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

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.