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Sveta_85 [38]
2 years ago
11

Which of the following elements would be a +2 cation?

Chemistry
1 answer:
Dafna1 [17]2 years ago
7 0

Answer:

helium

Explanation:

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one molecule of chlorophyll contains 137 atoms. how many of these atoms come from the metal magnesium?
Svetlanka [38]
The chemical formula for chlorophyll is C55H72O5N4<span>Mg. Only 1 of the 137 atoms comes from magnesium.

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7 0
2 years ago
5. Why doesn't the air at a frontal boundary mix?​
Vsevolod [243]

Answer:

At a front, the two air masses have different densities, based on temperature, and do not easily mix. One air mass is lifted above the other, creating a low pressure zone.

Explanation:

Hope this helps!

4 0
2 years ago
What are standards or attributes of a design that can be measured.
ankoles [38]

Explanation:

attribute of a person that often cannot be measured directly but can be assessed using numbers of indicators or manifest variables

4 0
3 years ago
If we react 5.4g of sodium chloride with an unknown amount of fluorine gas, we produce 4.9g of sodium fluoride and 3.7g chlorine
Bumek [7]

Answer:

4.43 g

Explanation:

The reaction between sodium chloride and flourine gas is given as;

NaCl + F2 --> NaF + Cl2

From the stochiometry of the equation;

1 mol of NaCl reacts eith 1 mol of F2 to form 1 mol of NaF and Cl2

Mass of 1 mol of F2 = 38g

Mass of 1 mol of sodium flouride, NaF = 42g

This means 38g of flourine reacted with NaCl to form 42g of NaF

xg of F2 would form 4.9g of NaF

38 = 42

x = 4.9

x = 4.9 * 38 / 42

x = 4.43 g

4 0
2 years ago
What is the maximum mass of P2I4 that can be prepared from 8.80g of P4O6 and 12.37g of iodine according to the reaction:
Dvinal [7]
The balanced chemical reaction would be as follows:

<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10

We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:

8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2

Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:

0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
8 0
3 years ago
Read 2 more answers
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