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Leviafan [203]
3 years ago
10

What is a net ionic equation?

Chemistry
1 answer:
jolli1 [7]3 years ago
7 0

Answer:

I think it will option D hope it helps

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How many different principal quantum numbers can be found in the ground-state electron configuration of nickel?A) 2.B) 3.C) 4.D)
Naddik [55]

Answer:

C) 4.

Explanation:

Hello!

In this case, since the electron configuration of nickel at its ground-state, considering 28 as its atomic number and the number of electrons it has in one atom, is:

Ni^{28}:1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^8

We can see it has four energy levels, 1, 2, 3 and 4, which are related to the following principal quantum number, that describes the energy of an electron in the atom and its most probable distance with respect to the nucleus.

Therefore, nickel has C) 4 different principal quantum numbers.

Best regards!

5 0
3 years ago
Is feeling warmth from a fireplace convection, radiation, or conduction?
Mashutka [201]
Am not exactly sure but i think ots conduction
4 0
3 years ago
Read 2 more answers
Determine the amount of both hydrogen and oxygen in a 500mL sample of Water.
alina1380 [7]
There would be about 1.67 x 10^25 oxygen atoms and about 3.34 x 10^25 hydrogen atoms.
3 0
3 years ago
Zinc carbonate dissolves in water to the extent of 1.12 x 10-4 g/L at 25 C. Calculate the solubility product Ksp for ZnCO3 at 25
sashaice [31]

Answer:

Ksp= 7.98 × 10^-13

Explanation:

According to the question, we are to calculate the solubility constant (Ksp) of Zinc carbonate (ZnCO3) in a dissolved solution.

The equilibrium of the reaction is:

ZnCO3 (aq) ⇌ Zn2+ (aq) + CO32- (aq)

According to this; 1 mole of Zinc carbonate (ZnCO3) dissolves to give 1 mole of Zinc ion (Zn2+) and 1 mole of carbonate ion (CO32-).

This illustrates that:

(Zn2+) = 1.12 x 10-4 g/L

(CO32-) = 1.12 x 10-4 g/L

However, 1.12 x 10-4 g/L is the solubility in mass concentration of ZnCO3, we need to convert it to molar concentration in mol/L by dividing by the relative molar mass of ZnCO3.

To calculate the molar mass of ZnCO3, we say:

Zn (65.4) + C (12) + 03 (16×3)

= 65.4+12+48

= 125.4g/mol.

Hence, molar concentration= 1.12 x 10-4 g/L / 125.4 g/mol

= 8.93 × 10^-7 mol/L.

Therefore;

Zn2+) = 8.93 x 10-7 mol/L

(CO32-) = 8.93 x 10-7 mol/L

Ksp = [Zn2+] [CO32-]

Ksp = (8.93 x 10-7) × (8.93 x 10-7)

Ksp = 7.98 × 10^-13

3 0
3 years ago
A buffer is prepared by dissolving 0.80 moles of NH3 and 0.80 moles of NH4Cl in 1.00 L of aqueous solution. If 0.10 mol of NaOH
Vikki [24]

Answer:

a. 10.54

Explanation:

reaction is

NH₃ +  NaOH   -----------------NH₄Cl + H₂O

0.10 mol NaOH will consume 0.10 mol NH₃ thereby decreasing the initial amount of moles NH₃ and increasing that of NH₄Cl

mol NH₃  = 0.80 - 010 = .70

mol NH₄Cl = 0.80 + .10 = 0.90

pH = pKₐ + log (( NH₃/NH₄Cl))

pH =  9.26 + log (( 0.90 + 0.70)) = 9.26 + 0.11 = 10.54

7 0
3 years ago
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