All categories

2 years ago

Propanoic acid fromula

Chemistry

1 answer:

Romashka [77]2 years ago

8 0

Answer:

C₃H₆O₂

Explanation:

Propionic acid is a colorless liquid with a sharp rancid odor. Produces irritating vapor. (USCG, 1999)

You might be interested in

WILL GIVE BRAINLIEST AND 10 POINTS!

GarryVolchara [31]

Student B because it requires a hypothesis

4 0

2 years ago

Read 2 more answers

Great amounts of atomic energy are released when a _______reaction occurs.

Mariulka [41]

Great amounts of atomic energy are released when a _______reaction occurs.

Great amounts of atomic energy are released when a chemical reaction occurs. The process can be an exothermic reaction or endothermic reaction depending on the substances involved in the reaction.

8 0

2 years ago

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

8 0

3 years ago

What type of Machine is a handrail

Elodia [21]

Hey There!

Here is your answer:

The answer is a compound machine!

Hope this helps!

4 0

2 years ago

Glucose is the monomer in the natural polymers ____________________ and cellulose.

Olin [163]

Answer:

Starch

I hope this helps

5 0

2 years ago

Propanoic acid fromula​

Propanoic acid fromula