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2 years ago

What is the pressure of 0.5 mol nitrogen (N2) gas in a 5.0 L container at 203 K?

Chemistry

1 answer:

Yuki888 [10]2 years ago

6 0

Answer:

The pressure of 0.5 mole of nitrogen gas= $168.78 k pa$

Given:

Moles of nitrogen gas=0.5mole

Volume of nitrogen gas=5L

Temperature of nitrogen gas=203K

To Find:

Pressure of 0.5g mole of nitrogen gas

Step by Step by Explanation:

Formula used for calculating the pressure of a gas is achieved through ideal gas equation

According to Ideal Gas;

$P V=n R T$

From this above equation, pressure can be calculated as

$P=\frac{n R T}{V}$

Where P=Pressure of the gas

N=number of moles=0.5moles

R=Gas constant $=8.3144598 L k p_{a} k^{-1} m o l^{-1}$

V=Volume of nitrogen gas in litres=5L

T=Temperature of nitrogen gas=203K

Substitute all the values in the above equation we get

$P=\frac{0.5 \times 8.3144598 \times 203}{5}$

$P=\frac{843.9177}{5}$

$P=168.78 k pa$

Result:

Thus the pressure of 0.5 mole of nitrogen gas is $168.78 k pa$

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A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calc

Sunny_sXe [5.5K]

Answer:

a) 101%

b)59.7%

Explanation:

The equation for the thermal decomposition of baking soda is shown;

2NaHCO3 → Na2CO3 + H2O + CO2

Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles

From the reaction equation;

2 moles of baking soda yields 1 mole of sodium carbonate

0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.

Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.

%yield = actual yield/theoretical yield ×100

% yield = 0.991/0.9805 ×100

%yield = 101%

Since ;

2NaHCO3 → Na2CO3 + H2O + CO2

And H2O + CO2 ---> H2CO3

Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3

Molar mass of H2CO3= 62.03 gmol-1

Molar mass of baking soda= 84 gmol-1

Therefore, mass of baking soda=

0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3

% of NaHCO3= 0.88/1.473 × 100 = 59.7%

7 0

2 years ago

Which of the following shows the correct rearrangement of the the heat equation q = mCpΔT to solve for specific heat?

strojnjashka [21]

When
q = m*Cp*ΔT

when q is Heat energy in Joules

and m is the mass of the substance in Kg

and Cp is the specific heat (J/Kg.K)

and Δ T is the change in temperature in Kelvin

so, by rearranging the formula we can get the specific heat Cp from:

∴Cp = q / m*ΔT

6 0

2 years ago

Read 2 more answers

How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge

Ivan

Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:- The balanced equation for the combustion of ethane is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

$36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})$

= 126 L $O_2$

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

$105.0LO_2(\frac{4LCO_2}{7L O_2})$

= 60.0 L $CO_2$

Similarly, let's calculate the volume of water vapors formed:

$105.0L O_2(\frac{6L H_2O}{7L O_2})$

= 90.0 L $H_2O$

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

$105.0LO_2(\frac{2LC_2H_6}{7LO_2})$