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dalvyx [7]
3 years ago
7

Each of the identical volumetric flasks contains the same solution at two different temperatures. There are two identical volume

tric flasks. The first volumetric flask is at 25 degrees Celsius and is filled with a solution to approximately 50% of the neck of the flask. The second volumetric flask is at 55 degrees Celsius and is filled with a solution to approximately 80% of the neck of the flask. What changes for the solution with temperature?
Chemistry
1 answer:
Montano1993 [528]3 years ago
8 0

Explanation:

We know that molarity is the number of moles of solute present in liter of solution.

Mathematically,   Molarity = \frac{\text{no. of moles}}{\text{volume in liter}}

As molarity is dependent on volume and volume of a solution or substance is dependent on temperature. So, with increase in temperature there will occur a decrease in volume of the solution. As a result, molarity will increase as it is inversely proportional to volume.

Hence, molarity of both the solutions will be different as temperature of both the solutions is different.

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Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 4.1 g of butane is m
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12.44 g

Explanation:

2C4H10 + 13O2 = 8CO2 + 10H2O

n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).

n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).

Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.

mass of CO2 produced =

M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol

M = 0.5656/2 * 44

M = 0.2828 * 44

M = 12.44 of CO2

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Which statements are true about Figure I and Figure II below? (Check all that apply)
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A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of
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<u>Answer:</u> The volume when the pressure and temperature has changed is 1.6\times 10^2mL

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

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P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

Let us assume:

P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K

Putting values in above equation, we get:

\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL

Hence, the volume when the pressure and temperature has changed is 1.6\times 10^2mL

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