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Pachacha [2.7K]
4 years ago
5

Identify the missing nuclide in the following nuclear equation:

Chemistry
1 answer:
ra1l [238]4 years ago
7 0

Answer:

B. Bi-214.

Explanation:

The equation shows beta particle emission of 214/82 Pb which result into 214/83 Bi, in which the mass remain same but the the atomic number increases by one.

During this emission neutron get split into an electron and a proton which are represented as 0/e/-1.

So, the final nuclear equation becomes : 214/82 Pb => 0 e -1 + 214/83 Bi

Hence, the correct answer is "B. Bi-214."

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What determines an atom's reactivity?
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Answer:

D) How many valence electrons the atom contains.

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Is tap water a heterogeneous or homogenous mixture? Explain your answer.
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To what family of the periodic table does this new element probably belong
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6 0
3 years ago
PLEASE HELP
swat32

Answer:

1461.7 g of AgI

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI

Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:

Mole of AgI = 6.22 moles

Molar mass of AgI = 108 + 127

= 235 g/mol

Mass of AgI =?

Mass = mole × molar mass

Mass of AgI = 6.22 × 235

Mass of AgI = 1461.7 g

Therefore, 1461.7 g of AgI were obtained from the reaction.

5 0
3 years ago
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