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vovangra [49]
3 years ago
14

Empirical formula for compound of 2.17 mol N and 4.35 mol O

Chemistry
1 answer:
balu736 [363]3 years ago
4 0

Answer:

Explanation:

ratio of  moles of N and O in molecule =

N / O = 2.17 / 4.35

1/2

empirical formula = NO₂

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An atom has 6 electrons in its outer orbit. It will most likely form an ion with a charge of:
GenaCL600 [577]

Answer:

c

Explanation:

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3 years ago
How can a person reduce consumption of natural resources when drinking water? Assume that you are not going to change the amount
miss Akunina [59]

Answer:

It will reduce the natural resources by not littering, putting trash in the recycle bin or trash bin. I will also keep drinking the normal amount a day which would be 3-4 bottles of water.  I will also not throw my trash on the ground or any thing like that it will stay with me until i'm near a trash can.   If you litter than you are ki**ing a poor innocent animal or mammal so please don't litter.

Explanation:

l'm sorry if its wrong but hope it helps you

4 0
2 years ago
A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent compos
Volgvan

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

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The calculation of the percentage composition of an element is made by:

percent composition element A=\frac{total mass of element A}{mass of compound} *100

In this case, the percent composition of fluorine is:

percent composition of fluorine=\frac{39.6 g}{60.3 g} *100

percent composition of fluorine= 65.67%

<u><em>The percent composition of fluorine is 65.67%</em></u>

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3 years ago
What are the reference points in thermometer with Celsius scale?
kupik [55]
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I hope this helps.  Let me know if anything is unclear. 
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