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geniusboy [140]
3 years ago
9

Suppose 40.8g of copper(II) acetate is dissolved in 200.mL of a 0.70 M aqueous solution of sodium chromate.

Chemistry
1 answer:
Contact [7]3 years ago
4 0

Answer:

0.42 M

Explanation:

The reaction that takes place is:

  • Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)

First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:

(200 mL = 0.200L)

  • 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄

Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:

  • 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂

Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:

  • 0.224 mol - 0.14 mol = 0.085 mol

Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:

  • 0.085 mol / 0.200 L = 0.42 M

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Part 1. A chemist reacted 12.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat
galben [10]

The mass of NaCl needed for the reaction is 91.61 g

We'll begin by calculating the number of mole of F₂ that reacted.

  • Volume (V) = 12 L
  • Temperature (T) = 280 K
  • Pressure (P) = 1.5 atm
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Number of mole (n) =?

PV = nRT

1.5 × 12 = n × 0.0821 × 280

18 = n × 22.988

Divide both side by 22.988

n = 18 / 22.988

n = 0.783 mole

Next, we shall determine the mole of NaCl needed for the reaction.

F₂ + 2NaCl —> Cl₂ + 2NaF

From the balanced equation above,

1 mole of F₂ reacted with 2 moles of NaCl.

Therefore,

0.783 mole F₂ will react with = 0.783 × 2 = 1.566 moles of NaCl.

Finally, we shall determine the mass of 1.566 moles of NaCl.

  • Mole = 1.566 moles
  • Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
  • Mass of NaCl =?

Mass = mole × molar mass

Mass of NaCl = 1.566 × 58.5

Mass of NaCl = 91.61 g

Therefore, the mass of NaCl needed for the reaction is 91.61 g

Learn more about stiochoimetry: brainly.com/question/25830314

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How to change 5 % W/V of NaCl to ppm , M ? molar mass = 58.5<br>please clear explain​
11Alexandr11 [23.1K]

Answer:

50000ppm and 0.855M.

Explanation:

ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters

A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.

To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:

<em>mg NaCl:</em>

5g * (1000mg / 1g) = 5000mg

<em>L Solution:</em>

100mL * (1L / 1000mL) = 0.100L

ppm:

5000mg / 0.100L = 50000ppm

To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:

5g * (1mol / 58.5g) = 0.0855moles NaCl

Molarity:

0.0855mol NaCl / 0.100L = 0.855M

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