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dybincka [34]
3 years ago
10

What is the complete ionic equation for H2SO4(aq) + Cal2(aq) -> CaSO4(s) + 2HI(aq)?

Chemistry
2 answers:
eimsori [14]3 years ago
6 0
2H(+) + SO4(2-) + Ca(2+) + 2I(-) -> CaSO4(s) + 2H(+) + 2I(-)
The signs in brackets are the subscripts for the charge of the ion. This is the complete ionic equation. The net ionic equation is:
Ca(2+) + SO4(2-) -> CaSO4
Leokris [45]3 years ago
4 0

Answer:

SO₄²⁻(aq) + Ca²⁺(aq) → CaSO₄(s)

Explanation:

Given equation:

H₂SO₄(aq) + Cal₂(aq) → CaSO₄(s) + 2HI(aq)

Ionic equation:

2H⁺(aq) + SO₄²⁻(aq) + Ca²⁺(aq) + 2I⁻(aq) → CaSO₄(s) + 2H⁺ (aq) + 2I⁻(aq)

Net ionic equation:

SO₄²⁻(aq) + Ca²⁺(aq) → CaSO₄(s)

2H⁺ (aq) and 2I⁻(aq) ions are common on both side of the equation hence they were eliminated. (aq) here means that the ions are in aqueous phase and (s) means solid or precipitate. CaSO₄ forms a white precipitate which does not ionize in the aqueous phase.

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3 years ago
A 0.67 gram sample of chromium is reacted with sulfur. The resulting chromium sulfide has a mass of 1.2888 grams. What is the em
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Answer:

Cr₂S₃

Explanation:

From the question given above, the following data were obtained:

Mass of chromium (Cr) = 0.67 g

Mass of chromium sulfide = 1.2888 g

Empirical formula =?

Next, we shall determine the mass of sulphur (S) in the compound. This can be obtained as follow:

Mass of chromium (Cr) = 0.67 g

Mass of chromium sulfide = 1.2888 g

Mass of sulphur (S) =?

Mass of S = (Mass of chromium sulfide) – (Mass of Cr)

Mass of S = 1.2888 – 0.67

Mass of S = 0.6188 g

Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:

Mass of Cr = 0.67 g

Mass of S = 0.6188 g

Divide by their molar mass

Cr = 0.67 / 52 = 0.013

S = 0.6188 / 32 = 0.019

Divide by the smallest

Cr = 0.013 / 0.013 = 1

S = 0.019 / 0.013 = 1.46

Multiply by 2 to express in whole number

Cr = 1 × 2 = 2

S = 1.46 × 2 = 3

Therefore, the empirical formula of the compound is Cr₂S₃

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Explanation:

 How many moles of C are in 2.25 x 1022 atoms of C?

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