Ask question

Ask question

All categories

4 years ago

5

Which of the following types of fossils is most commonly associated with wood?

1 answer:

DanielleElmas [232]4 years ago

5 0

Petrified fossils, they are made of wood that becomes petrified from pressure and lack of oxygen.

You might be interested in

What features form when rocks bend?

bezimeni [28]

When the sedimentary layers of rocks are soft and under pressure they can be bent into curved structures called geosynclines. One theory is that heat partially melts the sedimentary layers of rocks allowing them to be bent into the curved shapes.

Hope this helps :)!

3 0

3 years ago

13. Which of the following correctly provides the names and formulas of polyatomic ions?

sukhopar [10]

Mirror no :HCO3
Answer HCO3

4 0

3 years ago

Please help my class ends in a couple hours

marishachu [46]

Answer:

Homogeneous mixture

Solution

Liquid-gas

Hope this helps, hope its not late.

4 0

3 years ago

Read 2 more answers

At what phase of the cell cycle does the genetic material double?

natka813 [3]

There are the four phases of the cell cycle. M, G1, S and G3. And in S (Synthesis) phase genetic material is doubled.

3 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]

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

<u>Explanation:</u>

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:

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

<u>Initial:</u> 0.20

<u>At eqllm:</u> 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:

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

<u>Initial:</u> 0.083

<u>At eqllm:</u> 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