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3 years ago

From the choices provided below, list the reagent(s) in order that will react with 3-pentanone to form 3-pentanol

Chemistry

1 answer:

Soloha48 [4]3 years ago

5 0

3-pentanone to form 3-pentanol

Here 3-pentanone reacts with H2/Pt to given 3-pentanol

So here with one step we can convert given ketone to alcohol

The reaction will be

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1.Explain why the reactivity of group 7 decreases as you move down the group. Try to use the sentence starters here: When group

Kruka [31]

Answer: The reactivity of group 7 decreases as we move down the group because:

Explanation:

The elements of group 7 that is fluorine to iodine. The halogens are non metals and they react with metals to gain electrons. The metals loose electrons and the non metal gains it.

As we move down the group the atomic radius gets bigger( more electron and more proton) and as a result the outer shells move further away from the nucleus.

There is more distance between the negatively charged electrons and positively charged nucleus.

Therefore the force of attraction between the shells and nucleus is lesser or weaker.

This makes attracting an extra electron from metals very difficult which results in weaker reaction.

Consequently, the reactivity decreases as we move down the group 7

5 0

2 years ago

What mass of Cu(s) is electroplated by running 24.5A of current through a Cu2+(aq)solution for 4.00 h?Express your answer to thr

Masja [62]

Answer: 116 g of copper

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 24.5A

t= time in seconds = 4.00 hr = $4.00\times 3600s=14400s$ (1hr=3600s)

$Q=24.5A\times 14400s=352800C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500C=193000C$ of electricity deposits 63.5 g of copper.

352800 C of electricity deposits = $\frac{63.5}{193000}\times 352800=116g$ of copper.

Thus 116 g of Cu(s) is electroplated by running 24.5A of current

Thus remaining in solution = (0.1-0.003)=0.097moles

8 0

2 years ago

Assume that silver and gold form ideal, random mixtures. Calculate the mass of pure Ag needed to cause an entropy increase of 20

KengaRu [80]

Answer:

$m_{Ag}=2,265.9g$

Explanation:

Hello!

In this case, since the definition of entropy in a random mixture is:

$\Delta S=-n_TR\Sigma[x_i*ln(x_i)]$

For this silver-gold mixture we write:

$\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]$