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2 years ago

A) 24 g B) 36 g C) 48 g D) 60 g

Chemistry

1 answer:

Hitman42 [59]2 years ago

5 0

Answer:

I'm not sure if this right because i've only taken one year of Chemistry but the answer I got was

A. 24 g

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Fe(s) + CuSO4(aq) ⇒ Cu(s) + FeSO4(aq)

Alexxandr [17]

Answer:

The answer to your question is CuSO₄

Explanation:

To answer your question just remember the following information

- A chemical reaction is divided into sections

reactants and products

reactants on the left side of the reaction

products on the right side of the reaction

- All the symbols have a meaning

(s) means that that compound is in solid phase

(aq) means that that compound is dissolved in solution.

Then, the answer is CuSO₄

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2 years ago

Which buffer would be better able to hold a steady pH on the addition of strong acid, buffer 1 or buffer 2? Explain. Buffer 1: a

Talja [164]

Answer:

Buffer 1.

Explanation:

Ammonia is a weak base. It acts like a Bronsted-Lowry Base when it reacts with hydrogen ions.

$\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)$ .

$\rm NH_3$ gains one hydrogen ion to produce the ammonium ion $\rm {NH_4}^{+}$ . In other words, $\rm {NH_4}^{+}$ is the conjugate acid of the weak base $\rm NH_3$ .

Both buffer 1 and 2 include

the weak base ammonia $\rm NH_3$ , and
the conjugate acid of the weak base $\rm {NH_4}^{+}$ .

The ammonia $\rm NH_3$ in the solution will react with hydrogen ions as they are added to the solution:

$\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)$ .

There are more $\rm NH_3$ in the buffer 1 than in buffer 2. It will take more strong acid to react with the majority of $\rm NH_3$ in the solution. Conversely, the pH of buffer 1 will be more steady than that in buffer 2 when the same amount of acid has been added.

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2 years ago

What kind of glacier has pieces that breaks off as icebergs

forsale [732]

There is no specific name for a glacier that break off as an iceberg. However, the part of the glacier in which this happens is called the "zone of wastage". Chunks break off in a process called "calving".

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2 years ago

The equations given in the problem introduction can be added together to give the following reaction: overall: N2O4→2NO2 However

Murljashka [212]

Answer:

Reverse the $2 NO_2 \longrightarrow 2 NO + O_2$ reaction

Explanation:

Reactions:

$2 NO_2 \longrightarrow 2 NO + O_2$

$N_2O_4 \longrightarrow 2 NO + O_2$

Overall:

$N_2O_4 \longrightarrow 2 NO_2$

As can be seen, in the overall reaction we have $N_2O_4$ in the reactants like in the second reaction and $NO_2$ in the products. The $NO_2$ is in the first reaction but as a reactant so we need to reverse that reaction: