The chemical reaction is written as:
2Zn + O2 = 2ZnO
We are given the amount of the product to be produced from the reaction. We use this value and the relation of the substances in the reaction to calculate what is asked. We do as follows:
2.10 g ZnO ( 1 mol / 81.408 g ) ( 1 mol O2 / 2 mol ZnO ) ( 32 g / 1 mol ) = 0.414 g O2 is needed
The balanced chemical
reaction will be:
2H2O = 2H2 + O2
<span>We are given the amount of water used in the decomposition reaction. This will be our
starting point.</span>
<span>17.0 g H2O</span> (1 mol H2O/ 18.02 g H2O) (1 mol O2/2
mol <span>H2O</span>) ( 32.00 g O2/1mol O2) = 15.09 g O2
Percent yield = actual yield / theoretical yield x 100
<span>Percent yield =10.2 g / 15.09 g
x 100</span>
Percent yield = 67.58%
Answer:
0.4 M
Explanation:
Molarity is defined as moles of solute, which in your case is sodium hydroxide,
NaOH
, divided by liters of solution.
molarity
=
moles of solute
liters of solution
Notice that the problem provides you with the volume of the solution, but that the volume is expressed in milliliters,
mL
.
Moreover, you don't have the number of moles of sodium hydroxide, you just have the mass in grams. So, your strategy here will be to
determine how many moles of sodium hydroxide you have in that many grams
convert the volume of the solution from milliliters to liters
So, to get the number of moles of solute, use sodium hydroxide's molar mass, which tells you what the mass of one mole of sodium hydroxide is.
7
g
⋅
1 mole NaOH
40.0
g
=
0.175 moles NaOH
The volume of the solution in liters will be
500
mL
⋅
1 L
1000
mL
=
0.5 L
Therefore, the molarity of the solution will be
c
=
n
V
c
=
0.175 moles
0.5 L
=
0.35 M
Rounded to one sig fig, the answer will be
c
=
0.4 M
Explanation:
Answer:
M₂ = 0.0745 M
Explanation:
In case of titration , the following formula can be used -
M₁V₁ = M₂V₂
where ,
M₁ = concentration of acid ,
V₁ = volume of acid ,
M₂ = concentration of base,
V₂ = volume of base .
from , the question ,
M₁ = 0.0952 M
V₁ = 38.73 mL
M₂ = ?
V₂ = 49.48 mL
Using the above formula , the molarity of ammonia , can be calculated as ,
M₁V₁ = M₂V₂
0.0952 M * 38.73 mL = M₂* 49.48 mL
M₂ = 0.0745 M
