If you are using a thermometer, you are measuring the

Calculate the ph of an aqueous solution at 25°c that is 0.31 m in phenol (c6h5oh). (ka for phenol = 1.3 × 10−10.)

LekaFEV [45]

Given:

Concentration of phenol = 0.31 M

Ka (acid dissociation constant) = 1.3 * 10⁻¹⁰

To determine:

The pH of the solution

Explanation:

C6H5OH ↔ C6H5O⁻ + H⁺

Initial 0.31 M 0 0

change -x +x +x

Equlib 0.31-x x x

Ka = [C6H5O⁻][H⁺]/[C5H5OH]

1.31*10⁻¹⁰ = x²/0.31-x

x =[H+]= 6.37*10⁻⁶M

pH = -log[H+] = -log(6.37*10⁻⁶) = 5.19

Ans: pH of the solution is 5.19

6 0

4 years ago

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5. A study of the system, 4 NH3(g) + 7 O2(g) 2 N2O4(g) + 6 H2O(g), was carried out. A system was prepared with [NH3] = [O2] = 3.

emmasim [6.3K]

Answer: The value of the equilibrium constant, Kc, for the reaction is 0.013

Explanation:

Initial concentration of $NH_3$ = 3.60 M

Initial concentration of $O_2$ = 3.60 M

The given balanced equilibrium reaction is,

$4NH_3(g)+7O_2(g)\rightleftharpoons 2N_2O_4(g)+6H_2O(g)$

Initial conc. 3.60 M 3.60 M 0 M 0 M

At eqm. conc. (3.60-4x) M (3.60-7x) M (2x) M (6x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[N_2O_4]^2\times [H_2O]^6}{[NH_3]^4\times[Cl_2]}$

$[N_2O_4]=0.60M$

2x = 0.60 M

x= 0.30 M

Now put all the given values in this expression, we get :

$K_c=\frac{(2\times 0.30)^2\times (6\times 0.30)^6}{(3.60-3\times 0.30)^4\times (3.60-7\times 0.30)^7}$

$K_c=0.013$

6 0

3 years ago

Describe three different scenarios where a chemical process or reaction shows an increase in entropy.

Jet001 [13]

The request is characterized as knowing where things are and having the capacity to discover and utilize the things.
In a compound procedure, there is more issue, more entropy when the particles
1. warm up, increment in temperature. The atoms are more disorganized
2. get stirred up and must be isolated with exertion. Bedlam.
3. state changes, dissolves, vaporizes. The atoms are more turbulent
4. respond to frame a pack of various particles. More disorder

4 0

4 years ago

What is the ratio of the density of a substance to the density of a standard?

postnew [5]

Relative density, or specific gravity, is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material. Specific gravity usually means relative density with respect to water. The term "relative density" is often preferred in scientific usage.

8 0

3 years ago

How many grams of CO2 are produced by the combustion of 289 g of a mixture that is 28.6% CH4 and 71.4% C3H8 by mass?

serg [7]

845 grams of CO2 are produced by the combustion of 289 g of a mixture that is 28.6% CH4 and 71.4% C3H8 by mass.

Explanation:

The mixture of methane and propane is butane with molecular formula

C4H12

on combustion the balanced reaction is:

C4H12 + 7 O2 ⇒ 4CO2 + 6 H20

Molar mass of butane = 60.13 grams/mole, mass of C4H12 = 289 grams

number of moles of C2H12 is calculated by using the formula:

n = $\frac{mass}{molar mass of 1 mole}$

putting the values to know the moles of C4H12:

n = $\frac{289}{60.13}$

n = 4.8 moles of C4H12

from the balanced equation:

1 mole of C4H12 undergoes combustion to form 4 moles of CO2

thus, 4.8 moles when undergoes combustion will form x moles of CO2

$\frac{4}{1}$ = $\frac{x}{4.8}$

X = 4 X 4.8

x = 19.2 moles of CO2 is formed.

to convert mole into mass, formula used:

mass = number of moles x atomic mass

= 19.2 x 44.01

= 844.99 grams of CO2 is formed.

3 0

4 years ago