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Nastasia [14]
3 years ago
13

Please help! I don’t think my answer is right!

Chemistry
2 answers:
Ostrovityanka [42]3 years ago
4 0

Answer:

Thats right! Gj!

Explanation:

kirza4 [7]3 years ago
3 0

Answer:

You are actually correct!

Explanation:

<em>Base </em>turns Blue litmus to Red

<em>Acid </em>turns Red litmus to Blue

<em>Neuteral</em><em> </em><em>Solution </em><em>No </em><em>reaction!</em>

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Read 2 more answers
What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?
ioda

N₂O is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen.

Empirical formula can be calculated by

Suppose we have 100 g of the substance. That indicates that it has 36.69 grams of oxygen and 63.61 grams of nitrogen.

Masses transformed into moles:

Formula used

Given mass/ Molar mass

14.01 g contains 1 mol of N

So 63.61 g of N contains moles is equals to

(1 mol N / 14.01 g N) 63.61 g N = 4.540 mol N

Similarly

16 g of O contains 1 mole of O

36.69 g of O contains moles is equals to

(1 mol O / 16.00 g O) 36.69 g O = 2.293 mol O

Divide by the smallest to normalize:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

Therefore, there are roughly twice as many N as O atoms. N2O is the empirical formula as a result.

Ratio is basically 2:1

Hence, N₂O is the empirical formula of an oxide of nitrogen

Learn more about Empirical Formula here brainly.com/question/27873410

#SPJ4

7 0
2 years ago
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