Answer : The volume for 6.0m HCl solution required = 62.5 ml
Solution : Given,
Initial concentration of HCl solution = 6.0m
Final concentration of HCl solution = 1.5m
Final volume of HCl solution = 250 ml
Initial volume of HCl solution = ?
Formula used for dilution is,
where,
= initial concentration
= final concentration
= initial volume
= final volume
Now put all the given values in above formula, we get the initial volume of HCl solution.
= 62.5 ml
Therefore, the volume for 6.0m HCl solution required = 62.5 ml
3
One from fe
And two from cl2
Answer:
b
Explanation:
[H3O+] = 10-pH = 10-3.4 ≅ 3.981 x 10^-4 moles/liter
Answer:
0.6941 mg
Explanation:
First we <u>calculate how many LiNO₃ moles there are</u>, using the <em>given concentration and volume</em>:
- 1.0 mL * 0.10 M = 0.10 mmol LiNO₃
As 1 mol of LiNO₃ contains 1 mol of Li,<em> in the problem solution there are 0.10 mmol of Li</em> (the only metallic ion present).
Now we<u> convert Li milimoles into miligrams</u>, using its <em>atomic mass</em>:
- 0.10 mmol Li * 6.941 mg/mmol = 0.6941 mg
Answer:
27 liters of hydrogen gas will be formed
Explanation:
Step 1: Data given
Number of moles C = 1.03 moles
Pressure H2 = 1.0 atm
Temperature = 319 K
Step 2: The balanced equation
C +H20 → CO + H2
Step 3: Calculate moles H2
For 1 mol C we need 1 mol H2O to produce 1 mol CO an 1 mol H2
For 1.03 moles C we'll have 1.03 moles H2
Step 4: Calculate volume H2
p*V = n*R*T
⇒with p = the pressure of the H2 gas = 1.0 atm
⇒with V = the volume of H2 gas = TO BE DETERMINED
⇒with n = the number of moles H2 gas = 1.03 moles
⇒with R = the gas constant = 0.08206 L*Atm/mol*K
⇒with T = the temperature = 319 K
V = (n*R*T)/p
V = (1.03 * 0.08206 *319) / 1
V = 27 L
27 liters of hydrogen gas will be formed
