Answer:
Approximately 6.81 × 10⁵ Pa.
Assumption: carbon dioxide behaves like an ideal gas.
Explanation:
Look up the relative atomic mass of carbon and oxygen on a modern periodic table:
Calculate the molar mass of carbon dioxide
:
.
Find the number of moles of molecules in that
sample of
:
.
If carbon dioxide behaves like an ideal gas, it should satisfy the ideal gas equation when it is inside a container:
,
where
is the pressure inside the container.
is the volume of the container.
is the number of moles of particles (molecules, or atoms in case of noble gases) in the gas.
is the ideal gas constant.
is the absolute temperature of the gas.
Rearrange the equation to find an expression for
, the pressure inside the container.
.
Look up the ideal gas constant in the appropriate units.
.
Evaluate the expression for
:
.
Apply dimensional analysis to verify the unit of pressure.
Answer:
6.53g of K₂SO₄
Explanation:
Formula of the compound is K₂SO₄
Given parameters:
Volume of K₂SO₄ = 250mL = 250 x 10⁻³L
= 0.25L
Concentration of K₂SO₄ = 0.15M or 0. 15mol/L
Unknown:
Mass of K₂SO₄ =?
Methods:
We use the mole concept to solve this kind of problem.
>>First, we find the number of moles using the expression below:
Number of moles= concentration x volume
Solving for number of moles:
Number of moles = 0.25 x 01.5
= 0.0375mole
>>Secondly, we use the number of moles to find the mass of K₂SO₄ needed. This can be obtained using the expression below:
Mass(g) = number of moles x molar mass
Solving:
To find the molar mass of K₂SO₄, we must know the atomic mass of each element in the compound. This can be obtained using the periodic table.
For:
K = 39g
S = 32g
O = 16g
Molar mass of K₂SO₄ = (39x2) + 32 + (16x4)
= 78 +32 + 64
= 174g/mol
Using the expression:
Mass(g) = number of moles x molar mass
Mass of K₂SO₄ = 0.0375 x 174 = 6.53g
Answer:

Explanation:
Hello,
In this case, since nitric acid is HNO₃ and strontium hydroxide is Sr(OH)₂ we can represent the balanced chemical reaction by equaling the atoms of strontium, nitrogen, oxygen and hydrogen at both reactants and products as shown below:

Best regards.