Answer:
C₂H₇F₂P
Explanation:
Given parameters:
Composition by mass:
C = 24%
H = 7%
F = 38%
P = 31%
Unknown:
Empirical formula of compound;
Solution :
The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;
C H F P
% composition
by mass 24 7 38 31
Molar mass 12 1 19 31
Number of
moles 24/12 7/1 38/19 31/31
2 7 2 1
Dividing
by the
smallest 2/1 7/1 2/1 1/1
2 7 2 1
Empirical formula C₂H₇F₂P
Answer:
Keiko should mix 20 mL 1% solution and 80 mL 6% solution for to make 100 mL 5% solution
Explanation:
There are 2 unknown values X= mL 6% solution and Y=1% solution. So, we need 2 equations:
1. Equation acid concentration. X mL 6% + Y mL 1% = 100 mL 5%
2. Equation solvent concentration X mL 94% + Y mL 99% = 100 mL 95%
When clearing X and Y :
(X mL 6% + Y mL 1% = 100 mL 5%) (-15,7)
X mL 94% + Y mL 99% = 100 mL 95%
_______________________________
- Y 0.83 = 16.5
Y = 19.9 mL 1% solution
Replace Y in anyone equation and X = 80 mL 6% solution
I hope to see been helpful
Answer:
Explanation:
check the attachment for the propose neutral structure for each compound that is consistent with the data.
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:
The reagents are 
.
Explanation:
1-Methylenecyclopentene is treated with HBr form 1-bromo-1-methylcyclopentane, which is treated with strong base ethoxide ion and forms 1-methylcyclopent-1-ene.
This alkene is treated with osmium tetraoxide in the presence of sodium bisulfite to form target product.
The chemical reaction is as follows.
