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4 years ago

For each set of values, calculate the missing variable using the ideal gas law.

Chemistry

1 answer:

allsm [11]4 years ago

8 0

Answer:

1. n = 0.174mol

2. T= 26.8K

3. P = 1.02atm

4. V = 126.88L

Explanation:

1. P= 2.61atm

V = 1.69L

T = 36.1 °C = 36.1 + 273= 309.1K

R = 0.082atm.L/mol /K

n =?

n = PV / RT = (2.61x1.69)/(0.082x309.1)

n = 0.174mol

2. P = 302 kPa = 302000Pa

101325Pa = 1atm

302000Pa = 302000/101325 = 2.98atm

V = 2382 mL = 2.382L

T =?

n = 3.23 mol

R = 0.082atm.L/mol /K

T= PV /nR = (2.98x2.382)/(3.23x0.082) = 26.8K

3. P =?

V = 0.0250 m³ = 25L

T = 288K

n = 1.08mol

R = 0.082atm.L/mol /K

P = nRT/V = (1.08x0.082x288)/25 = 1.02atm

4. P = 782 torr

760Torr = 1 atm

782 torr = 782/760 = 1.03atm

V =?

T = 303K

n = 5.26 mol

R = 0.082atm.L/mol /K

V = nRT/P

V = (5.26x0.082x303)/1.03 = 126.88L

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1 year ago

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Anon25 [30]

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3 years ago

A 10.0-milliliter sample of NaOH(aq) is neutralized by 40.0 milliliters of 0.50 M HCl. What is the molarity of the NaOH(aq)?

Ludmilka [50]

$\frac{10 mL}{1000 mL} = 0.01 L$

$\frac{40mL}{1000mL} =0.04L$

$M _{1} V_{1}=M_{2}V_{2}$

$M_{1}(0.01)=(0.50)(0.04)$

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4 years ago

Carbon-14 is a radioactive isotope that decays according to first-order kinetics in a process that has a half-life of 5730 years

Sliva [168]

Answer : The time passed in years is $2.83\times 10^3\text{ years}$

Explanation :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

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

$k=\frac{0.693}{5730\text{ years}}$

$k=1.21\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the time passed.

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

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

where,

k = rate constant = $1.21\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = ?

a = let initial amount of the reactant = X g

a - x = amount left after decay process = $71\% \times (x)=\frac{71}{100}\times (X)=0.71Xg$

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{X}{0.71X}$

$t=2831.00\text{ years}=2.83\times 10^3\text{ years}$

Therefore, the time passed in years is $2.83\times 10^3\text{ years}$

8 0

3 years ago