas constant.

Now we will convert the temperature into kelvin.

100+273 = 373 K

V = nRT/P

V = 1.2 mol × 0.0821 atm.L/mol.K × 373 K / 3.675 atm

V = 447.68 / 3.675 atm

V = 121.82 L

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Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an

ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

Ti $Cl_{4(l)}$ ⇒ Ti $Cl_{4(g)}$

ΔH° $_{f}$ (Ti $Cl_{4(l)}$ ) = -804.2 kJ/mol

ΔH° $_{f}$ (Ti $Cl_{4(g)}$ ) = -763.2 kJ/mol

Therefore,

ΔH° $_{f}$ = ΔH° $_{f}$ (Ti $Cl_{4(g)}$ ) - ΔH° $_{f}$ (Ti $Cl_{4(l)}$ ) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(Ti $Cl_{4(l)}$ ) = 221.9 J/(mol*K)

s°(Ti $Cl_{4(g)}$ ) = 354.9 J/(mol*K)

Therefore,

s° = s° (Ti $Cl_{4(g)}$ ) - s°(Ti $Cl_{4(l)}$ ) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH° $_{f}$ /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

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3 years ago

Which of the following statements about the greenhouse effect is false? Greenhouse gases include carbon dioxide, chlorofluorocar

7nadin3 [17]

B is definitely correct

5 0

3 years ago

The mass of a 6.0 mL sample of kerosene is 4.92. The density of kerosene is

weqwewe [10]

4.92 grams / 6 mL = .82 grams / mL

<span>(A) 0.82 g/mL</span>

3 0

3 years ago

what is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0L

Mamont248 [21]

If x=0 then y=0 sorry dude I wish I could
Help but I’m dumb and I just want points but I’d defiantly recommend to look this up on safari or go into school and seek help

4 0

3 years ago

If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the of strontium-90 is ________ yr.

lesantik [10]

The question is incomplete, here is the complete question.

If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the half-life of strontium-90 is ________ yr.

a. 28.8 b. 30.9 c. 35.4 d. 32.2

Answer : The half-life of strontium-90 is 28.8 years.

Explanation :

This is a type of radioactive decay and all radioactive decays follow first order kinetics.

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken for decay process = 9.00 year

a = initial amount or moles of the reactant = 1.000 g

a - x = amount or moles left after decay process = 0.805 g

Putting values in above equation, we get:

$k=\frac{2.303}{9.00}\log\frac{1.00}{0.805}$

$k=0.0241\text{ year}^{-1}$

To calculate the half-life, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$0.0241\text{ year}^{-1}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=28.8\text{ years}$

Therefore, the half-life of strontium-90 is 28.8 years.

5 0

3 years ago