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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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How many grams N2F4 can be produced from 225 g F,?​
zavuch27 [327]

Answer:

308 g

Explanation:

Data given:

mass of Fluorine (F₂) = 225 g

amount of N₂F₄ = ?

Solution:

First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄

Reaction:

          2F₂ + N₂ -----------> N₂F₄

Now look at the reaction for mole ratio

          2F₂     +    N₂   ----------->  N₂F₄

        2 mole                              1 mole

So it is 2:1 mole ratio of Fluorine to N₂F₄

As we Know

molar mass of F₂ = 2(19) = 38 g/mol

molar mass of N₂F₄ = 2(14) + 4(19) =

molar mass of N₂F₄ = 28 + 76 =104 g/mol

Now convert moles to gram

                 2F₂          +       N₂   ----------->  N₂F₄

        2 mole (38 g/mol)                        1 mole (104 g/mol)

                 76 g                                           104 g

So,

we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.

Apply unity formula

                  76 g of F₂ ≅ 104 g of N₂F₄

                   225 g of F₂ ≅ X of N₂F₄

Do cross multiplication

                  X of N₂F₄ = 104 g x 225 g / 76 g

                  X of N₂F₄ = 308 g

So,

308 g N₂F₄ can be produced from 225 g F₂

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Explanation:

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