Answer:
pOH of resulting solution is 0.086
Explanation:
KOH and CsOH are monoacidic strong base
Number of moles of
in 375 mL of 0.88 M of KOH =
= 0.33 moles
Number of moles of
in 496 mL of 0.76 M of CsOH =
= 0.38 moles
Total volume of mixture = (375 + 496) mL = 871 mL
Total number of moles of
in mixture = (0.33 + 0.38) moles = 0.71 moles
So, concentration of
in mixture,
= 
Hence, ![pOH=-log[OH^{-}]=-log(0.82)=0.086](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E%7B-%7D%5D%3D-log%280.82%29%3D0.086)
H2S hydrogen sulfide gas has a higher lattice energy because
Formula: H2S
Molar mass: 34.1 g/mol
Boiling point: -76°F (-60°C)
Melting point: -115.6°F (-82°C)
Density: 1.36 kg/m³
Soluble in: Water, Alcohol
2H2O --> 2H2 + O2
The mole H2O:mole O2 ratio is 2:1
Now determine how many moles of O2 are in 50g: 50g × 1mol/32g = 1.56 moles O2
Since 1 mole of O2 was produced for every 2 moles of H2O, we need 2×O2moles = H2O moles
2×1.56 = 3.13 moles H2O
Finally, convert moles to grams for H2O:
3.13moles × 18g/mol = 56.28 g H2O
D) 56.28
A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .
It is given that,
pOH of solution = 7.1
Kw =2.93×10^(-15)
Firstly, we will calculate the value of pKw
The expression which we used to calculate the pKw is,
pKw=-log [Kw]
Now by putting the value of Kw in this expression,
pKw =−log{2.93×10^(-15)}
pKw =15log(2.93)
pKw=14.5
Now we have to calculate the pH of the solution.
As we know that,
pH+pOH=pKw
Now put all the given values in this formula,
pH+7.1=14.5
pH=7.4
Therefore, we find the value of pH of the solution is, 7.4.
learn more about pH value:
brainly.com/question/12942138
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The answer is C. <span>The wind removes surface particles, resulting in lower land surface.
Deflation is the process of lifting and moving of tiny particles of soil or sand, brought about by the wind. It can also be associated with soil erosion. This is a common phenomenon in deserts and sand dune areas near the coast.</span>