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tangare [24]
2 years ago
13

A car tire was inflated to 82.0 kPa in a repair shop where the temperature is 26.0 °c. What is the temperature of the gas in the

tire if the pressure in the tire increases to 87.3 kPa when it is out of the repair shop?
Chemistry
1 answer:
Vsevolod [243]2 years ago
4 0

Answer: The temperature of the gas in the tire if the pressure in the tire increases to 87.3 kPa is 45^0C

Explanation:

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

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=82.0kPa\\T_1=26.0^oC=(26.0+273)K=299K\\P_2=87.3kPa\\T_2=?

Putting values in above equation, we get:

\frac{82.0}{299}=\frac{87.3}{T_2}\\\\T_2=318K=(318-273)^0C=45^0C

Thus the temperature of the gas in the tire if the pressure in the tire increases to 87.3 kPa is 45^0C

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Explanation: see attachment below

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In a particular mass of kau(cn)2, there are 6.66 × 1020 atoms of gold. What is the total number of atoms in this sample?
Vinvika [58]

Total number of atoms in the sample is the sum of number of atoms of all the elements present in the sample.

Number of gold, Au atoms in KAu(CN)_2 = 6.66\times 10^{20} atoms     (given)

From the formula of compound that is KAu(CN)_2 it is clear that the number of potassium and gold are same whereas those of carbon and nitrogen are 2 times of them.

So, the number of atoms of each element is:

Number of potassium, K atoms in KAu(CN)_2 = 6.66\times 10^{20} atoms    

Number of carbon, C atoms in KAu(CN)_2 = 2\times6.66\times 10^{20} atoms = 13.32\times 10^{20}

Number of nitrogen, N atoms in KAu(CN)_2 = 2\times6.66\times 10^{20} atoms = 13.32\times 10^{20}

Total number of atoms in KAu(CN)_2 = Number of gold, Au atoms+Number of potassium, K atoms +Number of carbon, C atoms + Number of nitrogen, N atoms

Total number of atoms in KAu(CN)_2 = 6.66\times 10^{20}+6.66\times 10^{20}+13.32\times 10^{20}+13.32\times 10^{20}

Total number of atoms in KAu(CN)_2 = 39.96\times 10^{20} atoms

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7 0
2 years ago
A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resultin
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Answer:

C) 0.28 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Moles =Molarity \times {Volume\ of\ the\ solution}

Potassium hydroxide will furnish hydroxide ions as:

KOH\rightarrow K^{+}+OH^-

Given :

<u>For Potassium hydroxide : </u>

Molarity = 0.25 M

Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

Moles =0.25 \times {40.0\times 10^{-3}}\ moles

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

Ba(OH)_2\rightarrow Ba^{2+}+2OH^-

Given :

<u>For Barium hydroxide : </u>

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

Moles =2\times 0.15 \times {60.0\times 10^{-3}}\ moles

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

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Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity_{OH^-}=\frac{0.028 }{100\times 10^{-3}}

<u> The final concentration of hydroxide ion = 0.28 M</u>

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