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Artist 52 [7]
4 years ago
13

Please help me with this and explain! Thank you so much. Best answer gets Brainliest <3

Chemistry
1 answer:
saw5 [17]4 years ago
8 0
Answer: TO SACRIFICE MY OWN LIFE FOR PAKISTAN

add me on snap and ill help
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3 years ago
Arrange the following compounds from left to right in order of increasing percentage by mass of hydrogen: a) H2O, b) C12H26, c)
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Answer:

(A) N4H6  (B) H2O  (C) LiH  (D) C12H26

Explanation:

The given compounds have been arranged  from left to right in order of increasing percentage by mass of hydrogen.

The percent by mass of hydrogen can be calculated by mass of hydrogen in that compound divided by total mass of that compound and finally multiplying the result with 100 to obtain the required percentage.

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3 years ago
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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3 years ago
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