Answer : The excess reactant in the combustion of methane in opem atmosphere is
molecule.
Solution : Given,
Mass of methane = 23 g
Molar mass of methane = 16.04 g/mole
The Net balanced chemical reaction for combustion of methane is,

First we have to calculate the moles of methane.
=
= 1.434 moles
From the above chemical reaction, we conclude that
1 mole of methane react with the 2 moles of oxygen
and 1.434 moles of methane react to give
moles of oxygen
The Moles of oxygen = 2.868 moles
Now we conclude that the moles of oxygen are more than the moles of methane.
Therefore, the excess reactant in the combustion of methane in open atmosphere is
molecule.
Answer:
y or z one of those two i think its z but im so sorry if im wrong if not z then try y also I learned this last year.
Answer:
"1 M" will be the right solution.
Explanation:
The given values are:
Number of moles,
= 2 moles
Volume of solution,
= 2000 mL
or,
= 2 L
Now,
The molarity of the solution will be:
= 
On substituting the values, we get
= 
= 
<u>Answer:</u>
<u>For A:</u> The average molecular speed of Ne gas is 553 m/s at the same temperature.
<u>For B:</u> The rate of effusion of
gas is 
<u>Explanation:</u>
<u>For A:</u>
The average molecular speed of the gas is calculated by using the formula:

OR

where, M is the molar mass of gas
Forming an equation for the two gases:
.....(1)
Given values:

Plugging values in equation 1:

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.
<u>For B:</u>
Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

Where, M is the molar mass of the gas
Forming an equation for the two gases:
.....(2)
Given values:

Plugging values in equation 2:

Hence, the rate of effusion of
gas is 
KOH(s) ----> K⁺(aq)+OH⁻ (aq)
s-solid,
aq - aqueous