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3 years ago

What is the formula of the ionic compound expected to form between the elements chlorine and aluminum?

Chemistry

1 answer:

galben [10]3 years ago

3 0

Answer: AlCl3

Explanation:

Aluminum is in group 13 and has a valency of 3. Chlorine is in group 17 and has a valency of 1. When a compound is formed, it reflects the valencies of each of the combining elements. This can be simplified by saying that the combining elements exchange their valencies. Cl gets 3 and All gets 1 forming a compound with the formula shown above.

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Name the carnivores. Which one is not a top carnivore?

Strike441 [17]

The carnivores are the hawk, owl, and fox.

5 0

3 years ago

Read 2 more answers

Acid & Base Calculations Calculate the hydronium ion concentration for each. Tell whether it is an acid or a base. 1. pH = 5

Anuta_ua [19.1K]

Explanation:

pH is the negative logarithm of hydronium ion concentration present in a solution.

If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9
If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
The solution having pH equal to 7 is termed as neutral solution.

To calculate the pH of the solution, we use equation:

$pH=-\log[H_3O^+]$ ......(1)

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14 ........(2)

For 1:

We are given:

pH = 5.54

Putting values in equation 1, we get:

$5.54=-\log[H_3O^+]$

$[H_3O^+]=2.88\times 10^{-6}M$

Now, putting values in equation 2, we get:

14 = 5.54 + pOH

pOH = 8.46

The solution is acidic in nature.

For 2:

We are given:

pOH = 9.7

Putting values in equation 2, we get:

14 = 9.7 + pH

pH = 4.3

Now, putting values in equation 1, we get:

$4.3=-\log[H_3O^+]$

$[H_3O^+]=5.012\times 10^{-5}M$

The solution is acidic in nature.

For 3:

We are given:

pH = 7.0

Putting values in equation 1, we get:

$7.0=-\log[H_3O^+]$

$[H_3O^+]=1.00\times 10^{-7}M$

Now, putting values in equation 2, we get:

14 = 7.0 + pOH

pOH = 7.0

The solution is neither acidic nor basic in nature.

For 4:

We are given:

pH = 12.9

Putting values in equation 1, we get:

$12.9=-\log[H_3O^+]$

$[H_3O^+]=1.26\times 10^{-13}M$

Now, putting values in equation 2, we get:

14 = 12.9 + pOH

pOH = 1.1

The solution is basic in nature.

For 5:

We are given:

pOH = 1.2

Putting values in equation 2, we get:

14 = 1.2 + pH

pH = 12.8

Now, putting values in equation 1, we get:

$12.8=-\log[H_3O^+]$

$[H_3O^+]=1.58\times 10^{-13}M$

The solution is basic in nature.

For 6:

We are given:

$[H_3O^+]=1\times 10^{-5}M$

Putting values in equation 1, we get:

$pH=-\log(1\times 10^{-5})$

$pH=5$

Now, putting values in equation 2, we get:

14 = 5 + pOH

pOH = 9

The solution is acidic in nature.

5 0

3 years ago

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r

weqwewe [10]

Answer:

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by $\bigtriangledown H^\circ_{rxn}$

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ =?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1) $\bigtriangledown H^\circ_{rxn}$ = +157KJ

P₄(g)+6Cl₂(g)→ 4PCl₃(g).............(2) $\bigtriangledown H^\circ_{rxn}$ = -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3) $\bigtriangledown H^\circ_{rxn}$ = -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4) $\bigtriangledown H^\circ_{rxn}$ =4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ =( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ = - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

5 0

2 years ago

Không tính toán cụ thể mà lý luận để :

kotykmax [81]

Answer:

c) Tìm những chất có hàm lượng đồng bằng nhau trong 4 chất : CuO, Cu2O, CuS, Cu2S.

Explanation:

3 0

2 years ago

How many atoms of germanium are present in a sample containing 1.65 moles Ge?

otez555 [7]

Answer:

9.9363 x 10^23 atoms

Explanation:

1.65 x Avagadro's Number

1.65 * 6.022 x 10^23 =

5 0

2 years ago