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2 years ago

Why would it have been wrong to write Li2SO4 as lithium sulfur oxide?

Chemistry

1 answer:

Bess [88]2 years ago

5 0

Due to pyro-electric properties and molarity Li2SO4 cannot be written as lithium sulfur oxide.

Explanation:

Lithium sulfate is a white inorganic salt with the formula Li2SO4. It is the lithium salt of sulfuric acid.

Lithium sulfate has water solubility, though it does not follow the usual trend of solubility versus temperature — its solubility in water decreases with increasing temperature, as its dissolution is an exothermic process. This property is shared with few inorganic compounds, such as the lanthanoid sulfates.

Lithium sulfate has pyro-electric properties. When aqueous lithium sulfate is heated, the electrical conductivity also increases. The molarity of lithium sulfate also plays a role in the electrical conductivity optimal conductivity is achieved at 2M and then decreases.

Lithium sulfate has a rapid gastrointestinal absorption rate and complete following oral administration of tablets or the liquid form.

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Write an equation that shows the formation of a strontium ion from a neutral strontium atom

Pepsi [2]

Answer:

See explanation

Explanation:

Sr(s) + 2HCl(aq) -----> SrCl2(aq) + H2(g)

Ionically;

Sr(s) + 2Cl^-(aq) ----> SrCl2(aq)

If we look at the reaction above, strontium atom was dissolved in hydrochloric acid. The strontium atom is now oxidized by the acid to give Sr^2+ ion according to the equation shown above.

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2 years ago

3 ways of improving the quality of an indigenous soap

Monica [59]

Answer : Below are three ways described in brief for improving the quality of indigenous soap.

1) Add virgin coconut oil which serves as a softener and betel leaf extract. It is used to improve the quality of soap as an antimicrobial agent.

2) Add activated charcoal black granules or powder to the soap, it removes maximum dirt from the skin.

3) Adding a filler which will increase the shelf life of usage of the soap will also improve its quality.

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2 years ago

Determinación de pH Expresa las siguientes concentraciones de [H+ ] en función del pH • [H+] = 0.001 M • [H+] = 0.002 M • [H+] =

Pavel [41]

Answer:

• pH = 3.0

• pH = 2.70

• pH = 3.61

• pH = 8.28

• pH = 1.40

Explanation:

El pH es una medida en química usada para determinar el grado de acidez o basicidad en una solución.

Se define como:

pH = -log₁₀ [H⁺]

<em>El - logaritmo de la concentración molar de H⁺</em>

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Para las concentraciones de H⁺ dadas:

• [H+] = 0.001 M

pH = -log (0.001M) = 3

pH = 3.0

• [H+] = 0.002 M

pH = -log (0.002M)

pH = 2.70

• [H+] = 2.45X10-4 M

pH = -log (2.45X10-4 M )

pH = 3.61

• [H+] = 5.2X10-9 M

pH = -log (5.2X10-9 M)

pH = 8.28

• [H+] = 0.04 M

pH = -log (0.04M)

pH = 1.40

8 0

2 years ago

You are writing a safety contract for your class. List 10 things you would include in the contract.

andre [41]

1.always listen to teacher

2 no eating in class

3 always have attentive listening

4 always have proper safety material

5 wear eyeglasses when needed

6 let others be able to listen

sorry I could only think of 6

7 0

2 years ago

Read 2 more answers

1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl

kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0

2 years ago