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An oxide of nitrogen is 25.9% N by mass, has a molar mass of 108 g/mol, and contains no nitrogen-nitrogen or oxygen-oxygen bonds

Sonja [21]

Answer:

The compound is N2O4

Explanation:

We have certain important pieces of information about the compound;

1) it is an oxide (a binary compound of nitrogen and oxygen)

2) there are no N-N bonds present

3) there are no O-O bonds present

Since it contains only nitrogen and oxygen then nitrogen accounts for 25.9% of the molecule by mass then oxygen should account for (100-25.9) = 74.1% oxygen

Relative atomic mass of oxygen = 16

Relative atomic mass of nitrogen = 14

We now deduce the empirical formula

Nitrogen. Oxygen

25.9/14. 74.1/16

1.85/1.85. 4.6/1.85 (divide through by the lowest ratio)

1 2

Empirical formula is NO2

To find the molecular formula

(NO2)n = 108

(14+2(16))n= 108

46n=108

n= 108/46

n= 2

Therefore molecular formula= N2O4

4 0

3 years ago

What is the oxidation state of each element in FeBr2?

maxonik [38]

Fe: 2+
Br: 1-
Checked on the periodic table

3 0

3 years ago

Each of the diagrams below shows an electrically neutral atom. fill in the missing number of protons (n) neutrons (and), electro

vova2212 [387]

What is the diagram?

7 0

3 years ago

The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in

astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

$\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}$

For this, it is necessary to know the values in meters for any of these diameters:

$\\ 1m = 10^{3}mm = 1e+3mm$

$\\ 1m = 10^{12}pm = 1e+12pm$

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

$\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m$

<h3>Diameter of a biscuit in meters</h3>

$\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m$

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}$

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8$

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0

3 years ago

In which circumstance is displacement the same as distance traveled?

Ede4ka [16]

Distance travelled will be equal to displacement when the line drawn is completely straight.

Hopefully, this helps.

8 0

3 years ago

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