Answer:
1.89 × 10^-11 M
Explanation:
Recall that;
[H^+] [OH^-] = 1 × 10^-14
[H^+] = concentration of hydronium ions
[OH^-] = concentration of hydroxide ion
Where [OH^-] = 5.3 × 10−4 M
[H^+]= 1 × 10^-14/5.3 × 10−4
[H^+]= 1.89 × 10^-11 M
Answer:
165 ml
Explanation:
We are given;
Initial volume; V_a = 55 ml
Initial molarity; M_a = 3 M
Molarity of desired solution; M_b = 0.75 M
Volume of desired solution; V_b = (55 + x) ml
Where x is the volume of water to be added.
To solve for V_b, we will use the equation ;
M_a•V_a = M_b•V_b
V_b = (M_a•V_a)/M_b
V_b = (3 × 55)/0.75
V_b = 220 mL
Thus;
(55 + x) = 220
x = 220 - 55
x = 165 mL
It should be 1.13 x 10^25 atoms
Answer:
30.0 L.
Explanation:
- To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.
<em>(MV)before dilution = (MV)after dilution</em>
M before dilution = 5.0 M, V before dilution = 3.0 L.
M after dilution = 0.5 M, V after dilution = ??? L.
∵ (MV)before dilution = (MV)after dilution
∴ (5.0 M)(3.0 L) = (0.5 M)(V after dilution)
<em>∴ V after dilution = (5.0 M)(3.0 L)/(0.5 M) = 30.0 L.</em>
Answer:
The answer is "2.459".
Explanation:
The cell potential or emf norm = oxidation pot. of anode -oxidation pot. of cathode
When the pot oxidizer is the opposite of a red pot size it's been transformed to either the redox reaction so that anode was its higher oxide pot material that means (AL) so,