<span>6.12<span>(<span>1024</span>)</span></span><span>=<span><span>(6.12)</span><span>(<span><span>1e</span>+24</span>)</span></span></span><span>=<span><span>6.12e</span>+24</span></span>
<span>
=
</span>
Answer:
1.209g of MgO participates
Explanation:
In this problem, we have 0.030 moles of MgO that participates in a particular reaction.
And we are asked to solve for the mass of MgO that participates, that means, we need to convert moles to grams.
To convert moles to grams we need to use molar mass of the compound:
<em>1 atom of Mg has a molar mass of 24.3g/mol</em>
<em>1 atom of O has a molar mass of 16g/mol</em>
<em />
That means molar mass of MgO is 24.3g/mol + 16g/mol = 40.3g/mol
And mass of 0.030 moles of MgO is:
0.030 moles MgO * (40.3g/mol) =
<h3>1.209g of MgO participates</h3>
Answer:
MgCl2 + 2AgNO3 → 2AgCl + Mg(NO3)2
Explanation:
I'm assuming you want to balance it so...
The first thing I see is that there are two chlorines on the reactant side and one on the product side
Adding a coefficient of 2 would get 2AgCl2
Now there are two silvers on the reactant side, so add a 2 to AgNO3 on the products side. Now they are all balanced.
If that is not what you are looking for let me know!