The way scientists form a hypothesis is there is something called the scientific method. You established your question or problem first. Then you research it. You take your hypothesis, theory, or prediction of what is going to happen from your research.
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
Answer:
1s2 2s2 2p6 3s2 3p6
Explanation:
Calcium is the 20th element. It has 20 electrons in it's neutral state. The electronic configuration is given as;
1s2 2s2 2p6 3s2 3p6 4s2
However in the compound; calcium chloride, it does not exist in it's neutral state.
CaCl2 --> Ca2+ + Cl-
In Ca2+ state, it has lost two electrons, The total number of electrons becomes; 20 - 2 = 18 electrons.
The electronic configuration is given as;
1s2 2s2 2p6 3s2 3p6
<span>Many scientific investigations have provided evidence to support this as the best explanation of the data</span>
Answer:
Fe²⁺ = 80mol/Ls
H⁺ = 50mol/Ls
Mn²⁺ = 4.0x10²mol/Ls
Fe³⁺ = 80mol/Ls
H₂O = 1.0x10²mol/Ls
Explanation:
Based on the reaction:
MnO₄⁻ + 5 Fe²⁺ + 8H⁺ → Mn²⁺ + 5 Fe³⁺ + 4H₂O
<em>Where 1 mole of permangante with 5 moles of Fe²⁺ and 8 moles of H⁺ produce 1 mole Mn²⁺, 5 of Fe³⁺ and 4 of H₂O</em>
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If the diappearence of 1 mole has a rate of 4.0x10²mol/Ls, disappearence of 5 moles of Fe²⁺ and 8 moles of H⁺ have a rate of:
Fe²⁺ = 1 mole MnO₄⁻ × ( 4.0x10²mol/Ls / 5 moles of Fe²⁺) = 80mol/Ls
H⁺ = 1 mole MnO₄⁻ × ( 4.0x10²mol/Ls / 8 moles of H⁺) = 50mol/Ls
And rate of appearance of 1 mole Mn²⁺, 5 of Fe³⁺ and 4 of H₂O are:
Mn²⁺ = 1 mole MnO₄⁻ × ( 4.0x10²mol/Ls / 1 moles of Mn²⁺) = 4.0x10²mol/Ls
Fe³⁺ = 1 mole MnO₄⁻ × ( 4.0x10²mol/Ls / 5 moles of Fe³⁺) = 80mol/Ls
H₂O = 1 mole MnO₄⁻ × ( 4.0x10²mol/Ls / 4 moles of H₂O) = 1.0x10²mol/Ls