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3 years ago

Which noble gas does not have 8 electrons in its outer shell?

Chemistry

1 answer:

UkoKoshka [18]3 years ago

7 0

The answer is helium because it only has 2 electrons in its outer shell.

I hope this helps:)

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What is the correct formula for Triphosphorous hexachloride?

zlopas [31]

Answer:

Explanation:

Im pretty sure hope this helps

5 0

2 years ago

Which of the following requires an oxidizing agent?

laila [671]

Correct Answer: First Option

An oxidizing agent causes the oxidation of the other atom/element by itself being reduced. In simple words we can state that oxidizing agent gains electrons from the other atom/element i.e. the other atom loses electrons.

From the given options we have to find in which of them electrons are being removed. When the electrons are removed, the number of protons in the atom will be more than the number of electrons. As a result the net charge on the atom will be positive.

First option lists such a change. Initially charge on Al is neutral, 3 electrons are removed and it get +3 charge. This shows that Al is being oxidized, so it needs an oxidizing agent.

3 0

3 years ago

When the Sun provides heat energy to the water cycle it creates which of the following?

GenaCL600 [577]

Answer: Evaporation

Explanation:

When the heat of the sun touches the water, the water molecules heat up and move gradually faster. Over time molecules start to pop out the surface of the water individually.

And then eventually all the water is evaporated into water vapor.

7 0

3 years ago

When you straighten your leg, the muscles on top of your leg muscles on the bottom of your leg. Your answer

AveGali [126]

Answer:

your quad is the top part and your hamstring is the bottom part

Explanation:

fun fact the hamstring is one of the strongest muscles in the body

3 0

3 years ago

Set up a Hess’s law cycle, and use the following information to calculate ΔH∘f for aqueous nitric acid, HNO3(aq). You will need

Assoli18 [71]

Answer:

$\Delta H_{f}^{\circ } (HNO_{3}, aq)$ = (-207.39) kJ/mol

Explanation:

Given:

$\Delta H_{f}^{\circ } (NH_{3}, g)= -46.1 kJ/mol\\ \Delta H_{f}^{\circ } (H_{2}O, l)= -285.8 kJ/mol\\ \Delta H_{f}^{\circ } (HNO_{3}, aq)= ? kJ/mol$

3NO₂(g)+H₂O(l)→ 2HNO₃(aq)+NO(g); ΔH° = -137.3 kJ ....equation 1

2NO(g)+O₂(g)→ 2NO₂(g) ; ΔH° = -116.2 kJ ....equation 2

4NH₃(g)+5O₂(g)→ 4NO(g)+6H₂O(l) ;ΔH° = -1165.2 kJ ....equation 3

Multiplying equation 1 with (2/3), we get

2NO₂(g) + 2/3 H₂O(l) → 4/3 HNO₃(aq) + (2/3) NO(g) ....equation 4

Adding equation 2 and 4, we get

2NO₂(g) + 2/3 H₂O(l) + 2NO(g) + O₂(g) → 4/3 HNO₃(aq) + (2/3) NO(g) + 2NO₂(g)

⇒ (4/3) NO(g) + O₂(g) + 2/3 H₂O(l) → 4/3 HNO₃(aq) ....equation 5

Multiplying equation 3 with (1/3), we get

(4/3) NH₃(g) + (5/3) O₂(g) → (4/3) NO(g)+ 2 H₂O(l) ....equation 6

Now adding equation 5 and 6, we get

(4/3) NH₃(g) + (5/3) O₂(g) + (4/3) NO(g) + O₂(g) + 2/3 H₂O(l) → (4/3) NO(g)+ 2 H₂O(l) + 4/3 HNO₃(aq)

⇒ (4/3) NH₃(g) + (8/3) O₂(g) → (4/3) HNO₃(aq) + (4/3) H₂O(l) ....equation 7

Now multiplying equation 7 with (3/4), we get

NH₃(g) + 2 O₂(g) → HNO₃(aq) + H₂O(l) ...equation 8

Therefore, by Hess's law the standard enthalpy of formation is:

ΔH° = (3/4) [ (-137.3 kJ) × (2/3) + (-116.2 kJ) + (-1165.2 kJ) × (1/3)]

ΔH° = (3/4) [ - 91.53 - 116.2 - 388.4]

ΔH° = (3/4) [-596.13] = -447.09 kJ

Since the change in enthalpy of a reaction:

ΔH° = $[\Delta H_{f}^{\circ } (HNO_{3}, aq) + \Delta H_{f}^{\circ } (H_{2}O, l)] - [\Delta H_{f}^{\circ } (NH_{3}, g) + 2 \Delta H_{f}^{\circ } (O_{2}, g)]$

(-447.09 kJ) = $[\Delta H_{f}^{\circ } (HNO_{3}, aq)$ + (-285.8 kJ)] - [-46.1 kJ + 2 (0 kJ)]

⇒ $\Delta H_{f}^{\circ } (HNO_{3}, aq)$ = (-447.09) + 285.8 kJ - 46.1 kJ

⇒ $\Delta H_{f}^{\circ } (HNO_{3}, aq)$ = (-207.39) kJ/mol

6 0

3 years ago