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devlian [24]
3 years ago
8

What is charge on the cation SnCl4

Chemistry
2 answers:
Rasek [7]3 years ago
7 0
Its total charge is zero but for the elements:
Sn===> Sn4+ positive
Cl===> Cl- negative
Umnica [9.8K]3 years ago
6 0

<u>Answer:</u> The charge on the cation is +4

<u>Explanation:</u>

An ion is formed when a neutral atom looses or gains electrons.

  • When an atom looses electrons, it results in the formation of positive ion known as cation.
  • When an atom gains electrons, it results in the formation of negative ion known as anion.

We are given a chemical compound having chemical formula of SnCl_4

It is formed by the combination of tin ions and chloride ions.

Tin is the 50th element of the periodic table having electronic configuration of [Kr]4d^{10}5s^25p^2

This element will loose 4 electrons to form Sn^{4+} ion

Chlorine is the 17th element of the periodic table having electronic configuration of [Ne]3s^23p^5

This element will gain 1 electron to form Cl^{-} ion

Thus, the charge on the cation is +4

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3 years ago
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A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h
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Nitrogen combine with hydrogen to produce ammonia \text{NH}_3 at a 1:3:2 ratio:

\text{N}_2 \; (g) + 3 \;  \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. 3 \; \text{mol} of hydrogen gas would have been consumed while 2 \; \text{mol} of ammonia would have been produced. The final mixture would therefore contain

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  • \begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  1.926 \times 10^{3} \; \text{kPa} \end{array}
  • \begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  1.723 \times 10^{3} \; \text{kPa} \end{array}
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3 years ago
A metal sample has a mass of 64.2 g. What is the density of this metal?
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