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sdas [7]
3 years ago
9

The Law of Conservation of Mass states that mass is neither created nor destroyed during cher

Chemistry
2 answers:
dmitriy555 [2]3 years ago
8 0
The answer is true:)))))))))))))))
Vikki [24]3 years ago
5 0
The answer would be true
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The pressure of a sample of helium in a 200. ml container is 2.0 atm. If the helium is compressed to a pressure of 40. atm witho
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Answer:

V_2=10mL

Explanation:

Hello there!

In this case, according to the given information, it will be possible for us to solve this problem by using the Boyle's law as an inversely proportional relationship between pressure and volume:

P_2V_2=P_1V_1

In such a way, we solve for the final volume, V2, and plug in the initial volume and pressure and final pressure to obtain:

V_2=\frac{P_1V_1}{P_2} \\\\V_2=\frac{2.0atm*200.mL}{40.atm}\\\\V_2=10mL

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. Give an example of a salt that is less soluble when the temperature increases
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Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M
Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

  • 2AgNO₃(aq) + K₂CrO₄(aq)  → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to <em>identify the limiting reactant</em>:

We have:

  • 0.20 M * 50.0 mL = 10 mmol of AgNO₃
  • 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.

Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:

  • 4 mmol K₂CrO₄ * \frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4} = 1326.92 mg Ag₂CrO₄
  • 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
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2 years ago
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