First, we have to see how K2O behaves when it is dissolved in water:
K2O + H20 = 2 KOH
According to reaction K2O has base properties, so it forms a hydroxide in water.
For the reaction next relation follows:
c(KOH) : c(K2O) = 1 : 2
So,
c(KOH)= 2 x c(K2O)= 2 x 0.005 = 0.01 M = c(OH⁻)
Now we can calculate pH:
pOH= -log c(OH⁻) = -log 0.01 = 2
pH= 14-2 = 12
Answer:
The concentration of HI in the final solution is 1.3 M.
Explanation:
Dilution is the reduction in concentration of a chemical in a solution. It is achieved by adding more solvent to the same amount of solute.
In other words, in a dilution, the amount of solute does not change, but the volume of the solvent does: as more solvent is added, the concentration of the solute decreases, since the volume (and weight) of the solution increases.
When dealing with dilution you will use the following equation:
C1*V1= C2*V2
- C1 = initial concentration
- V1 = initial volume
- C2 = final concentration
- V2 = final volume
In this case:
- C1 = 3 M
- V1 = 65 mL= 0.065 L (being 1000 mL= 1 L)
- C2 = ?
- V2 = 0.15 L
Replacing:
3 M* 0.065 L= C2*0.15 L
Solving:
C2= 1.3 M
<u><em>The concentration of HI in the final solution is 1.3 M.</em></u>
Answer:
25.0%
A solution of benzene (C6H6) and toluene (C7H8) is 25.0% benzene by mass. At 25 °C the vapor pressures of pure benzene and pure toluene are 94.2 torr and 28.4 torr, respectively.
Considering the Charles's law, the sample of carbon dioxide gas will occupy 308.72 mL.
<h3>Charles's law</h3>
Charles's law establishes the relationship between the temperature and the volume of a gas when the pressure is constant. This law says that the volume is directly proportional to the temperature of the gas: for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases.
Mathematically, Charles's law states that the ratio between volume and temperature will always have the same value:
Considering an initial state 1 and a final state 2, it is fulfilled:
<h3>Final volume in this case</h3>
In this case, you know:
- V1= 250 mL
- T1= 25 C= 298 K (being 0 C=273 K)
- V2= ?
- T2= 95 C= 368 K
Replacing in Charles's law:
Solving:
<u><em>V2= 308.72 mL</em></u>
Finally, the sample of carbon dioxide gas will occupy 308.72 mL.
Learn more about Charles's law:
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