Answer:
i. n = 5
ii. ΔE = 7.61 × 
KJ/mole
Explanation:
1. ΔE = (1/λ) = -2.178 × 
(
- 
)
(1/434 × 
) = -2.178 × (
)
⇒ 434 × = (1/-2.178 × )
But, 
= 2
434 × = (1/2.178 × )
434 × × 2.178 × = 
⇒ 
= 5
Therefore, the initial energy level where transition occurred is from 5.
2. ΔE = hf
= (hc) ÷ λ
= (6.626 × 10−34 × 3.0 × 
) ÷ (434 × )
= (1.9878 × 
) ÷ (434 × )
= 4.58 × 
J
= 4.58 × 
KJ
But 1 mole = 6.02×
, then;
energy in KJ/mole = (4.58 × KJ) ÷ (6.02×)
= 7.61 × KJ/mole
Answer:
Upper F subscript 2 (g) plus upper C a (s) right arrow with delta above upper C a upper F subscript 2 (s).
Explanation:
This is a chemical reaction problem.
In expressing any chemical reaction, we need to understand that there are reactants and products.
- The reactants are the species on the left hand side that are combining.
- The products are the species on the right hand side that are formed.
- Every chemical reaction is obeys the law of conservation of matter i.e equal number of matter on both sides.
Using the statement of this problem, we can deduce that;
Reactants are Fluorine gas and Calcium metal
Product is Calcium Fluoride
Note: A metal is a solid(s) and powder is a solid(s). A gas is denoted as (g). They depict the state of the species reacting.
F₂
+ Ca
→ CaF₂
We can see that equal number of atoms are on both sides of the expression.
Answer : The results would show more amount of water in the hydrated sample.
Explanation :
The amount of water of crystallization can be found by taking the masses of hydrated copper sulfate and anhydrous copper sulfate.
The difference in masses indicates the mass of water lost during dehydration process.
If during dehydration process, some of the copper sulfate spatters out of the crucible, then this would give us less mass for anhydrous sample than the actual.
As a result, the difference in masses of hydrated sample and the anhydrous sample would be more.
Therefore the results would show more amount of water in the hydrated sample.
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
I don't know but I'm wasting 5 seconds of your time you can't take back sorry
