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3 years ago

A sample of krypton gas with a volume of 10.00 L has a temp of 303 K and exerts

Chemistry

1 answer:

marshall27 [118]3 years ago

7 0

Answer: The final volume will be 14.85 L.

Explanation:

To calculate the final volume when temperature increases, we use Charles' Law.

This law states that volume is directly proportional to the temperature of the gas if number of moles and pressure remains constant.

$V\propto T\\\\\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ = Initial volume and temperature

$V_2\text{ and }T_2$ = Final volume and temperature

We are given:

$V_1=10L\\
T_1=303K\\
V_2=?L\\
T_2=450K$

Putting values in above equation, we get:

$\frac{10L}{303K}=\frac{V_2}{450K}$

$V_2=14.85L$

Hence, the final volume of the gas is 14.85L

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A wave is a disturbance that occurs along a medium which transmits energy. Now we now that waves travel from place to place. The horizontal distance that is travelled by a wave is what we call the wavelength of the wave.

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1 year ago

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2 years ago

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By the reaction of carbon & oxygen , a mixture of CO &CO2 is obtained. What is the composition by mass of the mixture ob

Hatshy [7]

M(O₂)=20g
M(O₂)=32.0 g/mol
n(O₂)=20/32.0=0.625 mol

m(C)=12 g
M(C)=12.0 g/mol
n(C)=12/12.0=1.0 mol

2C + O₂ → 2CO
1 mol 0.625 mol 1 mol
0.625-0.5=0.125 mol

2CO + O₂ → 2CO₂
0.250 mol 0.125 mol 0.250 mol

n(CO)=1 mol - 0.250 mol = 0.750 mol
M(CO)=28.0 g/mol
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3 years ago

Identify which of the concentration expressions can also be used to describe a solution with a concentration of 1 mg/mL of solut

Alexxandr [17]

Answer: Parts per million (ppm)

Explanation:

Consider the units milligram per milliliter. This gives us one part of the solute per one million parts of solvent. That is 10^ -3/10^-3= 10^-6. This unit is commonly used in analytical chemistry to show very small concentration of analyte. A similar unit is parts per billion(ppb)

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3 years ago

Balance each of the following equations according to the half- reaction method:

lukranit [14]

Answer : The balanced chemical equation in a acidic solution are,

(a) $Sn^{2+}+2Cu^{2+}\rightarrow Sn^{4+}+2Cu^+$

(b) $H_2S+Hg_2^{2+}\rightarrow 2Hg+S+2H^+$

(c) $5CN^-+2ClO_2+H_2O\rightarrow 5CNO^-+2Cl^-+2H^+$

(d) $Fe^{2+}+Ce^{4+}\rightarrow Fe^{3+}+Ce^{3+}$

(e) $2HBrO\rightarrow 3Br^-+2Br+2O_2+H_2O+3H^+$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(a) The given chemical reaction is,

$Sn^{2+}+Cu^{2+}\rightarrow Sn^{4+}+Cu^+$

The oxidation-reduction half reaction will be :

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}+2e^-$

Reduction : $Cu^{2+}+1e^-\rightarrow Cu^+$

In order to balance the electrons, we multiply the reduction reaction by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation will be,

$Sn^{2+}+2Cu^{2+}\rightarrow Sn^{4+}+2Cu^+$

(b) The given chemical reaction is,

$H_2S+Hg_2^{2+}\rightarrow Hg+S$

The oxidation-reduction half reaction will be :

Oxidation : $H_2S\rightarrow S+2H^++2e^-$

Reduction : $Hg_2^{2+}+2e^-\rightarrow 2Hg$

The electrons in oxidation and reduction reaction are same. Now add both the equation, we get the balanced redox reaction.

The balanced chemical equation in a acidic solution will be,

$H_2S+Hg_2^{2+}\rightarrow 2Hg+S+2H^+$

(c) The given chemical reaction is,

$CN^-+ClO_2\rightarrow CNO^-+Cl^-$

The oxidation-reduction half reaction will be :

Oxidation : $CN^-+H_2O\rightarrow CNO^-+2H^++2e^-$

Reduction : $ClO_2+4H^++5e^-\rightarrow Cl^-+2H_2O$

In order to balance the electrons, we multiply the oxidation reaction by 5 and reduction reaction by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a acidic solution will be,

$5CN^-+2ClO_2+H_2O\rightarrow 5CNO^-+2Cl^-+2H^+$

(d) The given chemical reaction is,

$Fe^{2+}+Ce^{4+}\rightarrow Fe^{3+}+Ce^{3+}$

The oxidation-reduction half reaction will be :

Oxidation : $Fe^{2+}\rightarrow Fe^{3+}+1e^-$

Reduction : $Ce^{4+}+1e^-\rightarrow Ce^{3+}$

The electrons in oxidation and reduction reaction are same. Now add both the equation, we get the balanced redox reaction.

The balanced chemical equation will be,

$Fe^{2+}+Ce^{4+}\rightarrow Fe^{3+}+Ce^{3+}$

(e) The given chemical reaction is,

$HBrO\rightarrow Br^-+O_2$

The oxidation-reduction half reaction will be :

Oxidation : $HBrO+H_2O\rightarrow O_2+Br+3H^++3e^-$

Reduction : $HBrO+H^++2e^-\rightarrow Br^-+H_2O$

In order to balance the electrons, we multiply the oxidation reaction by 2 and reduction reaction by 3 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a acidic solution will be,

$2HBrO\rightarrow 3Br^-+2Br+2O_2+H_2O+3H^+$

5 0

3 years ago