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Rashid [163]
3 years ago
7

What is he empirical formula for the compound that is of 1.85 moles of nitrogen and 4.63 miles of oxygen

Chemistry
1 answer:
BartSMP [9]3 years ago
4 0

The empirical formula is N₂O₅.

The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.  

The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of N:O</em>.  

I like to summarize the calculations in a table.  

<u>Element</u> <u>Moles</u>  <u>Ratio¹ </u>  <u> ×2²  </u>  <u>Integers</u>³

     N        1.85    1             2             2

     O        4.63    2.503   5.005     5

¹To get the molar ratio, you divide each number of moles by the smallest number (1.85).

²Multiply these values by a number (2) that makes the numbers in the ratio close to integers.

³Round off the number in the ratio to integers (2 and 5).

The empirical formula is N₂O₅.

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3. What is the formula for the compound formed from the magnesium ion and nitrate?
bija089 [108]

Answer:

D.) Mg(NO₃)₂

Explanation:

Since magnesium (Mg) is a metal and nitrate (NO₃) is a polyatomic anion, they would combine to form an ionic compound. Magnesium would form the cation Mg²⁺ because it generally has 2 valence electrons. Nitrate always has a -1 charge.

In order for the overall compound to be neutral (have a charge of 0), there must be one Mg²⁺ and two NO₃⁻ ions in the compound (+2 + (-1) + (-1) = 0).

Therefore, the formula for the compound formed is Mg(NO₃)₂.

8 0
2 years ago
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Natasha2012 [34]
Please upload a photo of your question!
6 0
3 years ago
How do you solve (1 x 1025) =3 x 10
Mrrafil [7]

Hey There!!~

Your best answer choice is 10280 or 10,280.

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Hope This Helps....!!

4 0
3 years ago
Read 2 more answers
By the reaction of carbon &amp; oxygen , a mixture of CO &amp;CO2 is obtained. What is the composition by mass of the mixture ob
Hatshy [7]
M(O₂)=20g
M(O₂)=32.0 g/mol
n(O₂)=20/32.0=0.625 mol

m(C)=12 g
M(C)=12.0 g/mol
n(C)=12/12.0=1.0 mol

   2C     +     O₂      →    2CO
1 mol    0.625 mol        1 mol
         0.625-0.5=0.125 mol

      2CO    +         O₂       →        2CO₂
0.250 mol       0.125 mol       0.250 mol

n(CO)=1 mol - 0.250 mol = 0.750 mol
M(CO)=28.0 g/mol
m(CO)=0.750*28.0=21.0 g

n(CO₂)=0.250 mol
M(CO₂)=44.0 g/mol
m(CO₂)=0.250*44.0=11.0 g
4 0
3 years ago
6
kumpel [21]

Answer:

Explanationrr

8 0
3 years ago
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