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Bas_tet [7]
3 years ago
8

Which of the following sets are isoelectronic (i.e., have the same number of electrons)? i. Br-, Kr, Sr2+ ii. C, N-, O2- iii. Mg

2+, Ca2+, Sr2+ iv. O2-, O, O2+ v. Ag+, Cd2+, Pd
Chemistry
1 answer:
il63 [147K]3 years ago
8 0

Answer:

ii. C,\ N^-,\ O^{2-}

Explanation:

Isoelectric species are the species which have same number of electrons.

From the given options:-

C,\ N^-,\ O^{2-} are the isoelectric species.

All the three species have same number of electrons which is equal to 6.

Carbon has 6 electrons.

N^- has 6 electrons.

O^{2-} has 6 electrons.

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C should be the answer to the following formula
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All radioactive nuclides undergo what
kotykmax [81]
They undergo nuclear fission.
6 0
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Magnesium (average atomic mass = 24.3050 amu) consists of three isotopes with masses 23.9850 amu, 24.9858 amu, and 25.9826 amu.
iris [78.8K]

Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%

Explanation: Given that the average atomic mass(M) of magnesium

= 24.3050amu

Mass of first isotope (M1) = 23.9850amu

Mass of middle isotope (M2)=24.9858amu

Mass of last isotope(M3)= 25.9826amu

Total abundance = 1

Abundance of middle isotope = 0.10

Let abundance of first and last isotope be x and y respectively.

x+0.10+y =1

x = 0.90-y

M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope

24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y

Substitute x= 0.90-y

Then

y = 0.11

Since y=0.11, then

x= 0.90-0.11

x=0.79

Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%

5 0
3 years ago
The region in which an electron is likely to be found is known as a(n) ____________.
Zinaida [17]

Answer:

orbital

Explanation:

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3 0
3 years ago
Can someone solve this for me I'm confused.
Artemon [7]

Answer:

310.53 g of Cu.

Explanation:

The balanced equation for the reaction is given below:

CuSO₄ + Zn —> ZnSO₄ + Cu

Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:

Molar mass of CuSO₄ = 63.5 + 32 + (16×4)

= 63.5 + 32 + 64

= 159.5 g/mol

Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g

Molar mass of Cu = 63.5 g/mol

Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g

Summary:

From the balanced equation above,

159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.

Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:

From the balanced equation above,

159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.

Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.

Thus, 310.53 g of Cu were obtained from the reaction.

6 0
3 years ago
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