How many electrons does the isotope oxygen-18 have? 6 8 10 18

What kind of graph would use for this data? The months of the year and the amount of rainfall, and why

Digiron [165]

Answer:

A Desert Climatogram is a graph that shows the temperature and rainfall, for each month in the year, for the Desert.

Explanation:

6 0

3 years ago

How do optical telescopes work ?

inn [45]

They allow us to see further and are able to focus on light from distant object. This is done by refracting or reflecting the light using lenses or mirrors.

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

Neon-20 and neon-22 are called

natita [175]

Isotopes:) hope this helps

7 0

3 years ago

What is the total number of joules released when a 5.00-gram sample of water changes from liquid to solid at 0°C?

Marrrta [24]

Energy released from changing the phase of a substance from the liquid phase to solid phase can be calculated by using the specific latent heat of fusion. The heat of fusion of water at 0 degrees Celsius is 334 J/g. Calculation are as follows:

Energy = 5 grams x 334 J/g
Energy = 1670 J

4 0

3 years ago