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MaRussiya [10]
3 years ago
15

A 100.0 mL sample of a 0.200 M aqueous solution of K2CrO4 was added to 100.0 mL of a 0.100 M aqueous solution of BaCl2. The mixt

ure was stirred and the precipitate was collected, dried carefully, and weighed. How many grams of precipitate should be obtained? The reaction is shown below: K2CrO4(aq) + BaCl2(aq) ď‚® BaCrO4(s) + 2 KCl(aq) A. 2.05 g B. 2.53 g C. 5.07 g D. 6.16 g E. 7.60 g
Chemistry
1 answer:
stepan [7]3 years ago
5 0
The correct answer is 2.53 g of precipitate, BaCrO4.

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How many MOLES of boron tribromide are present in 3.20 grams of this compound ?
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Answer:

1) There are 0.0128 moles of BBr3 present in 3.20 grams of this compound

2) There are 909.35 grams of BBr3 present in 3.63 moles of this compound

Explanation:

<u>Step 1: </u>Given data

Boron tribromide = BBr3

Molar mass of Boron = 10.81 g/mole

Molar mass of Bromide = 79.9 g/mole

Molar mass of Boron tribromide = 10.81 + 3*79.9 = 250.51 g/mole

<u>Step 2:</u> Calculating number of moles

Number of moles = mass / molar mass

Number of moles of BBr3 = 3.20 grams / 250.51 g/mole

Number of moles of BBr3 = 0.0128 moles

<u>Step 3:</u> Calculating mass

If we have 3.63 moles of boron tribromide (= BBr3)

Mass of BBr3 = number of moles of BBr3 * Molar mass of BBr3

Mass of BBr3 = 3.63 moles * 250.51 g/mole

Mass of BBr3 = 909.35 grams

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Answer:

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Explanation:

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They allow maintenance of the high energy phosphate stores in its normal concentration ranges which discard the problems associated with ATP-consuming reactions in these tissues as against the presence of adenosine triphosphate.

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