Answer:
The mass of water to be added is 2 pounds
Explanation:
The given parameters are;
The mass of the given solution = 2 pounds
The concentration of the given solution = 30%
The desired concentration of the solution = 15%
The mass, m of the acetic acid in the given solution = 30% × 2 pounds
m = 30/100 × 2 pounds = 0.6 pounds
To make a 15% acetic acid solution of acetic acid, the mass X of the required volume, is given as follows;
15% of X = 0.6 pounds
15/100 × X = 3/20 × 0.6 pounds
∴ The mass of the solution required X = 0.6 × 20/3 = 4 pounds
The mass of the solution that will contain 0.6 pounds of acetic acid giving a 15% acetic acid solution is 4 pounds
Therefore, the mass of water to be added to the original solution to make the a 15% acetic acid solution is 2 pounds.
An alkyne contains four carbon atoms.... so if you do 26 multiplied by 4 it equals 104... I do not know if that’s the answer so I apologize if it’s wrong :,)
Answer:
30.62 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 55 L
Initial pressure (P₁) = 3.2 atm
Initial temperature (T₁) = 520 K
Final temperature (T₂) = 760 K
Final pressure (P₂) = 8.4 atm
Final volume (V₂) =?
The final volume of the gas can be obtained as follow:
P₁V₁ / T₁ = P₂V₂ / T₂
3.2 × 55 / 520 = 8.4 × V₂ / 760
176 / 520 = 8.4 × V₂ / 760
Cross multiply
520 × 8.4 × V₂ = 176 × 760
4368 × V₂ = 133760
Divide both side by 4368
V₂ = 133760 / 4368
V₂ = 30.62 L
Therefore, the new volume of the gas is 30.62 L
Answer:
450. g of 0.173 % KCN solution contains 779 mg of KCN.
Explanation:
Mass of the solution = m
Mass of the KCN in solution = 779 mg
Mass by mass percentage of KCN solution = 0.173%
1 mg = 0.001 g
m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g
450. g of 0.173 % KCN solution contains 779 mg of KCN.
Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that
As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M
