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Ket [755]
3 years ago
5

When chlorine and magnesium react chemically, how many chlorine atoms will combine with one magnesium atom?

Chemistry
1 answer:
Lesechka [4]3 years ago
8 0
Magnesium is divalent. This means that one magnesium atom needs to lose 2 electrons in order to become stable.

Chlorine, on the other hand, is monovalent. This means that one chlorine atom needs to gain one electron in order to become stable.

Based on this, one magnesium atom will combine with two chlorine atoms, where the magnesium loses two electrons, one for each chlorine.

The formula of the compound formed is: MgCl2
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What is the mass number of chlorine?
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Explanation:

35.453 u

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explain how one and two letters chemical symbols are used to identify elements whose names begin with the same letter, for examp
Nadusha1986 [10]

The symbol of helium is He and this distinguishes it from the symbol for hydrogen.

<h3>What are chemical symbols?</h3>

Chemical symbol are the means by which we designate elements on paper. The symbols could be one letter or two letter as the case may be. There is no chemical symbol that has more than two letters that are used to designate the element.

We know that hydrogen has the symbol H. There is to way that another element will also have the symbol H as that would be tantamount to confusion. As a result of that, the symbol of helium is He and this distinguishes it from the symbol for hydrogen.

Learn kore about chemical symbol:brainly.com/question/9249660

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8 0
1 year ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
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