Which element is likely to have the highest thermal conductivity?

Calculate the mass of ZnCl2 produced by the reaction of 49.8 grams of H2. Zn + 2HCl ZnCl2 + H2

tester [92]

Answer:

3300.85 g

Explanation:

Given data:

Mass of ZnCl₂ produced = ?

Mass of H₂ produced = 49.8 g

Solution:

Chemical equation:

Zn + 2HCl → ZnCl₂ + H₂

Number of moles of H₂:

Number of moles = mass/molar mass

Number of moles = 49.8 g/ 2.056 g/mol

Number of moles = 24.22 mol

Now we will compare the moles of H₂ with ZnCl₂ form balance chemical equation.

H₂ : ZnCl₂

1 : 1

24.22 : 24.22

Mass of ZnCl₂:

Mass = number of moles × molar mass

Mass = 24.22 × 136.286 g/mol

Mass = 3300.85 g

6 0

3 years ago

Please help I will mark Brainly!

IgorLugansk [536]

It's NF3 (one Nitrogen and 3 Flouride)

6 0

3 years ago

An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Pb\rightarrow Pb^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

$Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)$

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

6 0

3 years ago

A square chunk of plastic has a length of 5 cm, width of 5 cm and height of 5 cm. It has a mass of 200 g. What is it’s density

vichka [17]

Answer:

1.6g/mL

Explanation:

Density equation is D=m/v

Density = g/mL

m=mass of sample in grams

v = volume of sample in mL

The volume of a square can be calculated by V=l*w*h.

In this case it is 5cm*5cm*5cm = 125cm^3

Since we know that 1cm^3 ~ 1mL we can convert the volume to mL as so:

125cm^3 (1mL/(1cm^3)) = 125mL

Then simply plug into the density equation:

D=200g/125mL = 1.6g/mL

3 0

3 years ago

A water molecule consists of an oxygen atom and two hydrogen atoms covalently bonded in a bent shape. Oxygen however, attracts t

77julia77 [94]

Answer:

D.

Explanation:

Water is polar, for one thing. Polar mixes with polar, nonpolar mixes wih nonpolar. This leaves D.

6 0

3 years ago