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xenn [34]
2 years ago
15

Na3X, Enter the group number of X?

Chemistry
2 answers:
ICE Princess25 [194]2 years ago
8 0

Answer : The group number of X is 15.

Explanation :

In Na_3X, 'Na' is a sodium metal and 'X' is an unknown element.

First we have to determine the oxidation state of element 'X'.

Let the oxidation state of element 'X' be, 'a'.

As we know that sodium is an alkali metal which belongs to group 1 and has (+1) charge.

3(+1)+a=0\\\\+3+a=0\\\\a=-3

That means, the oxidation state or charge on 'X' is (-3) and the group 15 elements (nitrogen, phosphorous, arsenic, antimony and bismuth) have five electrons in valence shell and thus require three more electrons to acquire the nearest noble gas configuration. Thus, the group 15 elements have the tendency to show (-3) oxidation state.

Hence, the group number of X is 15.

Katen [24]2 years ago
4 0
X will be in group 5, since if you exchange the valencies of Na with any element on group 5, you will get Na3X
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2 years ago
A mixture of Na2CO3 and MgCO3 of mass 7.63 g is reacted with an excess of hydrochloric acid. The CO2 gas generated occupies a vo
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58.6 % by mass of Na₂CO₃

Explanation:

This is the reaction:

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Let's find out the moles of CO₂ produced, by the Ideal Gases Law

1.24 atm . 1.67 L = n . 0.082 . 299K

(1.24 atm . 1.67 L / 0.082 . 299K) = n

0.0844 moles = n

Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:

2 moles of CO₂ were produced by 1 mol of Na₂CO₃

Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2  = 0.0422 moles of Na₂CO₃.

Let's convert this moles into mass (mol . molar mass)

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3 0
2 years ago
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Answer:

The answer to your question is below

Explanation:

There are two kinds of mixtures

- Homogeneous is a mixture of two or more elements or compounds and its components can not be distinguished visually.

- Heterogeneous is a mixture of two or more elements or compounds and its components can be distinguished visually.

a cup of tea and sugar homogeneous

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3 0
2 years ago
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