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2 years ago

For each of these terms, give an example and a non-example.

1 answer:

4 0

Given what we know, we can confirm that an example is a situation given that corroborates the information shown, while a non-example is one that does not fall in line with the information provided.

<h3>What are examples of the situations given?</h3>

A number that is a multiple of 10 is 40, since 10 times 4 equals 40.
In order to get a product of 10, we can multiply two and five.
To result in a quotient of 10, we can divide one hundred by ten.

<h3>What are non-examples of the situations given?</h3>

one non-example of a multiple of 10 would be to multiply three and seven.
A non-example of a product of 10 is to multiply the number fifty by twenty-five.
To result in a non-example for a quotient of 10, we can divide the number fifteen by three.

Therefore, given the definition of an example as a situation given that corroborates the information shown, while a non-example is one that does not fall in line with the information provided, we can confirm that the ones listed above are correct.

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S(s)+3F2(g)->SF6(g) how many mol of F2 are required to react completely with 2.30 mol of S?

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Answer: There are 6.9 mol of $F_{2}$ are required to react completely with 2.30 mol of S.

Explanation:

The given reaction equation is as follows.

$S(s) + 3F_{2}(g) \rightarrow SF_{6}(g)$

Here, 1 mole of S is reaction with 3 moles of $F_{2}$ which means 1 mole of S requires 3 moles of $F_{2}$ .

Therefore, moles of $F_{2}$ required to react completely with 2.30 moles S are calculated as follows.

$1 mol S = 3 mol F_{2}\\2.30 mol S = 3 mol F_{2} \times 2.30 \\= 6.9 mol F_{2}$

Thus, we can conclude that there are 6.9 mol of $F_{2}$ are required to react completely with 2.30 mol of S.

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