What is the ideal gas constant?

Match these items. 1. kg measurement of area 2. cm 2 measurement of length 3. L measurement of mass 4. m measurement of volume

Phoenix [80]

Answer:

Kg measurement of volume.

cm measurement of area

L measurement of length

m measurement of maes

4 0

3 years ago

Calculate the equilibrium concentration of H 3 O H3O in a 0.20 M M solution of oxalic acid. Express your answer to two significa

Black_prince [1.1K]

Answer: The equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

Explanation:

We are given:

Molarity of oxalic acid solution = 0.20 M

Oxalic acid $(H_2C_2O_4)$ is a weak acid and will dissociate 2 hydrogen ions.

The chemical equation for the first dissociation of oxalic acid follows:

$H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)$

Initial: 0.20

At eqllm: 0.20-x x x

The expression of first equilibrium constant equation follows:

$Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}$

We know that:

$Ka_1\text{ for }H_2C_2O_4=0.059$

Putting values in above equation, we get:

$0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = x = 0.083 M

The chemical equation for the second dissociation of oxalic acid:

$HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)$

Initial: 0.083

At eqllm: 0.083-y 0.083+y y

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}$

We know that:

$Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}$

Putting values in above equation, we get:

$6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = y = 0.0000639 M

Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M

Hence, the equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

7 0

3 years ago

Write the chemical equation for the reaction of ammonia with hydrogen chloride

raketka [301]

If you mix Ammonia as well as Hydrogen chloride, you will get ammonium chloride. It is given by NH3 (g) + HCl (g) → NH4Cl(s).

3 0

3 years ago

Ethanol is added to gasoline because the oxygen it contains improves gasoline's burning efficiency. Its combustion reaction is g

ruslelena [56]

Answer:

A) The reaction is exothermic reaction

B) 46 gm CH3CH2OH = 1236KJ

15.3 gm ch3ch2OH = 1236/46 x 15.3

= 411.10 KJ..........released

therefore 1 KJ = 0.239 K cal

so, 411.10 x 0.239 kcal

= 98.2529 .................. released

c) 54 gm of H2O produced = 1236KJ

so, 42.7 gm H2O produced = 1236/54 x 42.7KJ

= 977.35 KJ released.

Explanation:

7 0

3 years ago

Aluminum has a density of 2.70g/cm what is the volume occupied by a block of aluminum which weights 4.32kg

romanna [79]

Answer:

Explanation:

0.000625 cm3

0.625 cm3

1.60 cm3

4 0

3 years ago