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konstantin123 [22]
3 years ago
8

Which one of the following substances is the product of this combination reaction? Al (s) + I2 (s) → ________(A) AlI2(B) AlI(C)

AlI3(D) Al2I3(E) Al3I2
Chemistry
1 answer:
Slav-nsk [51]3 years ago
7 0

Answer:

AlI_3

Explanation:

Combination reactions are the reactions in which two or more atoms/ molecules combine to give a single product.

Aluminum metal combines with iodine to given aluminium iodide.

Aluminium has a valency of 3 and iodine has valency of 1, so the formula is AlI_3

The reaction is shown below as:

Al_{(s)}+I_2_{(s)}\rightarrow AlI_3_{(s)}

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H2c+o
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Answer:

19.4 g of alum, will be its theoretical yield

Explanation:

The reaction is:

2 Al + 2 KOH + 4 H₂SO₄ + 22H₂O → 3H₂ + 2KAl(SO₄)₂•12H₂O

Let's determine the amount of acid.

M are the moles contained in 1 L of solution or it can be mmoles that are contained in 1 mL of solution

M = mmol /mL

M . mL = mmol

We replace: 8.3 mL . 9.9 M = 82.17 mmoles

We convert to moles: 82.17 mmol . 1 mol / 1000mmol = 0.082 moles

Ratio is 4:2

4 moles of sulfuric acid can make 2 moles of alum

By the way, 0.082 moles of acid may produce ( 0.082 . 2) /4 = 0.041085 moles.

We convert moles to mass:

Molar mass of alum is: 473.52 g/mol.

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6 0
2 years ago
For acidic solutions, which element is added to balance half-reactions
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For balancing acidic solutions, we would need to add H+ ions to the correct side of the equation to balance the total number of atoms and the overall charge.
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When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H
Taya2010 [7]

Answer:

- Empirical:

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- Molecular:

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Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC

Moreover, the moles of hydrogen:

n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

C_3H_7

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

C_6H_{14}

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Answer:

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