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Murljashka [212]
2 years ago
11

Which Anion is this?: CO2 – 3 (a) Nitrate (b) Carbonate (c) Nitrite (d) Sulphate

Chemistry
1 answer:
aalyn [17]2 years ago
4 0

Answer:

Carbonate

Explanation:

Carbonate is a combination of a carbon atom and two oxygen atoms for form a compound.

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What may explode in a combustible atmosphere?
shusha [124]

Explanation:

<h2><em><u>Dust explosions can occur where any dispersed powdered combustible material is present in high-enough concentrations in the atmosphere or other oxidizing gaseous medium, such as pure oxygen</u></em></h2>
7 0
3 years ago
How cholestrol transport fatty acids in body ?​
sleet_krkn [62]

In order to use the energy stored in fat, the body breaks dow triglycerides into fatty acids, which individual cells burn for energy

4 0
3 years ago
What is the osmotic pressure of a solution of FeCl3(aq) in which 69.5 g of FeCl3(s) (molar mass = 162.20 g mol-1) are dissolved
ivolga24 [154]

The osmotic pressure of the solution is 25.48atm

Data;

  • Mass = 69.5g
  • Molar mass = 162.20g/mol
  • Volume = 2.35L
  • Temperature = 35.0^oC
<h3>Osmotic Pressure</h3>

The osmotic pressure of the solution is calculated as

\pi = iMRT

  • i = Vant Hoff Factor
  • M = molarity
  • R = gas constant
  • T = Temperature

The Vant hoff factor of the factor = 4

Let's calculate the molarity of the solution;

M = \frac{number of moles}{volume}

Number of moles = mass / molar mass

n = \frac{mass}{molar mass} \\

substitute the values into the formula and solve for it

n = \frac{69.5}{162.20} \\n = 0.428 moles

The molarity of the solution is calculated as

M = \frac{number of moles}{volume} \\M = \frac{0.428}{2.35} = 0.182 mol/L

The osmotic pressure of the solution is calculated as

\pi = iMRT\\\pi = 4 * 0.182 * 35\\\pi = 25.48 atm

The osmotic pressure of the solution is 25.48atm

Learn more on osmotic pressure here;

brainly.com/question/8195553

5 0
2 years ago
CaSO4 +<br> Al(NO3)3<br> +<br> Al; (507)<br> Ca(NO3)2
Zigmanuir [339]

Answer:

what are you looking for?

Explanation:

4 0
3 years ago
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
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