Answer:
Explanation:
The solution of known concentration is expressed as molarity. Molarity is the mole fraction of solute (i.e. the dissolved substance) per liter of the solution, Molarity is also commonly called molar concentration.
Mathematically;
To copy and complete the road map from the given question, we have the following array:
Volume A (L)
↓
d. multiplied by the molarity of A
↓
moles A
↓
b. multiplied by the moles of B / moles of A
↓
moles B
↓
c. divided by the molarity of B
↓
volume B (L)
Answer:
0.535 g
Explanation:
The reaction that takes place is:
- NaCl + AgNO₃ → AgCl + NaNO₃
First we <u>calculate how many AgNO₃ moles are there in 25.0 mL of a 0.366 M solution</u>, using the <em>definition of molarity</em>:
- Molarity = moles / liters
- moles = Molarity * liters
<em>Converting 25.0 mL to L </em>⇒ 25.0 / 1000 = 0.025 L
- moles = 0.366 M * 0.025 L = 0.00915 mol AgNO₃
Then we <u>convert AgNO₃ moles into NaCl moles</u>:
- 0.00915 mol AgNO₃ * = 0.00915 mol NaCl
Finally we<u> convert NaCl moles into grams</u>, using its <em>molar mass</em>:
- 0.00915 mol NaCl * 58.44 g/mol = 0.535 g
If you are doing electronic configuration for Bromine atom; then you need to put the symbol of Argon (Ar) inside the brackets because the noble gas that comes before Bromine (Br) is Argon (Ar).
When PH = -㏒[H3O+]
and we have PH = 5.061
by substitution:
∴ [ H3O+] = 10^-5.061 = 8.7x10^-6
when we have Ka = [CH3COO-][H3O+] / [CH3COOH]
we have Ka = 1.8x10^-5 & [H3O+] = 8.7x10^-6m & [CH3COOH] = 0.5 m
by substitution in Ka formula:
1.8x10^-5 = [CH3COO-]*(8.7x10^-6) / 0.5
∴[CH3COO-] = 1.034
∴we need to add 1.034 mol of sodium acetate