How does an electron emit light?

Evgesh-ka [11]

Cp stands for specific heat.

8 0

3 years ago

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. Calculate the pressure when the volume is 1.41 L and the

Vlad1618 [11]

A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

<h3>What is Combined Gas Law ?</h3>

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.

It is expressed as

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where,

P₁ = first pressure

P₂ = second pressure

V₁ = first volume

V₂ = second volume

T₁ = first temperature

T₂ = second temperature

Now put the values in above expression we get

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

$\frac{1.02\ atm \times 2.30\ L}{281\ K} = \frac{P_2 \times 1.41\ L}{298\ K}$

$P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}$

P₂ = 1.76 atm

Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

Learn more about the Combined gas Law here: brainly.com/question/13538773

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4 0

1 year ago

You need to prepare 150 mL of 0.1 M solution of silver chloride. How much silver chloride is required?

Rudiy27

Answer:

amount of silver chloride required is 0.015 moles or 2.1504 g

Explanation:

0.1M AgCL means 0.1mol/dm³ or 0.1mol/L

1L = 1000mL

if 0.1mol of AgCl is contained in 1000mL of solution

then x will be contained in 150mL of solution

cross multiply to find x

x = (0.1*150)/1000

x= 0.015 moles

moles of silver chloride present in 150 mL of solution is 0.15 moles

To convert this to grams, simply multiply this value by the molar mass of silver chloride

molar mass of silver chloride AgCl =107.86 + 35.5

=143.36 g/mol

mass of AgCl = moles *molar mass

=0.015*143.36

=2.1504g

=

4 0

3 years ago

Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --

frosja888 [35]

Answer :

(a) The anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

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

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

Thus, the anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) Now we have to calculate the potential of a cell.

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

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

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

Reduction half reaction (cathode): $2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}$

The overall cell reaction will be,

$2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}$

$E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V$

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

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

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

$E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=1.54V-(0.34V)=1.20V$

Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}$

where,

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

$E_{cell}$ = emf of the cell = ?

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

$E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}$

$E_{cell}=1.022V$

Therefore, the emf of cell potential is 1.022 V

5 0

3 years ago

5. You are investigating an arson scene and you find a corpse in the rubble, but you suspect that the victim did not die as a re

Marina CMI [18]

Answer:

See the answer below.

Explanation:

Fire has three major components:

Heat
Smoke
Gases ( in form of CO, CO2 etc)

If the victim had died as a result of the fire, he/he would have inhaled smoke and hot gases from the fire. These components would have resulted in traces of burns and soot deposition in the trachea and lungs as well as traces of CO in the blood of the victim.

If the analysis of the victim's corpse does not reflect some of the results above, it can be effectively concluded that the victim has been dead before the fire.

<em>The single most important indicator of death by the fire would be the presence of CO in the blood of the victim's corpse. All others might be to a less significant degrees.</em>

8 0

2 years ago