Can someone help me with problem 21.105? Many thanks!

Please help! 35 points for a valid answer!

Furkat [3]

Answer:

Oxidation half reaction is written as follows when using using reduction potential chart

example when using copper it is written as follows

CU2+ +2e- --> c(s) +0.34v

oxidasation is the loos of electron hence copper oxidation potential is as follows

cu (s) --> CU2+ +2e -0.34v

Explanation:

6 0

3 years ago

Read 2 more answers

I am a nonmetal.I am in the Oxygen family and in row 3.I have 6 valence electrons.I am yellow and have a stinky smell.Who am i

Dimas [21]

Answer:

sulfur

Explanation:

In oxygen family sulfur has yellow color and also having stinky smell. Thus given statements are about sulfur.

It is present in oxygen family.

It has six valance electrons.

Its atomic number is 16.

Its atomic weight is 32 amu.

The electronic configuration of sulfur is given below,

S₁₆ = 1s² 2s² 2p⁶ 3s² 3p⁴

We can see the valance shell is third shell and it have six electrons thus sulfur have six valance electrons. (3s² 3p⁴ )

Sulfur is used in vulcanisation process.

It is used in bleach and also as a preservative for many food.

it is used to making gun powder.

8 0

2 years ago

balance the following reaction. a coefficient of "1" is understood. choose option "blank" for the correct answer if the coeffici

suter [353]

Balanced chemical reaction: 2KCl + Pb(NO₃)₂ → PbCl₂ + 2KNO₃.

According to principle of mass conservation, number of atoms must be equal on both side of balanced chemical reaction.

KCl is potassium chloride.

Pb(NO₃)₂ is lead(II) nitrate.

KNO₃ is potassium nitrate.

PbCl₂ is lead(II) chloride.

3 0

3 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

What kind of liquid promote corrosion

MaRussiya [10]

Is this the whole answer?

7 0

3 years ago