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Nezavi [6.7K]
1 year ago
8

Type the correct answer in the box. a pharmacist works with a 1.75 m solution of sodium bromide (nabr) and water. the volume of

the solution is 84.0 milliliters. if the pharmacist dilutes the solution to 1.00 m, what is the volume of the new solution? express your answer to three significant figures. the volume of the new solution is milliliters.
Chemistry
1 answer:
Elodia [21]1 year ago
8 0

Answer: 147 mL

Explanation:

<u>Given:</u>

Molarity of the sodium bromide (NaBr) solution (M1)  = 1.75 M

Volume of the solution (V1) = 84 mL

Molarity of the diluted NaBr solution (M2) = 1 M

Using the dilution formula to solve for V2:

\begin{gathered}M_{1} V_{1}=M_{2} V_{2} \\V_{2}=(1.75 \mathrm{M} \times 84 \mathrm{~mL}) / 1 \mathrm{M} \\V_{2}=147 \mathrm{~mL}\end{gathered}

Therefore, the new volume of the solution is 147 mL

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The speed of light passing from air into the substance is calculated as follows;

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speed of light in the substance = (3 x 10⁸) / (1.65)

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Learn more about speed of light here: brainly.com/question/104425

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In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water
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The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

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We can use the following equation to calculate the volume of more concentrated solution required:

\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

Thus, we would need 5.00 mL of the more concentrated solution.

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\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}

Thus, we would need 45.0 mL of water to prepare the solution.

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<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

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the total mass of reactants is equal to the total mass of products.

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