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Ivanshal [37]
2 years ago
6

At 4.00 L, an expandable vessel contains 0.864 mol of oxygen gas. How many liters of oxygen gas must be added at constant temper

ature and pressure if you need a total of 1.36 mol of oxygen gas in the vessel? Express the volume to three significant figures and include the appropriate units.
Chemistry
1 answer:
fgiga [73]2 years ago
4 0

Answer:

We have to add 2.30 L of oxygen gas

Explanation:

Step 1: Data given

Initial volume = 4.00 L

Number of moles oxygen gas= 0.864 moles

Temperature = constant

Number of moles of oxygen gas increased to 1.36 moles

Step 2: Calculate new volume

V1/n1 = V2/n2

⇒V1 = the initial volume of the vessel = 4.00 L

⇒n1 = the initial number of moles oxygen gas = 0.864 moles

⇒V2 = the nex volume of the vessel

⇒n2 = the increased number of moles oxygen gas = 1.36 moles

4.00L / 0.864 moles = V2 / 1.36 moles

V2 = 6.30 L

The new volume is 6.30 L

Step 3: Calculate the amount of oxygen gas we have to add

6.30 - 4.00 = 2.30 L

We have to add 2.30 L of oxygen gas

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If your lungs were filled with air containing this level of lead, how many lead atoms would be in your lungs? (Assume a total lu
kolbaska11 [484]

Answer:

1.505×10^23 atoms of lead

Explanation:

Volume of lead in the lungs = total volume of lungs = 5.60L

1 mole = 22.4L

5.6L of lead = 5.6/22.4 = 0.25 mole

From Avogadro's law

1 mole of lead contains 6.02×10^23 atoms of lead

0.25 mole of lead = 0.25×6.02×10^23 = 1.505×10^23 atoms of lead

6 0
2 years ago
It takes energy to ionize any atom. It takes progressively more and more energy for each successive electron that is removed and
IrinaVladis [17]

Answer:

Removal of Third Electron

Explanation:

a major jump is required to remove the third electron. In general, successive ionization energies always increase because each subsequent electron is being pulled away from an increasingly more positive ion.

Ionization energy increases from bottom to top within a group, and increases from left to right within a period.

5 0
2 years ago
Which statement correctly describes a sample of gas confined in a sealed container? (1) It always has a definite volume, and it
Luda [366]
A gas does not have a specific shape or volume (so we reject option A), instead it adjusts itself to the container (which is further influenced by other forces such as gravity and temperature) and it with time, willl fill the whole container evenly, so the correct answer is:
(2) It takes the shape and the volume of any container in which it is confined
5 0
2 years ago
Which of the following mixtures, with each component present at a concentration of 0.1 M, has a pH closest to 7? O A. HCIO(aq) a
eduard

for HClO, pKa = 7.54

for HNO_2, pKa = 3.15                                    

for CH3 COOH, pKa = 4.74    

Explanation:

The concentration of the solution given is 0.1 M has a pH closest to 7

The mixtures are weak acids and their salts except

HNO_3 and NaNO_3                                         pH = pH is near to '1'

for buffers( acidic)                                               pH = pKa + log [salt] / [acid]

therefore [salt] = [acid] = 0.1                               pH = pKa + log 0.1 / 0.1 = pKa

                                                     pH = pKa

for HClO, pKa = 7.54

for HNO_2, pKa = 3.15                                     therefore HClO and NaclO

                                                                         mixture hs a pH closest to '7'

for CH3 COOH, pKa = 4.74                

5 0
2 years ago
Read 2 more answers
A cell consists of a gold wire and a saturated calomel electrode (S.C.E.) in a 0.150 M AuNO 3 solution at 25 °C. The gold wire i
almond37 [142]

The half reaction that occurs at the Au electrode is  1.64

<u>Explanation:</u>

Half cell reaction at 'Au' electrode

We have the equation,

Au + (aq) + e-  ---> Au(s)

Given the concenration of AuNO3=0.150 M

[Au +] =0.150 m

From the equation,

Au + (aq) + e-  ---> Au(s)

Standard electric potential= Eo= 1.69 volt

Solving the problem using the Nerst equation

E cell= E0 cell - 2.303 RT/ nF  log Qc

Where,

T = 298 K

n= no of electron lost or gained

F= faraday's constant= 965000/mole

R= universal constant= 8.314 J/ K/ Mole

Substitue the values we get

E cell = 1.69 volt - 0.05 g/n log (1/0.150M)

E cell = 1.69 volt - 0.05 g/1  0.824

E cell= 1.64

The half reaction that occurs at the Au electrode is  1.64

<u />

6 0
2 years ago
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