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Vika [28.1K]
2 years ago
13

Calculate the equilibrium constant for the reaction below if a tank was found to contain 0.106M O2, 0.00652M SO3, and 0.00129M S

O2. Calculate the equilibrium constant for the reaction below if a tank was found to contain 0.106 M O2, 0.00652 M SO3, and 0.00129 M SO2.
2 SO3 (g) ⇌ 2 SO2 (g)+O2(g)

(a)1.34 × 10^-2
(b)6.78 × 10^-2
(c)4.15 × 10^-3
(d)4.35 × 10^-2
Chemistry
1 answer:
malfutka [58]2 years ago
6 0

equilibrium constant K = 4.15 × 10⁻³

Explanation:

We have the following chemical equilibrium:

2 SO₃ (g) ⇄ 2 SO₂ (g) + O₂ (g)

equilibrium constant K = ( [SO₂]² × [O₂] ) / [SO₃]²

equilibrium constant K = ( 0.00129² × 0.106 ) / 0.00652²

equilibrium constant K = 0.414 × 10⁻² = 4.14 × 10⁻³ ≈ 4.15 × 10⁻³

Learn more about:

equilibrium constant

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2. A compound with the following composition by mass: 24.0% C, 7.0% H, 38.0% F, and 31.0% P. what is the empirical formula
Svetach [21]

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

5 0
2 years ago
Keiko needs 100mL of a 5% acid solution for a science experiment. She has available a 1% solution and a 6% solution. How many mi
4vir4ik [10]

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

7 0
2 years ago
Read 2 more answers
Exactly 2.00 g of an ester A containing only C, H, and O was saponified with 15.00 mL of a 1.00 M NaOH solution. Following the s
Anna11 [10]

Answer:

Explanation:

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5 0
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Acrylonitrile () is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. If 1
pychu [463]

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

3 0
2 years ago
The target diol is synthesized in one step from 1-methylcyclopentene, but your lab partner exhausted the supply of that alkene.
andrezito [222]

Answer:

The reagents are CH_{3}CH_{3}O^{-},OsO_{4},NaHSO_{3}and H_{2}O.

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.

4 0
2 years ago
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