Explanation:
Charges on both magnesium and oxygen is 2. Though opposite in sign, they have equal charges so, both of them will be cancelled by each other.
As a result, formula of magnesium oxide is MgO and not
.
The student write the equation as
, it is not correct.
Therefore, given equation will be balanced as follows.

Since, number of atoms on both reactant and product side are equal. Hence, this equation is completely balanced.
The average atomic mass of your mixture is 1.03 u
.
The average atomic mass of H is the weighted average of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its % abundance).
Thus,
0.99 × 1.01 u = 0.998 u
0.002 × 2.01 u = 0.004 u
0.008 × 3.02 u = <u>0.024 u</u>
TOTAL = 1.03 u
Answer:
Percent yield = 90.8%
Explanation:
The reaction of Fe with S to produce FeS is:
Fe + S → FeS
<em>Where the moles of Fe added in excess of S are the moles of FeS</em>
<em />
Now, percent yield is defined as 100 times the ratio between actual yield (13.8g of FeS) and theoretical yield (15.2g FeS):
Percent yield = 13.8g / 15.2g * 100
<h3>Percent yield = 90.8%</h3>
Answer:
0.42 g
Explanation:
<u>We have: </u>
pH = 12.10 (25 °C)
V = 800.0 mL = 0.800 L
To find the mass of sodium hydroxide (NaOH) we can use the pH:


![pOH = -log ([OH^{-}])](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%20%28%5BOH%5E%7B-%7D%5D%29%20)
![[OH]^{-} = 10^{-pOH} = 10^{-1.90} = 0.013 M](https://tex.z-dn.net/?f=%5BOH%5D%5E%7B-%7D%20%3D%2010%5E%7B-pOH%7D%20%3D%2010%5E%7B-1.90%7D%20%3D%200.013%20M)
Now, we can find the number of moles (η) of OH:
Since we have 1 mol of OH in 1 mol of NaOH, the number of moles of NaOH is equal to 1.04x10⁻² moles.
Finally, with the number of moles we can find the mass of NaOH:

<em>Where M is the molar mass of NaOH = 39.9 g/mol </em>

Therefore, the mass of sodium hydroxide that the chemist must weigh out in the second step is 0.42 g.
I hope it helps you!