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
The answer is 6 because the number of chlorine is 2 so if you have 6 moles of chlorine the answer is 6
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Hope it helps you...
Explanation:
Start with a balanced equation.
2H2 + O2 → 2H2O
Assuming that H2 is in excess, multiply the given moles H2O by the mole ratio between O2 and H2O in the balanced equation so that moles H2O cancel.
5 mol H2O × (1 mol O2/2 mol H2O) = 2.5 mol O2
Answer: 2.5 mol O2 are needed to make 5 mol H2O, assuming H2 is in excess.
