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NeX [460]
2 years ago
13

1) how would you explain the plateaus in your heating and heating curves?

Chemistry
1 answer:
dem82 [27]2 years ago
4 0
Do u wanna go out for that or no


Explanation: I did the quiz
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Daniel is conducting an experiment on magnetic objects. He completes more than one trial. Why is it important for Daniel to do t
Luba_88 [7]

Answer:The conclusion is less accurate.

Explanation:

Feb 17, 2021 — He completes more than one trial. Why is it important for Daniel to do this?

4 0
2 years ago
Balance the following equation and find molar ratios from the equation,
Vikki [24]
To balance it, it would be N2 + 3H2 ------> 2NH3. 

for c) it would be 2N2 + 6H2 -------> 4NH3
7 0
3 years ago
Which actions represent endothermic reactions? Check all that apply.
marishachu [46]

Answer:

Well, one is if heat is absorbed by the system from the surroundings, the system gains heat from the surroundings and so the temperature of the surroundings decreases.

Explanation:

8 0
3 years ago
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Explain how stagnant sulfureted water may greatly accelerate metallic corrosion. (See the Email Link in Moodle) | (7 marks)
slavikrds [6]

Answer:

sulfur promotes oxide-reduction reactions.

Explanation:

In stagnant water, some solutes tend to precipitate. When Sulfur precipitate and touch a metal, Sulfur is being reduced and the metal is oxidated. This depends of potential redox of each element.  

6 0
3 years ago
Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE
nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let "\alpha is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

                  HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}

Initial                  C                         0            0

Equilibrium        c(1- \alpha)              c\alpha c\alpha

Dissociation constant = Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}

=\frac{c\alpha^{2}}{(1-\alpha)}

In this case weak acids \alpha is very small so, (1-\alpha ) is taken as 1.

Ka=C\alpha^{2}

\alpha=\sqrt\frac{ka}{c}

From the given the concentration = 0.050 M

Substitute the given value.

\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}

[H_{3}O^{+}]=c\alpha

[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}

pH= -log[H_{3}O^{+}]

=-log[4.9\times10^{-6}]

=6-log 4.9= 5.31

Therefore, The pH of the solution is 5.31.

7 0
3 years ago
Read 2 more answers
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