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3 years ago

12

What is the term for the smallest bit into which a chemical substance can be divided and still have the properties of that subst

ance?.

1 answer:

ziro4ka [17]3 years ago

3 0

Answer:

Molecule

Explanation:

molecule of the substance. You can break the molecule down further, into the atoms that make it up, but those don't have the properties of the original 'compound'.

Here's an example:

-- Sodium is a soft, slippery metal, that explodes when water touches it.

-- Chlorine is a poisonous green gas.

When an atom of Sodium and an atom of Chlorine combine, they make one molecule of a substance called "Sodium Chloride". That's SALT ! It isn't green, it isn't a gas, it isn't poisonous, it isn't soft and slippery, and it doesn't explode when water touches it.

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Is dew forming on a leaf a physical or chemical change?

solong [7]

No, the formation of dew is condensation, which is a physical change.

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3 years ago

What is true of a scientific law?

Minchanka [31]

Answer: d

Explanation:

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3 years ago

A sample of gallium Bromide GaBr2,weighing 0.165 g was dissolved in water and treated with silver nitrate AgNO3, and resulting t

tresset_1 [31]

<u>Answer:</u> The percent gallium in gallium bromide is 30.30 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of gallium bromide = 0.165 g

Molar mass of titanium gallium bromide = 229.53 g/mol

Putting values in equation 1, we get:

$\text{Moles of gallium bromide}=\frac{0.165g}{229.53g/mol}=0.00072mol$

The chemical equation for the reaction of gallium bromide and silver nitrate follows:

$GaBr_2+2AgNO_3\rightarrow 2AgBr(s)+Ga(NO_3)_2$

By Stoichiometry of the reaction:

1 moles of gallium bromide produces 1 mole of gallium nitrate

So, 0.00072 moles of gallium bromide will produce = $\frac{1}{1}\times 0.00072=0.00072moles$ of gallium nitrate

Now, calculating the mass of gallium nitrate from equation 1, we get:

Molar mass of gallium nitrate = 193.73 g/mol

Moles of gallium nitrate = 0.00072 moles

Putting values in equation 1, we get:

$0.00072mol=\frac{\text{Mass of gallium nitrate}}{193.73g/mol}\\\\\text{Mass of gallium nitrate}=0.139g$

Calculating the mass of gallium in the reaction, we use unitary method:

In 1 mole of gallium nitrate, 1 mole of gallium atom is present.

In 193.73 grams of gallium nitrate, 69.72 g of gallium atom is present.

So, in 0.139 grams of gallium nitrate, the mass of gallium present will be = $\frac{69.72}{193.73}\times 0.139=g$

To calculate the percentage composition of gallium in gallium bromide, we use the equation:

$\%\text{ composition of gallium}=\frac{\text{Mass of gallium}}{\text{Mass of gallium bromide}}\times 100$

Mass of gallium bromide = 0.165 g

Mass of gallium = 0.050 g

Putting values in above equation, we get:

$\%\text{ composition of gallium}=\frac{0.050g}{0.165g}\times 100=30.30\%$

Hence, the percent gallium in gallium bromide is 30.30 %.

3 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.

<u>Answer:</u> The empirical formula for the given compound is $CaCN_2$

<u>Explanation:</u>

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

<u>For calculating the mass of carbon:</u>

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.

<u>For calculating the mass of nitrogen:</u>

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.

<u>For calculating the mass of calcium:</u>

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:

<u>Step 1:</u> 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 = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

<u>Step 2:</u> 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 = $\frac{0.0852}{0.0426}=2$

<u>Step 3:</u> 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

Which would be a good source to find weather information about your town from last month?

VLD [36.1K]

Answer:

a reference book about weather

Explanation:

6 0

3 years ago