20. You are approaching a downhill. Which configuration woulo

Balance the equation :

Sati [7]

The balanced equation is 2 AlI 3 ( a q ) + 3 Cl 2 ( g ) → 2 AlCl 3 ( a q ) + 3 I 2 ( g ) .

<u>Explanation:</u>

Aluminum has a typical oxidation condition of 3+ , and that of iodine is 1- . Along these lines, three iodides can bond with one aluminum. You get AlI3. For comparable reasons, aluminum chloride is AlCl3.

Chlorine and iodine both exist normally as diatomic components, so they are Cl2( g ) also, I2( g ), individually. In spite of the fact that I would anticipate that iodine should be a strong.

Balancing the equation, we get:

2AlI 3( aq ) + 3Cl2 ( g ) → 2AlCl3 ( aq ) + 3 I 2 ( g )

Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the AlCl 3 on the right.

Normally, presently we have two Al on the right, so I multiplied the AlI 3 on the left. Hence, I have 6 I on the left, and I needed to significantly increase I 2 on the right.

We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.

3 0

3 years ago

Calculate the number of Na ions and Cl ions and total number of ions in 14.5g of NaCl.

Elanso [62]

Answer:

Number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

Number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.

Explanation:

1 mole of any compound contains 6.023 × 10²³ molecules.

molecular weight of NaCl is 23 + 35.5 = 58.5 g.

so, 58.5 grams of NaCl makes 1 mole

⇒ 14.5 g of NaCl = $\frac{14.5}{58.5}$ = 0.248 moles.

⇒ 0.248 mole contains 0.248 × 6.023×10²³ molecules

= 1.49 × 10²³ molecules.

And 1 molecule contains 1 Na ion and 1 Cl ion.

⇒ number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.

⇒ number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.

Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.

6 0

3 years ago

Help please I’m confused

Debora [2.8K]

Answer:1 answer number 1

Explanation:

5 0

3 years ago

The table shows a chemical equation and the dimensional analysis used by a student to calculate the number of moles of Ba3N2 req

UNO [17]

This problem is providing the chemical reaction whereby barium nitride reacts with water to produce barium hydroxide and ammonia, so the number of moles of barium nitride are required in order to produce 8.3 moles of ammonia. It asks for us to evaluate the student's setup, so we conclude the answer is C. "1 mol of NH3 should be replaced with 2 mol of NH3", according to:

<h3>Mole ratios:</h3>

In chemistry, stoichiometric calculations are used to figure out the moles or mass of a substance, given information about another one in the reaction. In this case, for the given chemical equation:

$Ba_3N_2 + 6H_2O \rightarrow 3Ba(OH)_2 + 2NH_3$