Answer: 162.8 grams
Explanation:
Magnesium nitrate has a chemical formula of Mg(NO3)2.
Given that:
Number of moles of Mg(NO3)2 = 1.1 moles
Mass in grams of Mg(NO3)2 = ?
For Molar mass of Mg(NO3)2, use atomic mass of magnesium = 24g, nitrogen = 14g, oxygen = 16g
Mg(NO3)2 = 24g + (14g + 16gx3) x 2
= 24g + (14g + 48g) x 2
= 24g + (62g) x 2
= 24g + 124g
= 148g/mol
Now, apply the formula:
Number of moles = Mass in grams / molar mass
1.1 moles = Mass / 148g/mol
Mass = 1.1 moles x 148g/mol
Mass = 162.8 grams
Thus, there are 162.8 grams of magnesium nitrate.
'cause alphe-particle which was +ve charge, get repulsion from the atom, so he deducted that.......
<u>Answer:</u> The equilibrium concentration of CO is 0.243 atm
<u>Explanation:</u>
We are given:
Initial partial pressure of carbon dioxide = 0.902 atm
As, carbon dioxide is present initially. This means that the reaction is proceeding backwards.
For the given chemical equation:

<u>Initial:</u> 0.902
<u>At eqllm:</u> 3x (0.902-3x)
The expression of
for above equation follows:

We are given:

Putting values in above equation, we get:

So, equilibrium concentration of CO = 3x = (3 × 0.0810) = 0.243atm[/tex]
Hence, the equilibrium concentration of CO is 0.243 atm