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djverab [1.8K]
3 years ago
12

ANSWER ASAP please I need help thank you I’ll give you brainly

Chemistry
1 answer:
Vilka [71]3 years ago
5 0

Answer:

See the attached notation!

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kifflom [539]
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8 0
3 years ago
The ionic charge of an element in Group 6A is 2–.<br><br> -True<br><br> -False
natita [175]

True. These ions are of a 2- charge. Oxygen is an example. It will form a 2- charge if ionize.  

4 0
3 years ago
An unknown compound has the following chemical formula: Mg_xCl_2
Art [367]

Answer:

Explanation:

Given parameters:

number of moles magnesium = 6.80mol

number of moles of chlorine = 13.56mol

To find the complete chemical formula, we should obtain the formula of the compound.

                                Mg                            Cl

Number of

moles                      6.8                           13.56

Dividing

by the smallest      6.8/6.8                    13.56/6.8

                                    1                                2

  The formula of the compound is MgCl₂

This is an ionic compound in which Magnesium loses two electrons that would be gained by Cl atoms requiring just an electron each to complete their octet.

3 0
3 years ago
Which of the following are examples of types of mixtures?
aev [14]
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6 0
3 years ago
How many H2O molecules are in 183.2 grams of H20 gas?
jek_recluse [69]

Answer: There are 61.24 \times 10^{23} molecules present in 183.2 grams of H_{2}O gas.

Explanation:

Given: Mass = 183.2 g

Number of moles is the mass of substance divided by its molar mass.

As molar mass of water is 18 g/mol. Therefore, moles of H_{2}O are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{183.2 g}{18 g/mol}\\= 10.17 mol

According to the mole concept, there are 6.022 \times 10^{22} molecules present in one mole of a substance.

Hence, molecules present in 10.17 moles are calculated as follows.

10.17 mol \times 6.022 \times 10^{23}\\= 61.24 \times 10^{23}

Thus, we can conclude that there are 61.24 \times 10^{23} molecules present in 183.2 grams of H_{2}O gas.

6 0
3 years ago
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