Answer:
3
Explanation:
From the question:
Barium phosphate reacts with sulfuric acid to form barium sulfate and phosphoric acid.
The equation for the above reaction can be written as follow:
Ba3(PO4)2 + H2SO4 —> BaSO4 + H3PO4
Now, let us balance the equation in order to obtain the coefficient of H2SO4. This is illustrated below below:
Ba3(PO4)2 + H2SO4 —> BaSO4 + H3PO4
There are 3 atoms of Ba on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of BaSO4 as shown below:
Ba3(PO4)2 + H2SO4 —> 3BaSO4 + H3PO4
There are 3 atoms of SO4 on the right side and 1 atom on the left side. It can be balance by putting 3 in front of H2SO4 as shown below:
Ba3(PO4)2 + 3H2SO4 —> 3BaSO4 + H3PO4
Now, we can see that there 6 atoms of H on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:
Ba3(PO4)2 + 3H2SO4 —> 3BaSO4 + 2H3PO4
Now the equation is balanced.
The coefficient of sulphuric acid (H2SO4) is 3
Answer:
B. When it's night time, your part of the Earth is facing away from the sun.
Explanation:
First, we need to get the concentration of [NaH2PO4]:
[NaH2PO4] =( mass / molar mass ) * volume L
when we have mass NaH2PO4 = 6.6 g & molar mass = 120g/mol & V = 0.355 L
So by substitution:
[NaH2PO4] = (6.6g / 120g/mol) * 0.355 L = 0.0195 M
then, we need to get the concentration of [Na2HPO4]:
[Na2HPO4]= (mass / molar mass ) * volume L
So by substitution:
[Na2HPO4] = (8g/ 142g/mol) * 0.355 L = 0.02 M
and when Pka of the 2nd ionization of phosphoric acid = 7.21
So by substitution in the following formula, we can get the PH:
PH = Pka + ㏒[A]/[AH]
∴PH = 7.21 + ㏒[0.02]/[0.0195]
∴ PH = 7.2
