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1 year ago

How much volume (in cm^3)is gained by a person who gains 11.8 lb of our fat? Human fat has a density of 0.918 g/cm^3

Chemistry

1 answer:

Zinaida [17]1 year ago

3 0

The volume (in cm³) gained by a person who gains 11.8 lb of fat is 5830.49 cm³

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

<h3>How to convert pounds to grams </h3>

1 lb = 453.592 g

Therefore,

11.8 lb = 11.8 × 453.592

11.8 lb = 5352.3856 g

<h3>How to determine the volume </h3>

Mass = 5352.3856 g
Density = 0.918 g/cm³
Volume =?

Volume = mass / density

Volume = 5352.3856 / 0.918

Volume = 5830.49 cm³

Learn more about density:

brainly.com/question/952755

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Suppose 2.4 g of Mg reacts with 10.0 g of O2, to create magnesium oxide. How much magnesium oxide is produced?

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Answer:

3.98 g

Explanation:

Step 1. Write the balanced chemical reaction. In this case, magnesium reacts with oxygen to produce magnesium oxide:

$2~Mg(s) + O_2 (g)\rightarrow 2~MgO (s)$

Step 2. Calculate the number of moles of magnesium:

$n_{Mg} = \frac{2.4~g}{24.305~g/mol} = 0.0987~mol$

Step 3. Calculate the number of moles of oxygen:

$n_{O_2} = \frac{10.0~g}{32.00~g/mol} = 0.3125~mol$

Step 4. Identify the limiting reactant comparing the equivalents. Equivalent of Mg:

$eq_{Mg} = \frac{0.0987~mol}{1} = 0.0987~mol$

Equivalent of oxygen:

$eq_{O_2} = \frac{0.3125~mol}{2} = 0.15625~mol$

Therefore, Mg is the limiting reactant.

Step 5. According to the stoichiometry of this reaction:

$n_{Mg} = n_{MgO} = 0.0987~mol$

Step 6. Convert the number of moles of MgO into mass:

$m_{MgO} = 0.0987~mol\cdot 40.304~g/mol = 3.98~g$

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