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olga nikolaevna [1]
4 years ago
10

A sample of oxygen gas has a volume of 3.24 L at 29°C. What volume will it occupy at 104°C if the pressure and number of mol are

constant? Enter your answer in the provided box.
Chemistry
1 answer:
ohaa [14]4 years ago
8 0

<u>Answer:</u> The final volume of the oxygen gas is 4.04 L

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=3.24L\\T_1=29^oC=(29+273)K=302K\\V_2=?\\T_2=104^oC=(104+273)K=377K

Putting values in above equation, we get:

\frac{3.24L}{302K}=\frac{V_2}{377K}\\\\V_2=4.04L

Hence, the final volume of the oxygen gas is 4.04 L

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How old is a bone if it now .3125 of C-14 when it originally had 80.0g of C-14
Taya2010 [7]
<h3>Answer:</h3>

42960 years

<h3>Explanation:</h3>

<u>We are given;</u>

  • Remaining mass of C-14 in a bone is 0.3125 g
  • Original mass of C-14 on the bone is 80.0 g
  • Half life of C-14 is 5370 years

We are required to determine the age of the bone;

  • Using the formula;
  • Remaining mass = Original mass × 0.5^n , where n is the number of half lives.

Therefore;

0.3125 g = 80.0 g × 0.5^n

3.90625 × 10^-3 = 0.5^n

  • Introducing logarithm on both sides;

log 3.90625 × 10^-3 = n log 0.5

Solving for n

n = log 3.90625 × 10^-3 ÷ log 0.5

   = 8

  • Therefore, the number of half lives is 8
  • But, 1 half life is 5370 years
  • Therefore;

Age of the rock = 5370 years × 8

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Thus, the bone is 42960 years old

7 0
4 years ago
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Marina CMI [18]

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6 0
3 years ago
Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "
Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

<em>As Mass = density × volume</em>

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

<em>PV = nRt</em>

<em>V = nRT/P</em>

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

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