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1 month ago

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A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr

essure is 0.600 atm. The gas is heated with a flame to a temperature of 72.4 C, what is it’s pressure at this temperature

1 answer:

KiRa [2.9K]1 month ago

3 0

Answer:

Explanation:

Given data:

Initial temperature T₁ = 25.2°C = 298.2K

Initial pressure P₁ = 0.6atm

Final temperature = 72.4°C = 345.4K

What we need to find:

Final pressure = ?

To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:

$\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} }$

Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.

Now we can substitute the known variables:

$\frac{0.6}{298.2} = \frac{P_{2} }{345.4}$

P₂ = 0.7atm

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A compound has a mass percentage of 53.46% C, 6.98% H, and 39.56% O. What is the empirical formula for this compound

lorasvet [2795]

Answer:

The empirical formula is: C₂H₃O

Explanation:

The empirical formula, often referred to as the “minimum formula,” provides the simplest way to represent a chemical compound, illustrating both the present elements and their minimum integer ratios in terms of atoms.

The mass percentage composition indicates the proportion by mass of each element in a compound.

Using 100 g of the compound as a reference, the percentage can be expressed in grams. Thus, presuming 100 g of the compound consists of 53.46 g of C, 6.98 g of H, and 39.56 g of O.

Considering the molecular masses of each component, one can derive the number of relative atoms for each element:

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H: $6.98 g *\frac{1 mol}{1.01 g } = 6.91 moles$

O: $39.56 g *\frac{1 mol}{16g } = 2.47 moles$

Subsequently, divide each value by the smallest one:

C: $\frac{4.45 moles}{2.47 moles}= 1.80$

H: $\frac{6.91 moles}{2.47 moles}= 2.8$

O: $\frac{2.47 moles}{2.47 moles}=1$

Approximating decimals to the nearest integer gives:

C: 2

H: 3

O: 1

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Molal-freezing-point-depression constant $(K_f)$ for water = $1.86^oC/m$

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First, we need to determine the mass of the water.

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$\text{Mass of water}=1.00g/mL\times 275mL=275g=0.275kg$

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Formula utilized:

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of }KNO_3}{\text{Molar mass of }KNO_3\times \text{Mass of water in Kg}}$

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i = Van't Hoff factor = 2 (for KNO₃ electrolyte)

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m = molality

Substituting all known values into this equation yields

$0.0^oC-(-14.5^oC)=2\times (1.86^oC/m)\times \frac{\text{Mass of }KNO_3}{101.1g/mol\times 0.275kg}$

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Consequently, the mass of KNO₃ that must be added is, $1.08\times 10^2g$

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