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3 years ago

What happened during the Renaissance that encouraged progress in the study of matter?

1 answer:

5 0

During the dark ages-All that wasn't around the christian religion were lost. Knowledge and discoveries before the Dark Ages were mainly in Greek ideas. Biblic scripts however were in Latin. Therefore all inventions up to the dark ages were lost and only religious scripts preserved in the churches. Outside the church, no one knew how to write, read or even communicate.

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When formic acid is heated, it decomposes to hydrogen and carbon dioxide in a first-order decay: HCOOH(g) →CO2(g) + H2 (g) The r

juin [17]

Answer : The formic acid pressure is, 99 torr

Explanation :

The given chemical reaction is:

$HCOOH(g)\rightarrow CO_2(g)+H_2(g)$

Initial pressure a 0 0

At time 't' (a-x) x x

According to the Dalton's law,

$P_{Total}=P_{HCOOH}+P_{CO_2}+P_{H_2}$

$P_{Total}=(a-x)+x+x=a+x$ .........(1)

As we are given that:

Initial pressure = a = 220 torr

$P_{Total}=319torr$

Now put the value of 'a' in equation 1, we get:

$P_{Total}=a+x$

$319torr=220torr+x$

$x=99torr$

Thus, the formic acid pressure is, 99 torr

3 0

3 years ago

The gas usually filled in the electric bulb is? 1.nitrogen 2.hydrogen 3.carbon dioxide 4.oxygen

nlexa [21]

Argon and nitrogen are usually filled in an electric bulb, so one is your answer.

3 0

3 years ago

What is the [H+] concentration of a solution with a pH of 3.78

lianna [129]

Answer:

[H+] = 1.66 x $10^{-4}$

Explanation:

To find the [H+] concentration of a solution, we can use the formula:

$[H+] = 10^{-pH}$

Let's plug in the pH.

$[H+] = 10^{-3.78}$

Evaluate the exponent.

[H+] = 1.66 x $10^{-4}$

Hope this helps!

6 0

3 years ago

Can you please find the answer key for this from internet?

aev [14]

Found it http://my.ccsd.net/userdocs/documents/uGKpi9UAEwFpSIgE.pdf

5 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.

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

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

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