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larisa86 [58]
3 years ago
12

Suppose 20.8 g of sodium iodide is dissolved in 250. mL of a 0.70 M aqueous solution of silver nitrate. Calculate the final mola

rity of ALL IONS in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Round your answer to 3 significant digits.
Chemistry
1 answer:
gtnhenbr [62]3 years ago
7 0

Answer:

[Ag+] = [NO3-] = 0.700M

0.555M = [Na+] = [I-]

Explanation:

To solve this question we need to find the moles of sodium iodide, NaI, using its molar mass -. With the moles and the volume we can find the molarity of Na+ and I-. The molarity of the ions of silver nitrate, AgNO3 doesn't change because we are assuming the volume doesn't change:

<em>Molarity Ag⁺ = Molarity NO₃⁻ = 0.700M</em>

<em>Moles NaI -Molar mass: 149.89g/mol-</em>

20.8g NaI * (1mol/149.89g) = 0.0139 moles NaI

<em>Molarity:</em>

0.0139 moles NaI / 0.250L = <em>0.555M = [Na+] = [I-]</em>

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4 years ago
How many grams N2F4 can be produced from 225 g F,?​
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308 g

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Solution:

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

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          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

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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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3 years ago
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