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chubhunter [2.5K]
3 years ago
9

A substance is tested with litmus paper. the blue litmus paper remains blue while the red litmus paper remains red. what can you

conclude about the substance?
Chemistry
1 answer:
kirill [66]3 years ago
6 0
The substance is neutral. it is neither an acid or a base.

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Products: 1-methylcyclohexene, 3-methylcyclohexene, methylenecyclohexane
Lostsunrise [7]

Answer:

See explanation

Explanation:

The reaction that we are considering here is quite a knotty reaction. It is difficult to decide if the mechanism is actually E1 or E2 since both are equally probable based on the mass of scientific evidence regarding this reaction. However, we can easily assume that the methylenecyclohexane was formed by an E1 mechanism.

Looking at the products, one could convincingly assert that the reaction leading to the formation of the two main products proceeds via an E1 mechanism with the formation of a carbocation intermediate as has been shown in mechanism attached to this answer. Possible rearrangement of the carbocation yields the 3-methylcyclohexene product.

8 0
3 years ago
In cellular respiration glucose is oxidized to three compounds
7nadin3 [17]
CO2 + H2O + Energy are the comments
7 0
3 years ago
The [-oh) in a sample of egg whites is 6.3 x 10-7 m. what is the [h3o+] in these<br>egg whites?
Kamila [148]

The hydrogen ion concentration [H3O+] in an egg white containing 6.3 × 10-⁷M of [OH-] is 1.5 × 10-⁸M.

<h3>How to calculate [H3O+]?</h3>

The hydrogen ion concentration of a solution can be calculated as follows:

pOH = - log [OH-]

pOH = - log [6.3 × 10-⁷M]

pOH = - [-6.2]

pOH = 6.2

Since pOH + pH = 14

pH = 14 - 6.2

pH = 7.8

pH = - log [H3O+]

7.8 = - log [H3O+]

[H3O+] = 10-⁷:⁸

[H3O+] = 1.5 × 10-⁸M

Therefore, the hydrogen ion concentration [H3O+] in an egg white containing 6.3 × 10-⁷M of [OH-] is 1.5 × 10-⁸M.

Learn more about hydrogen ion concentration at: brainly.com/question/15082545

7 0
2 years ago
A Y particle is a what ​
elena-s [515]

Answer:

The answer is <em>G</em><em>a</em><em>m</em><em>m</em><em>a</em><em> </em><em>r</em><em>a</em><em>y</em><em> </em><em>.</em>

(Correct me if I am wrong)

3 0
4 years ago
Calculate the percent ionization of nitrous acid in a solution that is 0.230 M in nitrous acid. The acid dissociation constant o
djyliett [7]
Ok first, we have to create a balanced equation for the dissolution of nitrous acid.

HNO2 <-> H(+) + NO2(-)

Next, create an ICE table

           HNO2   <-->  H+        NO2-
[]i        0.230M          0M       0M
Δ[]      -x                   +x         +x
[]f        0.230-x          x           x

Then, using the concentration equation, you get

4.5x10^-4 = [H+][NO2-]/[HNO2]

4.5x10^-4 = x*x / .230 - x

However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable, 

assume 0.230-x ≈ 0.230

4.5x10^-4 = x^2/0.230

Then, we solve for x by first multiplying both sides by 0.230 and then taking the square root of both sides.

We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.01M.

Then to find percent dissociation, you do final concentration/initial concentration.

0.01M/0.230M = .0434 or 

≈4.34% dissociation.
8 0
3 years ago
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