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3 years ago

A balloon has a volume of 3.0 L at room temperature (27oC). At what temperature would the balloon have a volume of 4.0L?

Chemistry

1 answer:

diamong [38]3 years ago

6 0

Answer: 400K

Explanation:

Given that,

Original volume of balloon V1 = 3.0L

Original temperature of balloonT1 = 27°C

Convert the temperature in Celsius to Kelvin

(27°C + 273 = 300K)

New volume of balloon V2 = 4.0L

New temperature of balloon T2 = ?

Since volume and temperature are given while pressure is constant, apply the formula for Charle's law

V1/T1 = V2/T2

3.0L/300K = 4.0L/T2

To get the value of T2, cross multiply

3.0L x T2 = 4.0L x 300K

3.0LT2 = 1200LK

Divide both sides by 3.0L

3.0LT2/3.0L = 1200LK/3.0L

T2 = 400K

Thus, at a temperature of 400 Kelvin, the balloon would have a volume of 4.0L.

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Show

SIZIF [17.4K]

Answer: 0.9375 g

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$ .....(1)

Molarity of $HCl$ solution = 0.75 M

Volume of $HCl$ solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Moles of} HCl={0.75}\times{0.025}=0.01875moles$

$CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(l)$

According to stoichiometry :

2 moles of $HCl$ require = 1 mole of $CaCO_3$

Thus 0.01875 moles of $HCl$ will require= $\frac{1}{2}\times 0.01875=0.009375moles$ of $CaCO_3$

Mass of $CaCO_3=moles\times {\text {Molar mass}}=0.009375moles\times 100g/mol=0.9375g$

Thus 0.9375 g of $CaCO_3$ is required to react with 25.0 ml of 0.75 M HCl

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3 years ago