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schepotkina [342]
3 years ago
6

7th grade science pls help

Chemistry
1 answer:
snow_tiger [21]3 years ago
3 0

Answer: The last answer.

Explanation:

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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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The buret clamp was used to stabilize the
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thermometer, calorimeter

Explanation:

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If the pressure of 50.0 mL of oxygen gas at 100°C increases from 735 mm Hg to 925 mm Hg, what is
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3 years ago
A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate
AlladinOne [14]

The given question is incomplete. The complete question is as follows.

A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 212 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)

Explanation:

When Ba^{2+} and Na_{2}SO_{4} are added  then white precipitate forms. And, reaction equation for this is as follows.

       Ba^{2+} + SO^{2-}_{4} \rightarrow BaSO_{4}

It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of BaSO_{4} is 233.43.

Now, we will calculate the number of moles as follows.

  No. of moles = mass × M.W

                        = \frac{0.212}{233.43}

                        = 0.00091 mol of BaSO_{4}

Hence, it means that 0.00091 mol of Ba^{2+}. Now, we will calculate the mass as follows.

       Mass = moles × MW

                 = 0.00091 \times 137.327

                 = 0.124 grams or 124 mg of barium

Thus, we can conclude that mass of barium into the original solution is 124 mg.

8 0
3 years ago
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