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3 years ago

Please help 15 points

Chemistry

1 answer:

Rus_ich [418]3 years ago

4 0

Nitrogen gained 4 electrons.

Because Nitrogen's redox number went from +6 to +2, it must have gained 4 electrons (-4) in order to achieve this number. Thus, Nitrogen is reduced.

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Which term describes this reaction? addition condensation elimination substitution

vlada-n [284]

The others are wrong it's B. Condensation

5 0

2 years ago

Read 2 more answers

When using 100ml or 50ml graduated cylinder to what decimal place can your volume be estimated?

Olegator [25]

Answer:

I know that the 100-mL graduated cylinders are always read to 1 decimal place.

I think for 50 mL graduated cylinders, it lets you measure volumes up to 50.0 mL to the nearest 0.1 or 0.2 mL, depending on your exact cylinder.

3 0

3 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

1. Of 100 animals on a farm, 20% are horses, 15 % are pigs, 30% are cows, and 35% are

ANEK [815]

B. a circle graph

circle graphs are the best to show percentages because they’re very easy to look at and get info from

7 0

3 years ago

How many liters are in 2.75 ounces? Use the conversion factor: 1 liter = 33.814 ounces Rounded to the result to 3 significant fi

Serggg [28]

1L = 33.814 oz
xL = 2.75 oz

so it's a proportion

1L / 33.814 oz = xL / 2.75

solve for x

(1/33.814) * 2.75 = 0.0813272609 on your calculator, but it's not the answer.

the number in your problem, 2.75 oz, has 3 significant figures. so you can only round this number to 3 significant figures too.

your equipment isn't accurate enough to give a reading to 10 significant figures if that makes sense. you have to give the answer in terms of the term you use with the lowest significant figures.

so with 3 significant figures,
0.0813272609 rounds to
0.0813 L

4 0

2 years ago