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3 years ago

What is Chronatography

Chemistry

2 answers:

lubasha [3.4K]3 years ago

5 0

Answer:

the right spelling is chromatography

Explanation:

nevsk [136]3 years ago

3 0

Answer:

plzz chat with me.....

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Which is always the first step in dealing with an accident in the lab?

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The answer is the letter C

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Why is amniocentesis a more widely used test than cvs?

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Higher probability of loss. Chorionic villus sampling (CVS) and Amniocentesis (AC). The prenatal diagnosis technique can be done earlier in fetal development CVS (first trimester --> 10-13 weeks). AC (second trimester --> 16-20 weeks)

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4 0

3 years ago

When a compound such as X2Z4 is created, what would be the correct simplified chemical formula?

Shtirlitz [24]

Answer:

XZ2

Explanation:

There are different ways in which compounds can be represented. Broadly, we have three different types of formula;

- Structural formular: This shows how th atoms in te compound are connected to each other.

- Molecular formular: This shows the actual number of atoms of element present in the compound

- Empirical Formular: This is the simplest formular of a compound. It basically shows the number of atoms in simple ration to each other.

This question requires us to input the empirical formular;

X2Z4

The ratio of the elements is; 2 : 4 which can be simplified into 1 : 2

This means the empirical formular is XZ2

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3 years ago

A graduated cylinder (approximate as a regular cylinder) has a radius of 1.045 cm and a height of 30.48 cm. What is the volume o

Stels [109]

Answer: 104.5 cm^3

Explanation:

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3 years ago

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

mylen [45]

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 $CaCN_2$

3 0

3 years ago