No - a precipitation will occur though. Potassium nitrate is soluble in water, so the potassium and nitrate ions will remain spectator ions and stay in solution. Lead (II) hydroxide is not soluble, and will precipitate out of solution to form a solid product.
Well thats a pretty difficult question, I think the answer might be 4
Answer:
9.07 L
Explanation:
<em>Calculate the volume occupied at s.t.p by 6.89 g of NH₃ gas [H = 1.0, N = 14.0].</em>
Step 1: Given and required data
- Molar mass of NH₃ (M): 17.0 g/mol
Step 2: Calculate the moles (n) of NH₃
We will use the following epxression.
n = m / M
n = 6.89 g / (17.0 g/mol) = 0.405 mol
Step 3: Calculate the volume occupied by 0.405 moles of NH₃ at STP
At STP, 1 mole of NH₃ occupies 22.4 L (assuming ideal behavior).
0.405 mol × 22.4 L/1 mol = 9.07 L
Answer:
520mL
Explanation:
Data obtained from the question include:
Molarity of stock solution (M1) = 12M
Volume of stock solution (V1) = 130mL
Molarity of diluted solution (M2) = 3M
Volume of diluted (V2) =..?
The volume of the diluted solution can be obtained as follow:
M1V1 = M2V2
12 x 130 = 3 x V2
Divide both side by 3
V2 = 12 x 130 / 3
V2 = 520mL.
Therefore, 520mL of the diluted solution can be prepared.
