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Semmy [17]
2 years ago
14

A 45ml of a 4M solution of CaBr2 contains how many grams of CaBr2?

Chemistry
1 answer:
natta225 [31]2 years ago
6 0

The required mass of calcium bromide is 35.98 grams.

<h3>What is molarity?</h3>

Molarity is any solution is define as the number of moles of solute present in per liter of solution as;

M = n/V, where

  • M = molarity = 4M
  • V = volume = 45mL = 0.045L

Moles will be calculated by using the above equation as:

n = (4)(0.045) = 0.18 mole

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

  • W = given mass
  • M = molar mass

Mass of CaBr₂ = (0.18mol)(199.89g/mol) = 35.98g

Hence required mass of CaBr₂ is 35.98 grams.

To know more about molarity, visit the below link:
brainly.com/question/22283918

#SPJ1

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How many particles are in 12.47 grams of NaCl?
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The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.376 g sample of e
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Answer:

The answer to your question is: C₄H₁₀O

Explanation:

Data

          CxHyOz

mass sample : 1.376 g

mass CO₂ = 3.268 g

mass H₂O = 1.672 g

Process

Reaction

                      CxHyOz  + O₂ ⇒   CO₂  +  H₂O

1.- Calculate the moles and mass of carbon

Molecular mass CO₂ = 44g

                      44 g of CO₂ --------------  12 g of C

                      3.268 g of CO₂  --------    x

                         x = (3.268 x 12) / 44

                        x = 0.891 g of Carbon

                       12 g of carbon -----------  1 mol

                       0.891 g of C     ----------   x

                       x = (0.891 x 1) / 12

                       x = 0.0743 moles of carbon

2.- Calculate the moles and mass of hydrogen

                      18 g of water --------------- 2 g of H

                      1.672 g of H₂O ------------  x

                      x = (1.672 x 2) / 18

                      x = 0.186 g of hydrogen

                      1 g of hydrogen ------------  1 mol of H

                      0.186 g of H       ------------  x

                      x = (0.186 x 1) / 1

                      x = 0.186 moles of H

3.- Calculate the mass of Oxygen and its moles

Mass of Oxygen = 1.376 - 0.891 - 0.186

                           = 0.299 g of O₂

Moles of Oxygen

                             16 g of Oxygen ---------------- 1 mol

                             0.299 g of O    -----------------  x

                             x = (0.299 x 1) / 16

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4.- Divide by the lowest number of moles

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Hydrogen     0.186/ 0.019 = 9.7 = 10

Oxygen         0.019/ 0.019 = 1

5.- Write the empirical formula

                              C₄H₁₀O                  

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Answer:

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Explanation:

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