Answer:
Approximately 
.
Explanation:
Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.
<h3>Formula mass of strontium hydroxide</h3>
Look up the relative atomic mass of 
, 
, and 
on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.
Calculate the formula mass of 
:
.
<h3>Number of moles of strontium hydroxide in the solution</h3>
means that each mole of formula units have a mass of 
.
The question states that there are 
of in this solution.
How many moles of formula units would that be?
.
<h3>Molarity of this strontium hydroxide solution</h3>
There are 
of formula units in this 
solution. Convert the unit of volume to liter:
.
The molarity of a solution measures its molar concentration. For this solution:
.
(Rounded to four significant figures.)
Given parameters:
Volume of CuSO₄ = 250mL
Concentration of CuSO₄ = 2.01M
Unknown:
Mass of CuSO₄.5H₂O = ?
To solve this problem, we must write the chemical relationship between both species.;
CuSO₄.5H₂O → CuSO₄ + 5H₂O
Now that we know the expression, it is possible to solve for the unknown mass.
First find the number of moles of CuSO₄;
Number of moles = Concentration x Volume
Take 250mL to L so as to ensure uniformity of units;
Volume = 250 x 10⁻³L
Input the parameters and solve for number of moles;
Number of moles = 250 x 10⁻³ x 2.01 = 0.5mol
From the equation;
1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O
So 0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O
Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16) = 249.6g/mole
Mass of CuSO₄.5H₂O = number of moles x molar mass
= 0.5 x 249.6
= 124.8g
The mass of CuSO₄.5H₂O is 124.8g
Answer:
The empirical formula is Ag2O.
The empirical formula is Ag2O.Explanation:
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of Ag to 2O.
do the steps ...
To get this into an integer ratio, we divide both numbers by the smaller value.
From this point on, I like to summarize the calculations in a table.
ElementAgMass/gXMolesXllRatiomllIntegers
—————————————————−———mAgXXXm7.96Xm0.07377Xll2.00mmm2
mlOXXXXl0.59mm0.0369Xml1mmmml1
There are 2 mol of Ag for 1 mol of O.
.
.
.
