Doesn’t require it but can be used with red meat and seafood
Molarity is expressed as
the number of moles of solute per volume of the solution. The mass of oxalic acid dihydrate needed for the solution is calculated as follows:
Amount in moles: (0.357 mol H2C2O4•2H2O / L) (.250 L ) = 0.0893 mol H2C2O4•2H2O
Amount in mass : 0.0893 mol H2C2O4•2H2O (126.08 g / mol ) = 11.2589 g H2C2O4•2H2O
Hope this answers the question. Have a nice day.
Option B: high pressure and low temperature
A gas is more soluble under high pressure and low temperature conditions.
On increasing temperature of a gas, its kinetic energy increases. The increase in kinetic energy increases the motion of particles of gas this causes most of the gaseous particles to escape from the gas phase. Thus, less particles are available to dissolve in liquid and solubility decreases.
The effect of pressure on solubility of gas can be explained with the help of Henry's law. According to the law, at constant temperature, solubility of gas and partial pressure of gas are related to each other as follows:
Here, p is the partial pressure of the gas, 
is Henry's law constant, and
c is the concentrate of the gas.
According to above relation, concentration of gas decreases on decreasing partial pressure. Thus, on increasing pressure, concentration of gas increases this increases the solubility of gas in liquid.
Therefore, solubility of gas is greatest at high pressure and low temperature.
Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
