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tester [92]
3 years ago
7

A nitride ion has 7 protons, 8 neutrons, and 10 electrons. What is the overall charge on this ion?

Chemistry
2 answers:
xz_007 [3.2K]3 years ago
7 0
The correct answer is: -3 (minus three)

Explanation:
To find the overall charge, we add electrons and protons. Make sure that you use the negative sign for an electron and positive sign for proton.

There are 10 electrons. It means the charge is "-10"
There are 7 protons. It means the charge is "+7"

Add both of them:

+7 - 10 = -3

Therefore the overall charge on this ion is -3.
Ne4ueva [31]3 years ago
7 0

Answer:

-3

Explanation:

You might be interested in
Which statement is true regarding methane and ammonia?
sukhopar [10]
The options for given question are as follow,

1) Methane molecules show hydrogen bonding. 
<span>2) Ammonia molecules show hydrogen bonding. </span>
<span>3) Methane has stronger hydrogen bonding than ammonia. </span>
<span>4) Both the compounds do not show hydrogen bonding. </span>
<span>5) Both the compounds have strong hydrogen bonding.
</span>
Answer:
            Correct answer is Option-2 (Ammonia molecules show hydrogen bonding).

Explanation:
                   Hydrogen bond interactions are formed when a partial positive hydrogen atom attached to most electronegative atom of one molecule interacts with the partial negative most electronegative element of another molecule. So, in Ammonia hydrogen gets partial positive charge as nitrogen is highly electronegative. While the C-H bond in Methane is non-polar and fails to form hydrogen bond interactions.
6 0
3 years ago
Read 2 more answers
Write the balanced NET IONIC equation for the reaction that occurs when ammonium nitrate and potassium hydroxide are combined. N
vova2212 [387]

Answer:

Net Ionic equation

NH₄⁺ + OH⁻ → NH₃ + H₂O

Option B is correct.

Weak Acid Strong Base

Check Explanation for the extent of the reaction.

Explanation:

Ammonium nitrate = NH₄NO₃

Potassium Hydroxide = KOH

Ammonium salts combine with alkalis to liberate NH₃ and form water.

The two reactants combine to give

NH₄NO₃ + KOH → KNO₃ + NH₃ + H₂O

In ionic form,

- NH₄NO₃ exists as NH₄⁺ and NO₃⁻

- KOH exists as K⁺ and OH⁻

- KNO₃ as K⁺ and NO₃⁻

And NH₃ and H₂O stay as they are, as per covalent compounds.

So, we have

NH₄⁺ + NO₃⁻ + K⁺ + OH⁻ → K⁺ + NO₃⁻ + NH₃ + H₂O

Eliminating the ions that exist on both sides, we have the net ionic equation to be

NH₄⁺ + OH⁻ → NH₃ + H₂O

which shows that this reaction is essentially a neutralization reaction in which the Bronsted Lowry acid, NH₄⁺, loses its proton to the base, OH⁻ and gives conjugate base, NH₃ and conjugate acid, H₂O.

This reaction is classified as a Weak acid versus Strong Base reaction as NH₄⁺ is from a Weak acid and OH⁻ is from a strong base.

Since this reaction is between a Weak base and a strong acid, the ionization isn't expected to be 100%, Hence, the extent of this reaction will be any option that is not 100%, a couple pieces of information might be required for the correct estimate, but above 50% seems correct.

Hope this Helps!!!

8 0
3 years ago
1 How is the density of a substance calculated? 
user100 [1]
Sorry for the scribbles lol

4 0
3 years ago
What experimental evidence did Thomson have for each statement?
Minchanka [31]
<span>Thomson studied electric discharge in a vacuum and found that the deflection of rays was evidence of atoms containing much smaller particles. He calculated that these particles would have a large charge in relation to their mass. While he did not name electrons, he knew they existed.</span>
5 0
3 years ago
Read 2 more answers
2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
azamat

Answer:

K=1.12x10^9

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

K=\frac{[NO_2]^2}{[NO]^2[O_2]}

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0
3 years ago
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