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Lena [83]
3 years ago
12

A tire at 21°C has a pressure of 0.82 atm. Its temperature decreases to –3.5°C. If there is no volume change in the tire, what i

s the pressure after the temperature change?
Chemistry
2 answers:
Alenkinab [10]3 years ago
8 0

Answer : The pressure after the temperature change is, 0.75 atm.

Solution :

According to the Gay-Lussac's Law, the pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T     (At constant volume and number of moles)

Or,

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

where,

P_1 = initial pressure = 0.82 atm

P_2 = final pressure = ?

T_1 = initial temperature = 21^oC=273+21=294K

T_2 = final temperature = -3.5^oC=273+(-3.5)=269.5K

Now put all the given values in the above formula, we get the final pressure of the gas.

\frac{0.82atm}{P_2}=\frac{294K}{269.5K}

P_2=0.75atm

Therefore, the pressure after the temperature change is, 0.75 atm

Damm [24]3 years ago
4 0

The correct answer is A 0.75 atm


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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

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Mass of H_2O = 78.0 g

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Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

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