Answer:
We need 7.5 mL of the 1M stock of NaCl
Explanation:
Data given:
Stock = 1M this means 1 mol/ L
A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL
Step 2: Calculate the volume of stock we need
The moles of solute will be constant
and n = M*V
M1*V1 = M2*V2
⇒ with M1 = the initial molair concentration = 1M
⇒ with V1 = the volume we need of the stock
⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L
⇒ with M2 = the concentration of the new solution = 0.15 M
1*V1 = 0.15*(50)
V1 = 7.5 mL
Since 0.0075 L of 1M solution contains 0.0075 moles
50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M
We need 7.5 mL of the 1M stock of NaCl
Answer:
HClO₃ /chloric acid /suffix -ic/ ClO₃⁻ (chlorate)
HClO₂/ chlorous acid/ suffix -ous/ ClO₂⁻ (chlorite)
HNO₃ /nitric acid /suffix -ic/ NO₃⁻ (nitrate)
HNO₂/ nitrous acid/ suffix -ous/ NO₂⁻ (nitrite)
Explanation:
Chlorine has 4 positive oxidation numbers to form oxyacids: +1, +3, +5 and +7.
- When it uses the oxidation number +5, it forms HClO₃, which is named chloric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion ClO₃⁻ (chlorate).
- When it uses the oxidation number +3, it forms HClO₂, which is named chlorous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion ClO₂⁻ (chlorite).
Nitrogen has 2 positive oxidation numbers to form oxyacids: +3 and +5.
- When it uses the oxidation number +5, it forms HNO₃, which is named nitric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion NO₃⁻ (nitrate).
- When it uses the oxidation number +3, it forms HNO₂, which is named nitrous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion NO₂⁻ (nitrite).
Your answer to your question is: 1s² 2s² 2p⁶
