D= M/V so, D=300/40 wich is 7.5 so D= 7.5g/m3
Answer:
- 40.66
- 9.91
Explanation:
For the first question:
Our theoretical compound is MR₂
1 mol of MR₂ contains 1 mol of M and 2 moles of R
Let's find out the molar mass:
9.45 g/mol + 18.12 g/mol . 2 = 45.69 g/mol
We can solve this, by an easy rule of three:
1 mol of MR₂ weighs 45.69 grams
Then, 0.89 moles may weigh 40.66 g
For the second question:
Our theoretical compound is D₂G
Let's determine the molar mass:
11.45 g/mol . 2 + 44.57 g/mol = 67.47 g/mol
1 mol of anything contains 6.02×10²³ molecules. By this definition we can say that 6.02×10²³ molecules weigh 67.47 grams. Let's solve by the rule of three:
6.02×10²³ molecules weigh 67.47 g
8.84×10²² molecules may weigh (8.84×10²² . 67.47 ) / 6.02×10²³ = 9.91 g
<span>Na (sodium) is highly electropositive. Its has 1 electron in its outermost orbit which is transferred to an electronegative atom to form an ionic bond.
It only needs to get rid of one valence electron to take part in a reaction. That's how it's highly reactive.</span>
3Mg + N₂= Mg₃N₂
n(Mg)=12,2g÷24,4g/mol=0,5mol-limiting reagentn
(N₂)=5,16g÷28g/mol=0,18mol
n(Mg₃N₂):n(Mg)=1:3, n(Mg₃N₂)=0,166mol, m(Mg₃N₂)=0,166·101,2=16,8g.
%(N)= 2·Ar(N)÷Mr(Mg₃N₂) = 2·14÷101,2=27,66%=0,2766
%(Mg) = 3·Ar(Mg)÷Mr(Mg₃N₂)= 3·24,4÷101,2=72,34% or 100% - 27,66%= 72,34%.
