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2 years ago

What is the atomic radius

Chemistry

1 answer:

dimaraw [331]2 years ago

5 0

Answer:

The atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.

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How does a chemist count the number of particles in a given number of moles of a substance?

jolli1 [7]

The chemist the count the number of particles (Atoms, Molecules or Formula Unit) in a given number of moles of a substance by using following relationship.

Moles = # of Particles / 6.022 × 10²³

Or,

# of Particles = Moles × 6.022 × 10²³

So, from above relation it is found that 1 mole of any substance contains exactly 6.022 × 10²³ particles. Greater the number of moles greater will be the number of particles.

8 0

3 years ago

Pls help asap!! plssss

kvv77 [185]

AgNO₃+NaCl⇒AgCl+NaNO₃

<h3>Further explanation</h3>

Double-Replacement reactions. Happens if there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product

Reaction

AB + CD⇒AD + CB

So for the option :

1. synthesis/combination reaction

2. decomposition reaction

3. double replacement reaction

4. single replacement reaction

4 0

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

A geological process (2 points)

densk [106]

They are all correct good job bud I hope it helps and good luck

4 0

3 years ago

Read 2 more answers

QUESTION 3

ddd [48]

Answer:

Different visible waves show different colors

5 0

3 years ago