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sleet_krkn [62]
3 years ago
9

Please I need help! It is timed!

Chemistry
2 answers:
frez [133]3 years ago
5 0

Answer:

From point D to G, the potential energy increases

Explanation:

CaHeK987 [17]3 years ago
3 0

Answer: I think your answer is correct

Explanation:

I haven't studied this yet but from what I see it's correct because from G to D it decreases

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How many h+ ions are present in 65. 5 mL of 0. 722 M sulfuric acid
tangare [24]

Answer:

are there answer options

Explanation:

8 0
2 years ago
Draw the major organic product that is expected when cyclopentanecarboxylic acid is treated with each of the following reagents:
son4ous [18]

Answer:

a. Sodium cyclopentanecarboxylate

b. No reaction

Explanation:

In this case, in the cyclopentanecarboxylic acid we have a <u>carboxylic acid</u> functional group. Therefore we have an "acid". The acids by definition have the ability to produce hydronium ions (H^+).

With this in mind, for molecule a. we will have an <u>acid-base reaction</u>, because NaOH is a base. When we put together an acid and a base we will have as products a <u>salt and water</u>. In this case, the products are  Sodium cyclopentanecarboxylate (the salt) and water.

For the second molecule, we have the hydronium ion  (H^+). This ion can not react with an acid. Because, the acid will produce the hydronium ion also, so <u>a reaction between these compounds is not possible.</u>

See figure 1

I hope it helps!

8 0
3 years ago
Consider the reaction below in a closed flask. At 200 o C, the equilibrium constant (Kp) is 2.40 × 103 . 2 NO (g)  N2 (g) + O2
olga55 [171]

Explanation:

Since, the given reaction is as follows.

       2NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g)

Initial:    36.1 atm                 0          0

Change:    2x                      x           x

Equilibrium: (36.1 - 2x)       x            x

Now, expression for K_{p} of this reaction is as follows.

            K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}

As the initial pressure of NO is 36.1 atm. Hence, partial pressure of O_{2} at equilibrium will be calculated as follows.

              K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}

        2.40 \times 10^{3} = \frac{x \times x}{(36.1 - 2x)^{2}}

                 x = 18.1 atm

Thus, we can conclude that partial pressure of O_{2} at equilibrium is 18.1 atm.

5 0
3 years ago
Read 2 more answers
I GIVE U BRAINLIEST!
FromTheMoon [43]

Acids give off H+ (Hydrogen) ions in water; bases give off OH- (Hydroxide) ions in water. Acids generally taste sour due to the sour H+ ion; bases taste bitter due to the OH- ion; but they may have other tastes depending on the other part of the molecule.

7 0
4 years ago
Read 2 more answers
I don't know how to get the volume and it is 10 point removed for it wrong so can someone help me
In-s [12.5K]
16.91


Add all the values for the volume added then divide by 3 because there are three values
7 0
3 years ago
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