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3 years ago

What happens when Mg forms an ionic bond?

Chemistry

2 answers:

makkiz [27]3 years ago

7 0

Answer:

it gains one electron and become a -2 anion...

it is the right answer..

above answer is right explanation

lara [203]3 years ago

6 0

When this reaction occurs, strong ionic bonds form between two oppositely charged particles (ions). Magnesium is a group II metal, and therefore has two electrons in it's highest energy level (or outermost electron shell). ... Therefore, they readily take up these two electrons to form negatively charged O2- ions

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Help ASAP will give brainilest

fiasKO [112]

Q is the heat added to the system.

5 0

3 years ago

Read 2 more answers

Calculate the volume of carbon dioxide at 20.0°C and 0.941 atm produced from the complete combustion of 4.00 kg of methane. Comp

tankabanditka [31]

Answer:

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

Explanation:

Methane

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Mass of methane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of methane = $\frac{4000 g}{16 g/mol}=250 mol$

According to reaction, 1 mole of methane gives 1 mole of carbon dioxide gas,then 250 moles of methane will give :

$\frac{1}{1}\times 250 mol=250 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 250 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{250 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6390.89 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

Propane

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Mass of propane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of propane = $\frac{4000 g}{44 g/mol}=90.91 mol$

According to reaction, 1 mole of propane gives 3 mole of carbon dioxide gas,then 90.91 moles of propane will give :

$\frac{3}{1}\times 90.91 mol=272.73 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 272.73 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{272.73 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6,971.95 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6,971.95 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of methane = 6390.89 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of propane = 6,971.95 Liters.

6390.89 Liters < 6,971.95 Liters

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

7 0

3 years ago

A sodium bromide solution is added to a beaker containing aqueous chlorine. What would happen?

Darya [45]

Answer:

See detailed explanation.

Explanation:

Hello!

In this case, for the described chemical reaction, we can proceed as follows:

A) For the complete chemical reaction we note down every reacted and produced species as well as the proper balancing process:

$2NaBr(aq)+Cl_2(aq)\rightarrow 2NaCl(aq)+Br_2(g)$