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Artist 52 [7]
2 years ago
8

What mass of barium sulfate (233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of

a 0.150 M solution of iron(III) sulfate
Chemistry
1 answer:
Rzqust [24]2 years ago
4 0

Answer:

4.37 g of barium sulphate

Explanation:

The reaction equation is;

3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)

From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles

To find the limiting reactant;

3 moles of barium chloride yields 3 moles of barium sulphate

0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate

1 mole of iron III sulphate yields 3 moles of barium sulphate

0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate

Hence,barium chloride is the limiting reactant

Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate

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A lone oxygen atom has 6 electrons in its outer shell which is not very stable, whereas as full octet (8 outer shell electrons) is stable. In order to achieve this two oxygen atoms will share 4 electrons, each contributing 2 electrons. Since these electrons exist within the orbitals of both atoms, to oxygen atoms essentially achieve a full octet.

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Glucose is the monomer in the natural polymers ____________________ and cellulose.
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Starch

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2 years ago
What process occurs as pieces of dry ice get smaller
fenix001 [56]
Sublimation. It's basically, in simple terms, when a solid changes to a gas without going into liquid form. 
8 0
3 years ago
Given the reaction below, if 0.00345 g of carbon dioxide is used up, how many grams of water will be produced?
zimovet [89]

From the  stoichiometry of the reaction, 1.4 * 10^-3 g is produced.

<h3>What mass of water is produced?</h3>

The equation of the reaction is written as; CO2 + 2LiOH → Li2CO3 + H2O. This can help us to apply the principle of stoichiometry here.

Thus;

Number of moles of CO2 = 0.00345 g/44 g/mol = 7.8 * 10^-5 moles

If 1 mole of CO2 produced 1 mole of water

7.8 * 10^-5 moles of CO2 produced  7.8 * 10^-5 moles of water

Mass of water produced =  7.8 * 10^-5 moles * 18 g/mol = 1.4 * 10^-3 g

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3 0
2 years ago
A sample of krypton has a volume of 6.00 L, and the pressure is 0.960 atm. If the final temperature is 55.0°C, the final volume
MissTica

193.38 K was the initial temperature of the krypton.

Explanation:

Data given:

Initial volume of the krypton gas = 6 litres

initial pressure of the krypton gas = 0.960 atm

initial temperature of the krypton gas = ?

final volume of the krypton gas  = 7.70 litres

final pressure of the Krypton gas  = 1.25 atm

final temperature of the krypton gas = 55 degrees or 273.25+55 = 323.15 K

Applying the  Combined Gas Laws:

\frac{P1V1}{T1} = \frac{P2V2}{T2}

Rearranging the equation:

T1  = \frac{P1V1T2}{P2V2}

Putting the value in the equation:

T1 = \frac{0.960 X 6 X 323.15}{1.25 X 7.70}

T1 = 193.38 K

Initial temperature of the krypton gas is 193.78 K

5 0
3 years ago
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