How many grams of CaBr2 would be needed to create 450. mL of a 2.00 M
solution? Round off your answer to the nearest gram. Record your answer in the
grid below.

Mass of CaBr₂ = 179.9 g

Explanation:

Given data:

Volume of CaBr₂ solution = 450.0  mL (450/1000 = 0.45 L)

Molarity of solution = 2.00 M

Mass of CaBr₂ = ?

Solution:

Formula:

Molarity = number of moles / volume of solution in L

Now we will put the values.

2.00 M = number of moles / 0.45 L

Number of moles = 2 M  × 0.45 L

Number of moles = 0.9 mol  ( M = mol/L)

Mass of CaBr₂:

Mass of CaBr₂ = number of moles × molar mass

Mass of CaBr₂ = 0.9 mol  × 199.89 g/mol

Mass of CaBr₂ = 179.9 g

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atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer