All categories

3 years ago

Which compound will be the most soluble in water?

Chemistry

1 answer:

Rudiy273 years ago

4 0

Answer:

c) HOCH2CH2CH2OH

Explanation:

Hope this helps

You might be interested in

If you have a gold brick that is 2cm by 3cm by 4cm and has density of 19.3g/cm3, what is its mass?

natulia [17]

Volume = a x a x a

V = 2 cm x 3 cm x 4 cm => 24 cm³

Density = 19.3 g/cm³

Mass = ?

Therefore:

m = D x V

m = 19.3 x 24

m = 463.2 g

4 0

3 years ago

Which part of the flower produces seeds? (992 points)

Nata [24]

Answer:

The pistil, B.

Explanation:

The seeds are formed from the ovule which is contained in the ovary; all in the base, the pistil.

5 0

3 years ago

Which of the following elements has the largest atomic radius

Olin [163]

Answer: Calcium

Explanation:

not 100% sure but i looked it up and got calcium. 99% sure.

8 0

3 years ago

Read 2 more answers

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

Both hydrogen sulfide (H2S) and ammonia (NH3)

faust18 [17]

Answer: The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster.

Effusion rate is inversely proportional to molar mass.

NH3 will have a higher average molecule velocity, so it will diffuse faster and will reach the other side of the room more quickly.

Explanation: change up your response a bit

4 0

3 years ago

Read 2 more answers