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Vera_Pavlovna [14]
3 years ago
13

Arun and Marcus are asked to compare the solubility of two salts in water at room temperature. The two salts are labelled X and

Y
1) name the independent variable in the investigation
2) name the dependent variable in the investigation
3) list the control variables in the investigation
Chemistry
1 answer:
oksian1 [2.3K]3 years ago
3 0

Answer:

1) Salts X and Y

2) The solubility of the salts

3) a) The solvent

b) The solvent temperature

Explanation:

1) The independent variable is the variable that is suspected to be the cause of the subject of the investigation

The given investigation is meant to investigate the solubility of different salts

Therefore, the solubility is expected to be dependent on the type of salt, and the independent variable is the type of salt, X or Y

2) The dependent variable is the effect meant to be observed in the investigation, which is the solubility of the salt in water at room temperature

3) The control variables are the variables which are held constant during the investigation, including;

a) The solvent used if the investigation; water

b) The temperature of the solvent; Room temperature

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The following reaction shows the products when sulfuric acid and aluminum hydroxide react.
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The correct answer is approximately 11.73 grams of sulfuric acid.

The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.

The balanced equation is:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g

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6 0
3 years ago
Ionic bonds are strong; therefore, ionic compounds
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Boron sulfide, B2S3(s), reacts violently with water to form dissolved boric acid (H3BO3) and hydrogen sulfide gas
Gwar [14]

The equation structure for the above mentioned reaction can be written as  

B_{2} S_{3}+6 H_{2} O \rightarrow 2 H_{3} B O_{3}+3 H_{2} S \uparrow

<u>Explanation:</u>

Considering the above reaction, When Boron sulfide, reacts with water more violently to form boric acid and hydrogen sulfide gas.

B_{2} S_{3}+H_{2} O \rightarrow H_{3} B O_{3}+H_{2} S \uparrow

In order to balance the equation, we can do as follows.There are 2 B - atoms on both sides of the equation, but only 2 H - atoms, and one O - atom on LHS, so we have to balance it by putting 6 in front of water and 2 in front of Boric acid and 3 in front of hydrogen sulphide gas, so that we have 2 B - atoms, 3 - S atoms, 12 H - atoms on both sides of the equation, and it is balanced. Balanced equation is given as,

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8 0
3 years ago
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Allushta [10]

Answer:

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