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Snowcat [4.5K]
3 years ago
15

If a substance is covalent, then it likely will

Chemistry
2 answers:
svet-max [94.6K]3 years ago
4 0

Answer : If a substance is covalent, then it is likely to melt easily and fast, it may be composed of two or more non-metals which share a covalent bonding between them and it will not conduct electricity.

It is seen that covalent substances are made up of two or more non-metals which share a covalent bond between them. They are found to melt easily and when dissolved in water they are seen to be insoluble. They are also not good conductors of electricity as there is no free electron available to conduct electricity.

12345 [234]3 years ago
4 0

Answer: water

Explanation:

Salt is ionic, and oil is covalent. If you add water, the salt will dissolve in the water, and the oil will float to the top. You could then skim the oil and start over with sugar. To double-check, you could even make sure the oil does not conduct electricity!

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It provides evidence that each organism is related and similar in each way, leading to the conclusion of a common ancestor.<span />
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3 years ago
A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958
Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

<u>Step 1</u>: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

<u>Step 2</u>: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

<u>Step 3:</u> Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

<u>Step 4:</u> Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

<u>Step 5:</u> Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

<u>Step 6:</u> Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

<u>Step 7: </u>Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol  

<u>Step 8:</u> Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

<u>Step 9</u>: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

<u>Step 10</u>: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

8 0
3 years ago
What color do acids turn litmus paper?
Ivanshal [37]

Answer:

red

Explanation:

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Biphenyl and benzene are common side products in this gringard reaction. explain at which points in the purification procedure t
mafiozo [28]

In Grignard reaction, Biphenyl and benzene are common side products which are removed during trituration.

In organic chemistry, a reaction in which the Grignard reagents or organometallic substances are added to organic compounds such as aldehydes and ketones to form alcohol is known as Grignard reaction.

These Grignard reagents are magnesium halides of alkyl, vinyl or allyl, which react with a carbonyl group to form alcohols.

During this reaction, primary, secondary and tertiary alcohols are formed.

While Biphenyl and benzene are common side products.

These are removed during trituration process in which cold petroleum ether is added to dissolve the biphenyl and benzene side products

If you need to learn more about Biphenyl and benzene click here:

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2 years ago
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