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Semenov [28]
3 years ago
9

How many atoms of nitrogen are in Fe2(NO4)2?

Chemistry
1 answer:
djverab [1.8K]3 years ago
5 0

<em>the</em><em> </em><em>number</em><em> of</em><em> </em><em>nitrogen</em><em> </em><em>atoms</em><em> </em><em>in</em><em> </em><em>the </em><em>compound</em><em> </em><em>is</em><em> </em><em>two</em>

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How is fractional distillation of crude oil related to stoichiometry
lora16 [44]

Answer:

Fractional distillation is the process by which oil refineries separate crude oil into different, more useful hydrocarbon products based on their relative molecular weights in a distillation tower.

8 0
3 years ago
TRUE or FALSE: The charge of an atom (ion) is determined by the ratio of electrons to protons.
Damm [24]
True The positive electric charge of a proton is equal in magnitude to the negative charge of an electron; therefore, the net electric charge of an ion is equal to its number of protons minus its number of electrons. Ions are highly reactive species.
8 0
3 years ago
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When 0.50 L of a 12 M solution is diluted to 1.0 L, what is the resulting molarity?
Lyrx [107]

Answer:

The resulting molarity is 6M.

Explanation:

A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).

We use the formula for dilutions:

C1 x V1 = C2 x V2

12 M x 0,5L = C2 x 1,0 L

C2= (12 M x 0,5 L)/1,0 L

<em>C2= 6 M</em>

5 0
3 years ago
H 2 O, CO 2 , and C 12 H 22 O 11 are all examples of chemical _____.
Reil [10]
They are all examples of chemical formulas
6 0
3 years ago
Read 2 more answers
1.15 g of a metallic element needs 300 cm3 of oxygen for complete reaction, at 298 K and 1 atm
sashaice [31]
1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm


Ideal gas equation: pV = nRT => n = pV / RT


R = 0.0821 atm*liter/K*mol

V = 300 cm^3 = 0.300 liter

T = 298 K

p = 1 atm


=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol


2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type


X (+) + O2 (g) ---> X2O          or   


2 X(2+) + O2(g) ----> X2O2 = 2XO     or


4X(3+) + 3O2(g) ---> 2X2O3


 
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)



So, lets probe those 3 cases.


3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol

=> x = 0.01226 moles of metal X


Now you can calculate the atomic mass of the hypotethical metal:

1.15 grams / 0.01226 mol = 93.8 g / mol


That does not correspond to any of the metal with valence 1+


So, now probe the case 2.



4) Case 2:


2moles X metal / 1 mol O2(g) = x / 0.01226 mol


=> x = 2 * 0.01226 = 0.02452 mol


And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol


That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.


4) Case 3


4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635 


atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol


That does not correspond to any metal.


Conclusion: the identity of the metallic element could be titanium.
5 0
4 years ago
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