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My name is Ann [436]
1 year ago
11

How many electrons does the nickel ion contain in the compound NiS?

Chemistry
1 answer:
Nostrana [21]1 year ago
5 0

There are 26 electrons in the nickel ion in the compound NiS.

<h3>How many electrons does the nickel ion contain in the compound NiS?</h3>

The number of electrons in a neutral Nickel atom is obtained from the atomic number of Nickel.

The atomic number of Nickel is 28, therefore, the neutral nickel atom has 28 electrons.

In  the compound NiS, nickel atom loses two electrons to to form the nickel ion.

Number of electrons left = 28 - 2 = 26

Therefore, there are 26 electrons in the nickel ion in the compound NiS.

In conclusion, nickel atom loses two electrons to form nickel ion in NiS.

Learn more about electrons at: brainly.com/question/860094

#SPJ1

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Answer: The solution will remain yellow.

Explanation:

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Imagine you need to explain to a friend how to convert a value on a food label to one that is measured in grams. Assume the pack
Mariulka [41]

Answer:

453.592 grams

Explanation:

Given

Mass = 1 lb

Required.

Convert to grams using dimensional analysis

Represent 1 lb with x g

In unit conversion, we have that.

1 lb = 453.592 g

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Getting the equivalent of lb in g, we have:

x g = 1 lb * (453.592 g/ 1 lb)

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The equivalent of 1 lb in grams is 453.592 grams

4 0
3 years ago
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3 years ago
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A mixture initially contains A, B, and C in the following concentrations: [A] = 0.650 M, [B] = 1.35 M, and [C] = 0.300 M. The fo
dimaraw [331]

<u>Answer:</u> The value of K_c for given reaction is 0.465

<u>Explanation:</u>

We are given:

Initial concentration of A = 0.650 M

Initial concentration of B = 1.35 M

Initial concentration of C = 0.300 M

Equilibrium concentration of A = 0.550 M

Equilibrium concentration of B = 0.400 M

For the given chemical equation:

                           A+2B\rightarrow C

<u>Initial:</u>                0.65     1.35     0.30

<u>At eqllm:</u>        0.65-x   1.35-2x   0.30+x

Evaluating the value of 'x'

0.650-x=0.550\\\\x=0.100

So, equilibrium concentration of B = 1.35 - 2x = [1.35 - 2(0.100)] = 1.15 M

Equilibrium concentration of C = (0.30 + x) = (0.300 + 0.100) = 0.400 M

The expression of K_c for above equation follows:

K_c=\frac{[C]}{[A][B]^2}

Putting values in above equation, we get:

K_c=\frac{0.400}{0.650\times (1.15)^2}\\\\K_c=0.465

Hence, the value of K_c for given reaction is 0.465

7 0
3 years ago
A 35.0-ml sample of 0.150 m acetic acid (ch3cooh) is titrated with 0.150 m naoh solution. calculate the ph after 17.5 ml of base
Gekata [30.6K]
When the moles of CH3COOH = volume of CH3COOH * no.of moles of CH3COOH
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moles of NaOH = volume of NaOH*no.of moles of NaOH
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SO the reaction after add the NaOH:
                           CH3COOH(aq) +OH- (aq) ↔ CH3COO-(aq) +H2O(l)
initial                  0.00525                0                         0
change             - 0.002625         +0.002625     +0.002625
equilibrium      0.002625             0.002625        0.002625
When the total volume = 35ml _ 17.5ml = 52.5ml = 0.0525L
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and [CH3COO-]= 0.002625/0.0525= 0.05 m
when PKa = -㏒Ka
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PH = Pka + ㏒[CH3COO-]/[CH3COOH]
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∴PH = 4.74
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