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Gekata [30.6K]
4 years ago
12

Calculate the mass of onygen gas whosevolume is 320mL at 17 degree celsius and 2 atm Pressure​

Chemistry
1 answer:
GREYUIT [131]4 years ago
5 0

Answer:

The mass of oxygen gas is 0.96 g.

Explanation:

Given data:

Mass of oxygen gas = ?

Volume of as = 320 mL (320mL × 1L /1000 mL = 0.32 L)

Temperature of gas = 17 °C

Pressure of gas = 2 atm

Solution:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

17+273 = 290 K

Now we will put the values in formula.

2 atm × 0.32 L = n × 0.0821 atm.L/ mol.K   × 290 K

0.64 atm.L =  n × 23.81 atm.L/ mol.

n = 0.64 atm.L / 23.81 atm.L/ mol.

n = 0.03 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.03 mol × 32 g/mol

Mass = 0.96 g

The mass of oxygen gas is 0.96 g.

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Emphasis on object vs woman simply means <u>s</u><u>-</u><u> </u><u>e-</u><u> </u><u>x</u><u>-</u><u> </u><u>ual</u><u> objectification</u>

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2 years ago
According to the fossil record found in these sedimentary layers, what conclusion can be drawn about the movement of life to lan
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<em>A conclusion that can be drawn about the movement of life to land is that;</em>

B.) As land plants became more complex, animal life did as well.

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3 years ago
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Consider the reaction to produce methanolCO(g) + 2H2 (g) &lt;-----&gt; CH3OHAn equilibrium mixture in a 2.00-L vessel is found t
MariettaO [177]

Answer : The value of K_c of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of CH_3OH at equilibrium = 0.0406 mole

Moles of CO at equilibrium = 0.170 mole

Moles of H_2 at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of CH_3OH,CO\text{ and }H_2 at equilibrium.

\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M

\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)

The expression of equilibrium constant K_c for the reaction will be:

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

Now put all the values in this expression, we get :

K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}

K_c=10.5

Therefore, the value of K_c of the reaction is, 10.5

There are 3 conditions:

When K_{c}>1; the reaction is product favored.

When K_{c}; the reaction is reactant favored.

When K_{c}=1; the reaction is in equilibrium.

As the value of K_{c}>1. So, the reaction is product favored.

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3 years ago
The triple point of nitrogen occurs at a temperature of 63.1 K and a pressure of 0.127 atm. Its normal boiling point is 77.4 K.
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Nitrogen has a normal boiling point of 77.4 K and a melting point (at 1 atm) of 63.2 K. Its critical temperature is 126.2 K, and its critical pressure is 2.55 * 104 torr. It has a triple point at 63.1 K and 94.0 torr.

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The temperature and pressure at which the solid, liquid, and vapour phases of a pure substance can coexist in equilibrium.

- Normal melting point: 63.2 K.

- Normal boiling point: 77.4 K.

- Triple point: 0.127 atm and 63.1 K.

- Critical point: 33.5 atm and 126.0 K.

In such a way:

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In cell notation, all species appearing to the left of the double vertical line represent:________.
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