Answer:
Explanation:
An oxidizing accepts an electron and becomes reduced while a reducing agent loses an electron and become oxidized.
Chemical equation:
1) 2 N₂H₄ + N₂O₄ → 3 N₂ + 4 H₂O
2) Hydrazine ( N₂H₄) is being oxidized
Dinitrogen tetroxide N₂O₄ is being reduced
3) The reducing agent is Hydrazine ( N₂H₄) and the oxidizing agent is dinitrogen tetroxide (N₂O₄)
A. The penny experienced a color change, and a gas was produced.
The penny was less tarnished (color change), and tiny bubbles formed around the penny (due to a gas being produced).
Answer is: molality of urea is 5.84 m.
If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
<span>Well... first, let's recognize that the chemical formula for chlorodifluoromethane is CHClF2. Count out how many valence electrons there are. C = 4, H = 1, Cl = 7, F (X2) = 14. Total is 26. Let's put C as the central atom, and put the other elements surrounding it. Draw a pair of electrons beach each element and the central atom. Then fill the halogen elements with 3 pairs of electrons each to fill their octets. Count out how many dots you have. There should be 26, making this the correct lewis structure!
Remember, hydrogen doesn't have a full octet, only a maximum of two electrons.</span>
2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)
First off.. not a chem board.. but n e way.
This is a limiting reagent problem.
set it up as a DA problem.(Dimension Analysis)
Start with what you want.
you want Grams of acrylonitrile (C3H3N)
so start with that (Using ACL in place of Acrylonitrile.. just for ease of typing)
(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)
solve that you wiill get grams of Acrylonitrile created by 15 grams oc C3H6 = 18.9g
Same setup for the two other reactants.
so i did it and for
oxygen I got 11.04 grams
and for Ammonia i got 15.29 grams
So the most you can make is 11.04 grams because if you have ot make any more .. you will have to get more O2 .. but since you have only 10 grams of it .. that is the most u can make in this reaction.
Both the other reactants are in excess.
rate brainliest pls