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lbvjy [14]
3 years ago
5

If 45 mL of water are added to 250 mL of a 0.75 M K2SO4 solution, what will the molarity of the diluted solution be?

Chemistry
1 answer:
krok68 [10]3 years ago
8 0

Answer:

\large\boxed{\large\boxed{0.64M}}

Explanation:

When you form a <em>diluted solution</em> from a mother (concentrated) solution, the moles of solute are determined by the mother solution.

The main equation is:

Molarity=\dfrac{\text{moles of solute}}{\text{volume of the solution in liters}}

Then, since the moles of solute is the same for both the mother solution and the diluted solution:

          \text{Molarity mother solution }\times\text{ volume mother solution}=\\\\=\text{Molarity diluted solution }\times\text{ volume diluted solution}

Substitute and solve for the molarity of the diluted solution:

           250mL\times 0.75M=(45mL+250mL)\times M\\\\\\M=\dfrac{250mL\times 0.75M}{295mL}=0.64M

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Nitrogen ion (Nitride) is ... Cation/Anion/Neither? # of protons? # of electrons? Charge (1-, 2-, 3-, 1+, 2+, 3+, or 0) Number o
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Explanation:

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What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
Margarita [4]

Answer:

108.6 g

Explanation:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:

  • P = 1.00 atm
  • V = 56.0 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 0 °C ⇒ 0 + 273.2 = 273.2 K

<u>Inputting the data</u>:

  • 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
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Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

  • 2.5 mol N₂ * \frac{2molNaN_3}{3molN_2} = 1.67 mol NaN₃

Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:

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