The molar mass of a gas that moves 1.25 times as fast as CO2 is 28.16 g.
<h3>
Molar mass of the gas</h3>
The molar mass of the gas is determined by applying Graham's law of diffusion.
R₁√M₁ = R₂√M₂
R₁/R₂ = √M₂/√M₁
R₁/R₂ = √(M₂/M₁)
where;
- R₁ is rate of the CO2 gas
- M₁ is molar mass of CO2 gas
- R₂ is rate of the second gas
- M₂ is the molar mass of the second gas
R₁/1.25R₁ = √(M₂/44)
1/1.25 = √(M₂/44)
0.8 = √(M₂/44)
0.8² = M₂/44
M₂ = 0.8² x 44
M₂ = 28.16 g
Thus, the molar mass of a gas that moves 1.25 times as fast as CO2 is 28.16 g.
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Answer:

Explanation:
Hello,
In this case, we are asked to compute the by mass percent for the given 3.2 g of ethylene glycol in 43.5 g of water. In such a way, since the by mass percent is computed as follows:

Whereas the solute is the ethylene glycol and the solvent the water, therefore we obtain:

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Explanation:
( a )
The decomposition reaction of KClO₃ to form diatomic oxygen gas is the following way -
2 KClO₃ ( s ) -------> 2 KCl ( s ) + 3O₂ ( g )
( b )
The reaction of solid metal of Aluminum with Iodine in the following way -
2 Al ( s ) + 3 I₂ ( s ) ------> Al₂I₆ ( s )
( c )
The reaction of sodium chloride with aqueous sulfuric acid is as follow -
2 NaCl ( s ) + H₂SO₄ ( aq ) ------> 2 HCl ( g ) + Na₂SO₄ ( aq )
( d )
The reaction of phosphoric acid with potassium hydroxide , in the following way -
H₃PO₄ ( aq ) + KOH ( aq ) ------> KH₂PO₄ ( aq ) + H₂O ( l )
Your question is incomplete. However, I found a similar problem fromanother website as shown in the attached picture.
To solve this problem, you must know that at STP, the volume for any gas is 22.4 L/mol. So,
Moles O₂: 156.8 mL * 1 L/1000 mL* 1 mol/22.4 L = 0.007 moles
Mass calcium: 0.007 mol O₂ * 2 mol Ca/1 mol O₂ * 40 g/mol Ca =
<em> 0.56 g Ca</em>