What mass of calcium carbonate is produced when 250 mL of 6.0 M sodium carbonate is added to 750 mL of 1.0 M calcium fluoride

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VMariaS [2998]

Response: The volume of the cube that is underwater is 29.8 mL

Clarification:

Initially, we need to find the mass of the ice.

Equation applied:

$\text{Mass of ice}=\text{Density of ice}\times \text{Volume of ice}$

Provided information:

Ice density = $0.9000g/cm^3=0.9000g/mL$

Volume of ice = 45.0 mL

$\text{Mass of ice}=0.9000g/mL\times 45.0mL$

$\text{Mass of ice}=40.5g$

The cube displaces 40.5 g of liquid to float.

Next, we will ascertain how much of the cube is submerged in the liquid.

$\text{Volume of ice}=\frac{\text{Mass of liquid}}{\text{Density of liquid}}$

$\text{Volume of ice}=\frac{40.5g}{1.36g/mL}$

$\text{Volume of ice}=29.8mL$

Therefore, the submerged portion of the cube is 29.8 mL

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2 months ago