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labwork [276]
3 years ago
14

A flexible vessel contains 32.00 L of gas at a pressure of 1.59 atm. Under conditions of constant temperature and moles of gas,

what is the volume of the gas when the pressure of the vessel is decreased by a factor of three
Chemistry
1 answer:
HACTEHA [7]3 years ago
4 0

Answer:

When the pressure of the vessel is decreased by a factor of three, the volume will increase by a factor 3. This means the new volume is 96.00 L

Explanation:

Step 1: Data given

Volume of the gas = 32.00 L

Pressure of the gas = 1.59 atm

Temperature and moles of gas stay constant

the pressure of the vessel is decreased by a factor of three

Step 2: Calculate the new volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.59 atm

⇒with V1 = the initial volume = 32.00 L

⇒with P2 = the decreased pressure = 1.59 / 3 = 0.53 atm

⇒with V2 = the new volume = TO BE DETERMINED

1.59 atm * 32.00 L = 0.53 atm * V2

V2 = (1.59 * 32.00 L) / 0.53 atm

V2 = 96.00 L

When the pressure of the vessel is decreased by a factor of three, the volume will increase by a factor 3. This means the new volume is 96.00 L

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3 What is the number located in the top right hand corner of the atomic symbol of an ion with 16 protons, 17 neutrons and 14 ele
siniylev [52]

The number located in the top right corner of the ion is +2

An ion is an atom or group of atoms which possess an electric charge.

The charge on an ion can be obtained by using the following formula:

<h3>Charge = Proton – Electron </h3>

With the above formula, we can obtain the charge of the ion given in the question above as illustrated below:

Proton = 16

Neutron = 17

Electron = 14

<h3>Charge =? </h3>

Charge = Proton – Electron

Charge = 16 – 14

<h3>Charge = +2</h3>

Therefore, the charge on the ion is +2

Learn more: brainly.com/question/3428265

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3 years ago
A miner develops cancer of the esophagus. ten years later
Free_Kalibri [48]

Answer:

a miner or minor like young?

Explanation:

4 0
3 years ago
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KCl and KBr are both ionic solids. A mixture of KCl and KBr has a mass of 3.595 g. When this mixture is heated in the presence o
monitta

Answer:

The percentage (by mass) of KBr in the original mixture was 33.1%.

Explanation:

The mixture of KCl and KBr has a mass of 3.595g, thus the sum of the moles of KCl (<em>x</em>) multiplied by it molar mass (74.5g/mol) and the moles of KBr (<em>y</em>) multiplied by it molar mass (119g/mol) is the total mass of the mixture:

x.74.5g/mol + y.119g/mol = 3.595g

Also, after the conversion of KBr into KCl, the total mass of 3.129 g is only from KCl moles, hence

\frac{3.129g}{74.5g/mol} = 0.042 moles

But the 0.042 moles came from the originals KCl and KBr moles, thus

x + y = 0.042moles

Now it is possible to propose a system of equations:

x.74.5g/mol + y.119g/mol = 3.595g

x + y = 0.042moles

Solving the system of equations,

x=0.032moles\\y=0.010 moles

0.010 moles of KBr multiplied it molar mass is

0.010molesx119g/mol = 1.19g

Therefore, the percentage (by mass) of KBr in the original mixture was:

\frac{1.19g}{3.595g}x100% = 33.1%%

4 0
3 years ago
The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:
german

<u>Answer:</u>

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

<u>Explanation:</u>

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

We are given:

Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

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

  • <u>Cell having 1st and 2nd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-0.337=0.462V

  • <u>Cell having 1st and 3rd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.28)=1.079V

  • <u>Cell having 1st and 4th half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.74)=1.539V

  • <u>Cell having 2nd and 3rd half reactions:</u>

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.337-(-0.28)=0.617V

  • <u>Cell having 3rd and 4th half reactions:</u>

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=-0.28-(-0.74)=0.46V

Hence,

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0
3 years ago
Which of the following chemical formulas represents a total of 12 atoms of oxygen?
9966 [12]

Answer:

3Sn(s04)2

Explanation:bc it cant be anything else

4 0
3 years ago
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