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1 year ago

A solid object has a mass of 1.62 lb and a volume of 190 mL.

Chemistry

1 answer:

Pachacha [2.7K]1 year ago

8 0

The density of a solid object that has a mass of 1.62 lb and a volume of 190 mL is 3.87g/mL.

<h3>How to calculate density?</h3>

Density is the measure of the mass of matter contained by a unit volume. It can be calculated by dividing the mass of the substance by its volume.

According to this question, a solid object is said to have a mass of 1.62 lb and a volume of 190 mL.

1 pound = 453.592 grams

1.62 Ibs is equivalent to 734.82 grams

Density of the solid object = 734.82 grams ÷ 190 mL

Density = 3.87g/mL

Therefore, the density of a solid object that has a mass of 1.62 lb and a volume of 190 mL is 3.87g/mL.

Learn more density at: brainly.com/question/20337365

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A medication 60 mg is ordered. the vial is labeled 1 gm/ 10 ml. How many ml would you give?

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Determine the possible traits of the calves if:

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Answer:

1. All red calves i.e. RR

2. All roan calves i.e RW

3. 2 red calves (RR) and two roan calves (RW)

Explanation:

According to this question, a gene coding for fur colour in cattle is involved. Red alleles (R) and white alleles (W) are co-dominant to produce a roan cattle (RW). The possible traits of the following crosses are (see attached punnet square):

1) A red bull (RR) is mated to a red (RR) cow: All red calves i.e. RR

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3 years ago

A sample of gas X occupies 10 mº at a pressure of 120 kPa.

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Answer:

The new pressure of the gas comes out to be 400 KPa.

Explanation:

Initial volume of gas = V = $10\textrm{ m}^{3}$

Initial pressure of gas = P = 120 KPa

Final volume of gas = V' = $3\textrm{ m}^{3}$

Assuming temperature to be kept constant.

Assuming final pressure of the gas to be P' KPa

$PV = P'V' \\120\textrm{ KPa}\times 10\textrm{ m}^{3} = \textrm{P'}\times 3\textrm{ m}^{3} \\\textrm{P'} = 400\textrm{ KPa}$

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2 years ago

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n

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The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

Answer: The empirical formula for the given compound is $CaCN_2$

Explanation:

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

For calculating the mass of nitrogen:

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

For calculating the mass of calcium:

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

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Explanation:

I hope this helps

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3 years ago