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Olenka [21]
2 years ago
11

Round off 318.90 to 2 significant figures

Chemistry
1 answer:
motikmotik2 years ago
6 0

Answer:

320

Explanation:

Because the question is asking for 2 significant figures, you want to try to get rid of all of the numbers to the right of the 1. That way, the only numbers that will be significant are the numbers in the position of the 3 and the 1.

Before you can make the rest of the numbers equal 0, you need to correctly round the number in the last significant position (the number 1).

If the number to the right of the 1 is from 0-4, the number stays the same. If the number is from 5-9, the 1 needs to be increased by a unit.

Since the number to the right is an 8, you need to change the 1 to a 2. Now, you can make all of the other numbers a 0. There should be no decimal place because it would mean that the zero to the left of it is significant.

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How many of the following species are paramagnetic? sc3+ br- mg2+ se?
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Answer:
           One: <u>Selenium</u> is Paramagnetic

Explanation:
                   Those compounds which have unpaired electrons are attracted towards magnet. This property is called as paramagnetism. Lets see why remaining are not paramagnetic.

Electronic configuration of Scandium;

Sc  =  21  = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹

Sc³⁺  =  1s², 2s², 2p⁶, 3s², 3p⁶ 

Hence in Sc³⁺ there is no unpaired electron.

Electronic configuration of Bromine;

Br  =  35  = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁵

Br⁻  =  1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶

Hence in Br⁻ there is no unpaired electron.

Electronic configuration of Magnesium;

Mg  =  12  = 1s², 2s², 2p⁶, 3s²

Mg²⁺  =  1s², 2s², 2p⁶

Hence in Mg²⁺ there is no unpaired electron.

Electronic configuration of selenium;

Se  =  34  = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴

Or,

Se  =  34  = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4px², 4py¹, 4pz¹

Hence in Se there are two unpaired electrons hence it is paramagnetic in nature.
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