1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Kisachek [45]
3 years ago
11

Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm8 at 27 ∘C. (R =

0.083 bar dm8 K−1 mol−1​
Chemistry
1 answer:
maxonik [38]3 years ago
7 0

Answer:

Total pressure =  56.77 bar

Explanation:

Given data:

Mass of dioxygen = 8 g

Mass of dihydrogen = 4 g

Volume of vessel = 1 dm³

Temperature = 27°C (27+273 = 300 K)

R = 0.083 bar.dm³ / mol.K

Total pressure = ?

Solution:

Number of moles of dioxygen:

Number of moles = mass/molar mass

Number of moles = 8 g/ 32 g/mol

Number of moles = 0.25 mol

Pressure of dioxygen:

PV = nRT

P = nRT/V

P = 0.25 mol × 0.083 bar.dm³ / mol.K × 300 K / 1 dm³

P = 6.97 bar.dm³  /1 dm³

P = 6.97 bar

Number of moles of dihydrogen:

Number of moles = mass/molar mass

Number of moles = 4 g/ 2 g/mol

Number of moles = 2 mol

Pressure of dihydrogen:

PV = nRT

P = nRT/V

P = 2 mol × 0.083 bar.dm³ / mol.K × 300 K / 1 dm³

P = 49.8 bar.dm³  /1 dm³

P = 49.8 bar

Total pressure of mixture in a vessel:

Total pressure = P (O₂) + P(H₂)

Total pressure =  6.97 bar + 49.8 bar

Total pressure =  56.77 bar

You might be interested in
Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha
elixir [45]

Answer:

0.269 g

Explanation:

Po_{210} ^{84}  ⟶  Pb_{206} ^{82}+  + He_{4} ^{2}

Data:

Half-life of  Po_{210} ^{84}  (T(1/2)) = 138.4 days

Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol  

Time = 338.8 days  

Isotopic masses  

Po_{210} ^{84} = 209.98 g/mol  

Pb_{206} ^{82} = 205.97 g/mol  

Concepts  

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.  

The radioactive decay law is  

N=Noe^(-λt)

Where:  

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days  

λ= radioactive decay constant  

The radioactive constant is related to the half-life by the next equation  

λ= \frac{ln 2}{t(1/2)}

so  

λ= \frac{ln2}{138.4 days}  =0,005008 days^(-1)

No (Atoms of  Po_{210} ^{84}  in t=0)

To get No we need to calculate the number of atoms of  Po_{210} ^{84}   in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of  Po_{210} ^{84}  in the initial sample.  

This is:

\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol} = 0,001599 mol of  PoCl4

This is the number of mol of  Po_{210} ^{84} in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23  

0,001599 mol of  Po_{210} ^{84} * 6.023*10^23 atoms/ mol of  Po_{210} ^{84} = 9.632 *10^20 atoms of  Po_{210} ^{84}

Atoms after 338.8 days

We use the radioactive decay law to get this value  

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of  Po_{210} ^{84} in the sample after 338.8 days has passed  

The number of atoms  Po_{210} ^{84} transformed is equal to the number of atoms of Pb_{206} ^{82}  produced.  

The number of atoms of Po_{210} ^{84} transformed is No - N  

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of Pb_{206} ^{82} produced  

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of Pb_{206} ^{82} ) / ( 6.023*10^23 atoms of Pb_{206} ^{82} / mol of Pb_{206} ^{82} =  0.001306 moles of Pb_{206} ^{82}

This number of moles have a mass of:

(0,001306 moles of Pb_{206} ^{82} )* (205.97 g of Pb_{206} ^{82} /mol of Pb_{206} ^{82} ) = 0.269 g  

3 0
3 years ago
Which two systems first respond to burning your finger?
Bess [88]

Answer:

B. is my guess

I'm sorry if I'm wrong

God bless :)

7 0
3 years ago
Read 2 more answers
1. Write the structural formulas and give IUPAC names for all isorneric alcohols of molecular formula
EastWind [94]

Explanation:

Butanol (С4Н9OH)

Structural formula is :-

CH3-CH2-CH2-CH2-OH

6 0
3 years ago
Covalent Bonds <br><br> How many covalent bonds can Nitrogen atoms make ?
notsponge [240]
A nitrogen atoms can make 3 covalent bonds because it has three unpaired electrons

6 0
3 years ago
Read 2 more answers
Use electron configurations to account for the stability of the lanthanide ions Ce⁴⁺ and Eu²⁺.
Citrus2011 [14]

Answer:

Explanation:

 The Ce metal has electronic configuration as follows

[Xe] 4f¹5d¹6s²

After losing 4 electrons , it gains noble gas configuration ,. So Ce ⁺⁴ is stable.

Eu  has electronic configuration as follows

[ Xe ] 4 f ⁷6s²

[ Xe ] 4 f ⁷

Its outermost orbit contains 2 electrons so  Eu²⁺ is stable. Its +3 oxidation state is also stable.

Ce⁺²

7 0
3 years ago
Other questions:
  • What is the difference between air in high pressure and low pressure areas
    7·1 answer
  • Complementary angles<br> supplementary angles<br> right angles<br> vertical angles
    10·2 answers
  • When making a fire, without gas, matches, or a lighter, what are two important things about the wood?
    10·2 answers
  • What is always true of an object with a lot of mass?
    5·1 answer
  • How to find the value of the equilibrium
    6·1 answer
  • Does a negative exponent mean that the number is less than 1​
    10·1 answer
  • Define pathogens.(1 point)
    6·1 answer
  • A gas has a volume of 4 liters at 50 oC. What will its volume be (in liters) at 100oC
    12·1 answer
  • Which of the following particles is the smallest in radius
    15·2 answers
  • A 50.0 g sample of liquid water at 0.0°c ends up as ice at -20.0°c. How much energy is involved in this change.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!