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Alisiya [41]
3 years ago
9

What is the empirical formula for a compound if 300.00 g of it is known to contain 82.46224 g of molybdenum, 45.741 g of chlorin

e and the rest is bromine
Chemistry
1 answer:
Marysya12 [62]3 years ago
8 0

Answer:

MoClBr₂

Explanation:

First we calculate the mass of bromine in the compound:

  • 300.00 g - (82.46224 g + 45.741 g) = 171.79676 g

Then we<u> calculate the number of moles of each element</u>, using their <em>respective molar masses</em>:

  • 82.46224 g Mo ÷ 95.95 g/mol = 0.9594 mol Mo
  • 45.741 g Cl ÷ 35.45 g/mol = 1.290 mol Cl
  • 171.79676 g Br ÷79.9 g/mol = 2.150 mol Br

Now we <u>divide those numbers of moles by the lowest number among them</u>:

  • 0.9594 mol Mo / 0.9594 = 1
  • 1.290 mol Cl / 0.9594 = 1.34 ≅ 1
  • 2.150 mol Br / 0.9594 = 2.24 ≅ 2

Meaning the empirical formula is MoClBr₂.

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What volume of 0.100 m sodium chloride must be added to 75.0 ml of 0.200 m lead(ii) nitrate to precipitate all of the lead ions?
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When a saturated solution of NH4Br dissolved in 100 grams of water is cooled from 60°C to 30°C, how much NH4Br will precipitate?
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Answer:

m_{precipitated}=24.8g

Explanation:

Hello,

In this case, since at 60 °C, 108 grams of ammonium bromide are completely dissolved in 100 grams of water for a saturated solution, once it is cooled to 30 °C, wherein only 83.2 grams are completely dissolved in 100 grams of water, the following mass will precipitate:

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Best regards.

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