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IgorC [24]
2 years ago
15

the atomic number of four elements p,q,r,s are 6,8,10 and 12 respectively. the two elements which can bond ionically are:

Chemistry
1 answer:
denis-greek [22]2 years ago
6 0

Answer:

The atomic numbers of four elements A, B, C, and D are 6,8,10, and 12 respectively. The two elements which can react to form ionic bonds (or ionic compounds) are B (8= 2,6) and D (12 =2,8,2). So D donates its two electrons to B to fulfill their octet.

Explanation:

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Polonium-218 undergoes beta decay, converting a neutron into a proton. Then the daughter isotope undergoes another beta decay, a
sergij07 [2.7K]

Answer: ^{218}_{84}Po \rightarrow ^{218}_{85}At + ^{0}_{-1}e is the correct equation for beta decay.

Explanation:

When a beta particle, that is, ^{0}_{-1}e is emitted in a radioactive decay then it is known as beta decay.

Therefore, beta decay of Polonium-218 is as follows.

^{218}_{84}Po \rightarrow ^{218}_{85}At + ^{0}_{-1}e

Therefore, we can conclude that ^{218}_{84}Po \rightarrow ^{218}_{85}At + ^{0}_{-1}e is the correct equation for the given beta decay.  

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3 years ago
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Explanation:

Ca 40.8

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3 0
3 years ago
How many grams of potassium chloride will be needed to produce<br> 829 grams of zinc chloride?
grandymaker [24]

Answer:

[tex]2KCl + Zn {}^{2 + } → 2K {}^{ + } + ZnCl _{2} \\ molecular \: mass \: of \: zinc \: chloride = 65 + (35.5 \times 2) = 136 \: g \\ molecular \: mass \: of \: potassium \: chloride = 39 + 35.5 = 74.5 \: g

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Calculate the volume in liters of a M barium chloride solution that contains of barium chloride . Round your answer to significa
Reptile [31]

Answer:

1.667L of a 0.30M BaCl₂ solution

Explanation:

<em>Of a 0.30M barium chloride, contains 500.0mmol of barium chloride.</em>

<em />

Molarity is an unit of concentration used in chemistry defined as the moles of solute present in 1 liter of solution.

In a 0.30M BaCl₂ solution there are 0.30 moles of BaCl₂ in 1 liter of solution.

Now, in your solution you have 500mmol of BaCl₂ = 0.500 moles of BaCl₂ (1000 mmol = 1 mol). Thus, 0.500 moles of BaCl₂ are present in:

0.500 moles * (1L / 0.30 moles) =

<h3>1.667L of a 0.30M BaCl₂ solution</h3>
5 0
3 years ago
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