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Rudiy27
3 years ago
10

Express the density 5.6 g/cm3

Chemistry
1 answer:
Irina-Kira [14]3 years ago
5 0
I think it's 5,600kg/m^3
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Which describes the number of protons in the atom of a particular element
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Atomic Number

Explanation:

The number of protons an atom has is unique to that element. The number of protons is noted as the atomic number for that element on the periodic table.

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Select all that apply. Which of the following statements are true? Significant figures are an indicator of the certainty in meas
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Do electromagnets or mechanical waves move faster
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A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt
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Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

<u>Step 1:</u> Data given

Sample 1: The 1.15 M sample  has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

<u>Step 2:</u> Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1:  1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

<u>Step 3:</u> Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams  ( in 55mL)

<u>Step 4</u>: Calculate moles of AgBr

AgNO3 reacts with KBr  

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

<u>Step 5:</u> Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

8 0
3 years ago
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