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Ahat [919]
3 years ago
15

Which must be the same when comparing 1 mol of oxygen gas, O2, with 1 mol of carbon monoxide gas, CO? the mass the volume the nu

mber of molecules the number of oxygen atoms
Chemistry
2 answers:
Phoenix [80]3 years ago
6 0

Answer : The correct option is, the number of molecules.

Explanation :

As we are given that 1 mole of oxygen gas, (O_2) and 1 mole of carbon monoxide gas, (CO).

From the given options, the number of molecules must be same when comparing 1 mole of oxygen gas, (O_2) and 1 mole of carbon monoxide gas, (CO) because there are 2 molecules present in both oxygen gas and carbon monoxide gas.

The mass of 1 mole of oxygen gas, (O_2) is, 32 g and the mass of 1 mole of carbon monoxide gas, (CO) is, 28 g. So, the mass will not be same.

The number of oxygen atoms in 1 mole of oxygen gas, (O_2) is, 2\times 6.022\times 10^{23} and the number of oxygen atoms in 1 mole of carbon monoxide gas, (CO) is, 1\times 6.022\times 10^{23}. So, the number of oxygen atoms will not be same.

The volume of 1 mole of oxygen gas, (O_2) and 1 mole of carbon monoxide gas, (CO) will not be same due to the difference in the masses.

Hence, the correct option is, the number of molecules.

jekas [21]3 years ago
3 0
The number of molecules because 1 mol is equal to 6.02x10^23 molecules, therefore they both have the same amount of molecules because they both have 1 mol
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4 0
4 years ago
For a single substance at atmospheric pressure, classify the following as describing a spontaneous process, a nonspontaneous pro
Maslowich

Answer and Explanation:

a. Solid melting below its melting point  ⇒ Nonspontaneous

The process is spontaneous <u>above</u> the melting point.

b. Gas condensing below its condensation point  ⇒ Spontaneous

Below the condensation point, the subtance is liquid (it condenses)

c. Liquid vaporizing above its boiling point  ⇒ Spontaneous

A liquid vaporizes at a temperature above the boiling point, it passes from liquido to the gas state

d. Liquid freezing below its freezing point  ⇒ Spontaneous

A liquid freezes at a temperature below its freezeing point, it passes from the liquid to the solid state.

e. Liquid freezing above its freezing point  ⇒ Nonspontaneous

The freezing is spontaneous below the freezing point.

f. Solid melting above its melting point  ⇒ Spontaneous

A solid melts at a temperature above the melting point

g. Liquid and gas together at boiling point with no net condensation or vaporization  ⇒ Equilibrium system

The systems is at equilibrium: there is no net change toward the liquid or towards the gas state.

h. Gas condensing above its condensation point  ⇒ Nonspontaneous

The condensation is spontaneous at a temperature below the condensation point.

i. Solid and liquid together at the melting point with no net freezing or melting ⇒ Equilibrium system

The system is at equilibrium: there is no net change towards the solid or the liquid state.

5 0
3 years ago
A helium filled weather balloon has a volume of 806 L at 20.9°C and 753 mmHg. it is released and rises to an altitude of 6.8 km,
OLEGan [10]
<h3>Answer:</h3>

1257.45 L

<h3>Explanation:</h3>

We are given;

  • Initial volume of Helium gas, V1 as 806 L
  • Initial temperature of Helium gas,T1 as 20.9°C
  • Initial pressure of Helium gas, P1 as 753 mmHg
  • Pressure of Helium at the altitude 6.8 km, P2 as 417 mmHg
  • Temperature of Helium gas at the altitude 6.8 Km, T2 as -19.1°C

But, K = °C + 273.15

Therefore, T1 = 294.05 K and T2 = 254.05 K

  • We are required to calculate the new volume of the balloon at 6.8 km.
  • To determine the new volume we are going to use the combined gas law.
  • According to the combined gas law, \frac{P1V1}{T1}=\frac{P2V2}{T2}

Thus, rearranging the formula;

V2=\frac{P1V1T2}{P2T1}

V2=\frac{(753)(806L)(254.05K)}{(417)(294.05)}

V2=1257.45L

Therefore, the volume of the balloon at an altitude of 6.8 km is 1257.45 L

7 0
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