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3 years ago

What is the correct way to show sodium chloride in aqueous solution? A.NaCl (s) B.NaCl (w) C.NaCl (as) D.NaCl (aq)

2 answers:

5 0

Your answer would be, Sodium Chloride in Aqueous Solution is represented as, Letter choice(D), NaCI(aq), or as HCI + NaOH + H2O(I) + NaCI(aq)

Hope that helps!!!

irina [24]3 years ago

5 0

Answer: D.NaCl (aq)

Explanation: In Sodium chloride

Sodium atom has oxidation number of +1.

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.

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

Chlorine atom has oxidation number of -1.

Electronic configuration of sodium:

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

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

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

In sodium chloride the one electron from sodium metal gets transferred to chlorine atom to form ionic compound.

Ionic compounds dissociate to give ions when dissolved in water and thus are written as NaCl (aq).

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Answer: Multiple covalent bonds may occur in atoms that contain carbon,nitrogen, or oxygen.

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2 years ago

100.0 g of 4.0°C water is heated until its temperature is 37.0°C. If the specific heat of water is 4.184 J/g°C, calculate the am

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Answer:

13.8072 kj

Explanation:

Given data:

Mass of water = 100.0 g

Initial temperature = 4.0 °C

Final temperature = 37.0°C

Specific heat capacity = 4.184 j/g.°C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 37.0°C - 4.0 °C

ΔT = 33.0°C

Q = 100.0 g ×4.184 j/g.°C × 33.0°C

Q = 13807.2 j

Joule to KJ:

13807.2 j × 1kj /1000 j

13.8072 kj

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2 years ago

What goes in the blank?

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3 years ago

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Answer:

1. First

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3 years ago

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