Question

question 26
fill in the blanks below to show how to use 5.00 m hbr solution to make 2.00 l of 2.75 m hbr solution.
combine blank of select an answer hbr solution with select an answer to make 2.00 l of 2.75 m hbr solution.
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question 27
you need to dilute a solution of 44 m hydrochloric acid. you need the dilute solution to be 11 m and need 160 ml of it. how many ml of the concentrated solution are needed?
blank ml
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question 28
how many ml of a 3.3 m hcl solution would it take to make a 250 ml of a 1.0 m solution?
253.30
825.00
75.76
75.45
76.06

Explanation:

Question 26

Step1: Recall dilution formula

The dilution formula is \( M_1V_1 = M_2V_2 \), where \( M_1 \) is the initial molarity, \( V_1 \) is the initial volume, \( M_2 \) is the final molarity, and \( V_2 \) is the final volume.

Step2: Identify values

Here, \( M_1 = 5.00 \, M \), \( M_2 = 2.75 \, M \), \( V_2 = 2.00 \, L \). We need to find \( V_1 \).

Step3: Solve for \( V_1 \)

Using \( V_1=\frac{M_2V_2}{M_1} \), substitute the values: \( V_1 = \frac{2.75 \, M \times 2.00 \, L}{5.00 \, M} = 1.10 \, L \) (or 1100 mL). The concentrated solution is 5.00 M, and the amount of solvent (water) to add is \( V_2 - V_1 = 2.00 \, L - 1.10 \, L = 0.90 \, L \) (or 900 mL of water).

Step1: Use dilution formula

The dilution formula is \( M_1V_1 = M_2V_2 \), where \( M_1 = 44 \, M \), \( M_2 = 11 \, M \), \( V_2 = 160 \, mL \). We need to find \( V_1 \).

Step2: Solve for \( V_1 \)

Rearrange the formula to \( V_1=\frac{M_2V_2}{M_1} \). Substitute the values: \( V_1=\frac{11 \, M \times 160 \, mL}{44 \, M} = 40 \, mL \).

Step1: Recall dilution formula

The dilution formula is \( M_1V_1 = M_2V_2 \), where \( M_1 = 3.3 \, M \), \( M_2 = 1.0 \, M \), \( V_2 = 250 \, mL \). We need to find \( V_1 \).

Step2: Solve for \( V_1 \)

Rearrange the formula to \( V_1=\frac{M_2V_2}{M_1} \). Substitute the values: \( V_1=\frac{1.0 \, M \times 250 \, mL}{3.3 \, M} \approx 75.76 \, mL \).

Answer:

Combine \(\boldsymbol{1.10 \, L}\) (or 1100 mL) of \(\boldsymbol{5.00 \, M}\) \( HBr \) solution with \(\boldsymbol{0.90 \, L}\) (or 900 mL) of water to make 2.00 L of 2.75 M \( HBr \) solution.

Question 27