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GarryVolchara [31]

3 years ago

8

A student performing this experiment weights out 0.6680 grams of an antacid tablet. It is was determined that this tablet contai

ned 0.4210 grams of CaCO3. What is the percent CaCO3 of this tablet?

1 answer:

Rufina [12.5K]3 years ago

3 0

Answer:

63 wt%

Explanation:

0.4210/0.6680

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Gold has a density of 1,200 lb./ft. What is the density of gold in g/em? For conversion factors use I lb. 453.6 g, and l inch-2.

Jlenok [28]

<u>Answer:</u> The density of gold in $g/cm^3$ is $19.22g/cm^3$

<u>Explanation:</u>

Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

$\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}$

We are given:

Density of gold = $1200lb/ft^3$

Using conversion factors:

1 lb = 453.6 g

1 feet = 12 inches

1 inch = 2.54 cm

Converting given quantity into $g/cm^3$ , we get:

$\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3$

Hence, the density of gold in $g/cm^3$ is $19.22g/cm^3$

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What is mass????????????????????

lbvjy [14]

Answer:

Do you need the definition or the equation to find it?

Explanation:

Equation:

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