Answer:
1. pH = 1.23.
2. 
Explanation:
Hello!
1. In this case, for the ionization of H2C2O4, we can write:
It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the pKa is:
The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:
2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:
It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:
Which is also shown in net ionic notation.
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Answer:
sample A
Explanation:
the first one because of the ppm value
write the equation for the reaction
that is 6 F2 +P4 =4 PF3
find the theoretical mass that is
let the theoretical yield be represented by y
theoretical yield = 78.1/100 = 120/y
y= 153.6 grams
find the number of moles of PF3
moles = mass/molar mass
= 153.6/87.97 =1.746 moles
by use of mole ratio between F2 :PF3 which is 6:4 the moles of F2 is therefore= 1.746 x 6/4 = 2.62 moles
mass = moles x molar mass
= 1.746 moles x38 g/mol = 99.6 grams
Answer:
To release 7563 kJ of heat, we need to burn 163.17 grams of propane
Explanation:
<u>Step 1</u>: Data given
C3H8 + 5O2 -----------> 3CO2 + 4H2O ΔH° = –2044 kJ
This means every mole C3H8
Every mole of C3H8 produces 2044 kJ of heat when it burns (ΔH° is negative because it's an exothermic reaction)
<u>Step 2: </u>Calculate the number of moles to produce 7563 kJ of heat
1 mol = 2044 kJ
x mol = 7563 kJ
x = 7563/2044 = 3.70 moles
To produce 7563 kJ of heat we have to burn 3.70 moles of C3H8
<u>Step 3: </u>Calculate mass of propane
Mass propane = moles * Molar mass
Mass propane = 3.70 moles * 44.1 g/mol
Mass propane = 163.17 grams
To release 7563 kJ of heat, we need to burn 163.17 grams of propane
When temperature of liquid is increased, liquid gets thinner and thinner and hence it's viscosity decreases.
Density = mass/volume.
As we increase the temperature, volume of the liquid starts to increase but mass of the liquid remains constant. As a result, density of liquid decreases.
Hope this helps!
