The volume of the stock solution that has a concentration of 1.5 M SO2 and is diluted to a 0.54 M solution with a volume of 0.18 L is 0.065L.
<h3>How to calculate volume?</h3>
The concentration of a solution can be calculated using the following formula:
C1V1 = C2V2
Where;
- C1 = initial concentration = 1.5M
- C2 = final concentration = 0.54M
- V1 = initial volume = ?
- V2 = final volume = 0.18L
1.5 × V1 = 0.54 × 0.18
1.5V1 = 0.0972
V1 = 0.0972 ÷ 1.5
V1 = 0.065L
Therefore, the volume of the stock solution that has a concentration of 1.5 M SO2 and is diluted to a 0.54 M solution with a volume of 0.18 L is 0.065L.
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Answer:
290.82g
Explanation:
The equation for the reaction is given below:
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 now, let us obtain the masses of H2SO4 and Al2(SO4)3 from the balanced equation. This is illustrated below:
Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 +64 = 98g/mol
Mass of H2SO4 from the balanced equation = 3 x 98 = 294g
Molar Mass of Al2(SO4)3 = (2x27) + 3[32 + (16x4)]
= 54 + 3[32 + 64]
= 54 + 3[96] = 54 + 288 = 342g
Now, we can obtain the mass of aluminium sulphate formed by doing the following:
From the equation above:
294g of H2SO4 produced 342g of Al2(SO4)3.
Therefore, 250g of H2SO4 will produce = (250 x 342)/294 = 290.82g of Al(SO4)3
Therefore, 290.82g of aluminium sulphate (Al(SO4)3) is formed.
Answer: The pH of a 4.4 M solution of boric acid is 4.3
Explanation:
at t=0 cM 0 0
at eqm
So dissociation constant will be:
Give c= 4.4 M and
= ?
Putting in the values we get:
Also
Thus pH of a 4.4 M
solution is 4.3
Answer:
6, double
Explanation:
Hex- is a prefix for number 6.
Ene- is a suffix for a double bond.
Density of liquid=
.
so, density of liquid=
= 1.2 gm/cm³.