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SpyIntel [72]
3 years ago
15

Please answer asap MUCH NEEDED!!

Chemistry
1 answer:
gulaghasi [49]3 years ago
3 0

Answer:

I think its C I am sorry if I am wrong

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Which metal will react spontaneously with Cu2+ (aq) at 25°C?
Gwar [14]

Answer:

Mg  

Explanation:

The standard reduction potentials are

                                              <u>E°/V </u>

Au³⁺(aq ) + 3e⁻  ⟶  Au(s);     1.42

Hg²⁺(aq)  + 2e⁻  ⟶  Hg(l);     0.85

Ag⁺(aq)    +   e⁻  ⟶  Ag(s);    0.80

Cu²⁺(aq)   + 2e⁻ ⟶  Cu(s);   0.34

Mg2+(aq) + 2e- ⟶  Mg(s);   -2.38

The more negative the standard reduction potential, the stronger the metal is as a reducing agent.

Mg is the only metal with a standard reduction potential lower than that of Cu, so

Only Mg will react spontaneously with Cu²⁺.

 

5 0
3 years ago
Read 2 more answers
Calculate the ratio of diffusion rates for neon and helium.
Olin [163]
I cant that that :P xD
6 0
3 years ago
Please help with these two (last question is nwse)
Vlad1618 [11]

Answer:

your answer gonna be 20 miles

5 0
3 years ago
An irregularly-shaped piece of copper (Cu) has a mass of 55.0 grams. What is the volume in cm³ of this piece of copper if its de
vredina [299]

Answer:

<h2>6.14 cm³</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question we have

volume =  \frac{55}{8.96}  \\  = 6.138392...

We have the final answer as

<h3>6.14 cm³</h3>

Hope this helps you

8 0
3 years ago
If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?
Nata [24]

Answer:

The correct answer is option B.

Explanation:

3NaOH+2NBr_3\rightarrow 3HOBr+3NaBr+N_2

Moles of NBr_3 = 40 mol

Moles of NaOH = 48 mol

According to reaction, 3 moles of NaOH reacts with 2 moles NBr_3

Then ,48 moles of NaOH will reacts with:

\frac{2}{3}\times 48 mol=32 mol of NBr_3

Then ,40 moles of NaBr_3 will reacts with:

\frac{3}{2}\times 40 mol=60 mol of NaOH

As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.

Moles left after reaction = 40 mol - 32 mol = 8 mol

Hence, the NBr_3 is an excessive reagent.

6 0
3 years ago
Read 2 more answers
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