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Diano4ka-milaya [45]
3 years ago
11

When Na and Cl combine to form a compound, which atom(s) will become a cation?

Chemistry
2 answers:
vivado [14]3 years ago
8 0
Na only because cation is a positive ion whole cl is a negative ion in anion
KATRIN_1 [288]3 years ago
3 0

Answer: A. Na only

Explanation:

An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element and forms a positively charged ion called as cation. The element which accepts the electrons is known as electronegative element and forms a negatively charged ion called as anion.

Electronic configuration of sodium:

[Na]=1s^22s^22p^63s^1

Sodium atom will loose one electron to gain noble gas configuration and form sodium cation with +1 charge.

[Na^+]=1s^22s^22p^63s^0

Electronic configuration of chlorine:

[Cl]=1s^22s^22p^63s^23p^5

Chlorine atom will gain one electron to gain noble gas configuration and form chloride ion with -1 charge.

[Cl^-]=1s^22s^22p^63s^23p^6

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Protons/Electrons: 92

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When do plants use the process of cellular respiration
d1i1m1o1n [39]

Answer:

So the answer is every day

Explanation:

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3 years ago
H. Nitrogen gas and hydrogen gas combine to produce ammonia gas (NH3).
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Answer:

<u>27.3 L</u>

Explanation:

<u>Reaction</u>

  • N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
  • This is the basic reaction at STP

<u>Solving</u>

  • 10 g N₂ x 1 mol NH₃/28 g N₂ x 3 mol H₂/1 mole N₂
  • ⇒ 1.07 mol H₂

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4 0
2 years ago
Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actua
GrogVix [38]

The value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

The rust forms when 4.85X10³ kJ of heat is released is 888.916 g.

<h3>Chemical reaction:</h3>

4 Fe + 3O2 ------ 2Fe2O3

∆H = -1.65×10³kJ

A) Given,

mass of iron = 0.250kg = 250 g

<h3>Calculation of number of moles</h3>

moles = given mass/ molar mass

= 250/ 55.85 g/mol.

= 4.476 mol

As we know that,

For the rusting of 4 moles of Fe, ∆H = -1.65×10³kJ

For the rusting of 4.476 moles of Fe ∆H required can be calculated as

-1.65×10³kJ × 4.476 mol/ 4mol

∆H required = -1.846 × 10³kJ

Now,

when 2 mol of Fe2O3 formed, ∆H = - 1.65×10³kJ

It can be said that,

-1.65×10³kJ energy released when 2 mol of Fe2O3 formed

So, -4.6 × 10³kJ energy released when 2 mol of Fe2O3 formed

= 2 × -4.6 × 10³kJ / -1.65×10³kJ

= 5.57 mol of Fe2O3 formed

Now,

mass of Fe2O3 formed = 5.57 mol × 159.59 g/mol

= 888.916 g

Thus, we calculated that the rust forms when 4.85X10³ kJ of heat is released is 888.916 g. and the value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

learn more about ∆H:

brainly.com/question/24170335

#SPJ4

DISCLAIMER:

The given question is incomplete. Below is the complete question

QUESTION:

Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actual process requires water, but a simplified equation is 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1.65×10³kJ

a) What is the ∆H when 0.250kg iron rusts.

(b) How much rust forms when 4.85X10³ kJ of heat is released?

7 0
1 year ago
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