How many grams of Cabr2 would be needed to create 450 mL of a 2.00 M solution?

Mass = 179.9 g

Explanation:

Given data:

Volume of solution = 450 mL

Molarity of solution = 2.00 M

Mass in gram required = ?

Solution:

Volume of solution = 450 mL× 1 L / 1000 mL = 0.45 L

Molarity = number of moles of solute/ Volume of solution in L

2.00 M = number of moles of solute / 0.45 L

Number of moles of solute = 2.00 M × 0.45 L

M = mol/L

number of moles of solute = 0.9 mol

Mass of CaBr₂ in gram:

Mass = number of moles ×  molar  mass

Mass = 0.9 mol ×199.89 g/mol

Mass = 179.9 g

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be