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lubasha [3.4K]
3 years ago
7

An atom with atomic number of 6 would have how many protons?

Chemistry
2 answers:
aalyn [17]3 years ago
4 0

Answer:

6

Explanation:

Any atom with the atomic number 6 is carbon and has 6 protons

Oksanka [162]3 years ago
3 0

Answer: The answer is 6 protons

Explanation: The reason being is that for every atom there should be a proton

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Phosphorus pentachloride reacts with water to form hydrochloric acid and phosphoric acid. How many total moles of acid are forme
klio [65]

Answer:

0,13 moles of acid are produced

Explanation:

The reaction of the problem is:

PCl₅ + 4H₂O → 5HCl + H₃PO₄

Based on the reaction, 1 mole of PCl₅ produces 6 moles of acid (5 moles of HCl + 1 mole of H₃PO₄).

The molecular mass of PCl₅ is:

1P = 30,97g/mol + 5Cl = 5×35,45g/mol = <em>208,24 g/mol</em>

That means 4,5g of PCl₅ are:

4,5g PCl₅×(1mol / 208,24g) = 0,0216 moles of PCl₅. As 1 mole of PCl₅ produces 6 moles of acid, 0,0216 moles of PCl₅ produce:

0,0216 moles PCl₅ × (6 moles acid / mole of PCl₅) =

<em>0,13 moles of acid are produced</em>

I hope it helps!

3 0
3 years ago
Cacti and succulents are better adapted
masha68 [24]

Answer:

answer is C

Explanation:

encourage the release of carbon dioxide from the stems

8 0
3 years ago
Now liters at the same temperature and pressure: liters of H2 + liters of O2 → liters of H2O
iragen [17]
The answer is 2H2 + O2----> 2H2O
5 0
3 years ago
For the following electron-transfer reaction:
creativ13 [48]

Answer:

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

Explanation:

Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)

In the oxidation half reaction, the oxidation number increases:

Mn changes from 0, in the ground state to Mn²⁺.

The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.

Silver changes from Ag⁺ to Ag.

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

To balance the hole reaction, we need to multiply by 2, the second half reaction:

Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2

2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)  

Now we sum, and we can cancel the electrons:

2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻

4 0
3 years ago
Match the scientists to their contributions to the development of the periodic table.
zloy xaker [14]

Answer:

1. Dmitri Mendeleev

2. Johann Dobhereiner

3. John Newlands

4. Henry Moseley

If you find this helpful please give me Brainliest award

4 0
3 years ago
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