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garik1379 [7]
3 years ago
6

Balance this equation: Ca(OH)2 + HCI → CaCl2 + H20

Chemistry
1 answer:
Verizon [17]3 years ago
5 0

Answer:

Double Displacement (Acid-Base)

Explanation:

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Extrapolation allows you to estimate values greater than those in your actual data.
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Extrapolation allows you to estimate values greater than those in your actual data. True
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In nuclear reactions, which of the following is likely to occur?
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It is both a and b
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4 0
3 years ago
Ethanol has a Kb of 1.22 °C/m and usually boils at 78.4 °C. How many mol of an nonionizing solute would need to be added to 48.8
Galina-37 [17]

Answer:

0.272 mol

Explanation:

∆Tb = m × Kb

∆Tb = 85.2°C - 78.4°C = 6.8°C

Kb = 1.22°C/m

mass of ethanol = 48.80 g = 48.80/1000 = 0.0488 kg

Let the moles of non-ionizing solute be y

m (molality) = y/0.0488

6.8 = y/0.0488 × 1.22

y = 6.8×0.0488/1.22 = 0.272 mol

6 0
2 years ago
What is the maximum mass of S8 that can be produced by combining 88.0 g of each reactant? 8SO2+16H2S⟶3S8+16H2O
Leni [432]

Answer:

124.24 g of S₈.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

8SO₂ + 16H₂S —> 3S₈ + 16H₂O

Next, we shall determine the masses of SO₂ and H₂S that reacted and the mass of S₈ produced from the balanced equation.

This is illustrated below:

Molar mass of SO₂ = 32 + (16×2)

= 32 + 32 = 64 g/mol

Mass of SO₂ from the balanced equation = 8 × 64 = 512 g

Molar mass of H₂S = (2×1) + 32

= 2 + 32 = 34 g/mol

Mass of H₂S from the balanced equation = 16 × 34 = 544 g

Molar mass of S₈ = 32 × 8 = 256 g/mol

Mass of S₈ from the balanced equation = 3 × 256 = 768 g

Thus,

From the balanced equation above,

512 g of SO₂ reacted with 544 g of H₂S to produce 768 g of S₈.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

512 g of SO₂ reacted with 544 g of H₂S.

Therefore, 88 g of SO₂ will react with = (88 × 544) / 512 = 93.5 g of H₂S.

From the calculation made above, we can see that it will take a higher mass (i.e 93.5 g) than what was given (i.e 88g) of H₂S to react completely with 88 g of SO₂.

Therefore, H₂S is the limiting reactant and SO₂ is the excess reactant.

Finally, we shall determine the maximum mass of S₈ produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of S₈ since all of it is consumed in the reaction.

The limiting reactant is H₂S and the maximum mass of S₈ produced can be obtained as follow:

From the balanced equation above,

544 g of H₂S reacted to produce 768 g of S₈.

Therefore, 88 g of H₂S will react to produce = (88 × 768) / 544 = 124.24 g of S₈.

Therefore, the maximum mass of S₈ produced from the reaction is 124.24 g.

7 0
2 years ago
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