All categories

4 years ago

How many valence electrons does aluminum (Al) have available for bonding? 1 2 3 4

2 answers:

6 0

The electronic configuration of Al is 2,8,3. So therefore Aluminum has 3 valance electrons available.

We know this because of the number of electrons in the outermost shell.

I hope this helps

zhuklara [117]4 years ago

5 0

Answer:

Explanation:

You might be interested in

Mrs. Mason does not want the ringing of the telephone to water up her baby. Which of the following would produce the greatest re

PilotLPTM [1.2K]

What are the options I could help

7 0

3 years ago

How many kilojoules (kJ) in 373 joules (J)?

Nonamiya [84]

Answer: There are 0.373 kJ in 373 joules.

Explanation:

According to the standard conversion units 1 kJ is equal to 1000 J. This means that 1 J is equal to 0.001 kJ.

Hence, 373 joules is converted into kilojoules as follows.

$1 J = 0.001 kJ\\373 J = 373 J \times \frac{0.001 J}{1 kJ}\\= 0.373 kJ$

Thus, we can conclude that there are 0.373 kJ in 373 joules.

4 0

3 years ago

Mg (s) + 2 HCl (aq) --> MgCl2 (aq) + H2 (g)

Lera25 [3.4K]

Answer:

m = 190. g MgCl2

Explanation:

Use a molar ratio to get the # of moles MgCl2 produced from 4.00 mol of HCl:

4.00 mol HCl × (1 mol MgCl2/2 mol HCl) = 2.00 mil MgCl2

So 2.00 moles of MgCl2 will be produced. To find the mass in grams, use the molar mass of MgCl2:

2.00 mol MgCl2 × (95.211 g MgCl2/1 mol MgCl2)

= 190. g MgCl2

8 0

3 years ago

Chanice drives her scooter 7 kilometers north she stops for lunch and then drives 5 kilometers east what distance did she cover?

Serhud [2]

Her displacement is 2 i think

6 0

3 years ago

Read 2 more answers

Calculate ΔHo for the following reaction ussing the given bond dissociation energiesCH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)BOND

Mazyrski [523]

Answer:

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

Explanation:

$CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,$ ΔH° = ?

We are given with:

$\Delta H_{O-O}=142 kJ/mol$

$\Delta H_{O=O}=498 kJ/mol$

$\Delta H_{H-O}=459 kJ/mol$

$\Delta H_{C-H}=411 kJ/mol$

$\Delta H_{C-O}=358 kJ/mol$

$\Delta H_{C=O}=799 kJ/mol$

ΔH° =

(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

$\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})$

$\Delta H^o=(1 mol\times 4\times 411 kJ/mol+2 mol\times 1\times 498 kJ/mol)-(1 mol\times 2\times 799 kJ/mol+2 mol\times 2\times 459 kJ/mol)$

$\Delta H^o=-794kJ$

$\Delta H^o>0$ endothermic reaction

$\Delta H^o$ exothermic reaction

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

4 0

4 years ago