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.
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The correct answer is + 32.5 J because heat is absorbed here by copper not evolved also after calculation we will find the H value is positive, lets calculate it:
H = m C Δt where:
m = 10 g
C = 0.13 J/g.°C
Δ t = final temperature - initial temperature = 50 - 25 = + 25 °C
so H = 10 x 0.13 x (+25) = + 32.5 J
Answer:
The enthalpy change in the the reaction is -47.014 kJ/mol.
Explanation:

Volume of water in calorimeter = 22.0 mL
Density of water = 1.00 g/mL
Mass of the water in calorimeter = m

Mass of substance X = 2.50 g
Mass of the solution = M = 2.50 g + 22 g = 24.50 g
Heat released during the reaction be Q
Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C
Specific heat of the solution is equal to that of water :
c = 4.18J/(g°C)


Heat released during the reaction is equal to the heat absorbed by the water or solution.
Heat released during the reaction =-1.433 kJ
Moles of substance X= 
The enthalpy change, ΔH, for this reaction per mole of X:

Answer:a) 11.34 g of ethane
can be formed
b)
is the limiting reagent
c) 3.44 g of the excess reagent remains after the reaction is complete
Explanation:
To calculate the moles :
1. 
2. 
According to stoichiometry :
1 mole of
require 1 mole of 
Thus 0.378 moles of
will require=
of 
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
moles of
left = (2.10-0.378) = 1.72 moles
mass of
left=
According to stoichiometry :
As 1 mole of
give = 1 mole of 
Thus 0.378 moles of
give =
of 
Mass of 
Thus 11.34 g of ethane is formed.