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lutik1710 [3]
3 years ago
12

The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure

Chemistry
1 answer:
AlekseyPX3 years ago
5 0

Answer : The temperature of the air in the tire is, 341 K

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

or,

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

where,

P_1 = initial pressure = 198 kPa

P_2 = final pressure = 225 kPa

T_1 = initial temperature = 27^oC=273+27=300K

T_2 = final temperature = ?

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

\frac{198kPa}{300K}=\frac{225kPa}{T_2}

T_2=340.9K\approx 341K

Therefore, the temperature of the air in the tire is, 341 K

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Determine the mass of CuSO4 • 5H20 that must be used to prepare 250mL of 2.01 M CuSO4(aq).
mario62 [17]

Given parameters:

Volume of CuSO₄ = 250mL

Concentration of CuSO₄ = 2.01M

Unknown:

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To solve this problem, we must write the chemical relationship between both species.;

             CuSO₄.5H₂O  →   CuSO₄ + 5H₂O

Now that we know the expression, it is possible to solve for the unknown mass.

First find the number of moles of CuSO₄;

         Number of moles  = Concentration x Volume

Take 250mL to L so as to ensure uniformity of units;

           Volume  = 250 x 10⁻³L

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From the equation;

             1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O  

So  0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O

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Mass of CuSO₄.5H₂O = number of moles x molar mass

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3 years ago
Sample A: 300 mL of 1M sodium chloride
yarga [219]

Answer:

sample B contains the larger density

Explanation:

Given;

volume of sample A, V = 300 mL = 0.3 L

Molarity of sample A, C = 1 M

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Molarity of sample B, C = 1.5 M

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The density of sample A  = \frac{mass}{volume} = \frac{17.55}{0.3} = 58.5 \ g/L

The density of sample B = \frac{mass}{volume} = \frac{12.72}{0.145} = 87.72 \ g/L

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