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notka56 [123]
4 years ago
15

I need somebody to help me with my homework pls i dont get it .

Chemistry
1 answer:
zubka84 [21]4 years ago
7 0
26. B
27. D
28. C
Hope this helped ☺️
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Brainly is so boring like upgrade it to be better or something
Nikitich [7]

Answer:

ya same here i hope so btw

7 0
3 years ago
fills a 500.mL flask with 3.6atm of carbon monoxide gas and 1.2atm of water vapor. When the mixture has come to equilibrium she
enot [183]

Answer:

The answer to the question is

The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits

Explanation:

To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure

At the first trial the mixture contains

3.6 atm CO

1.2 atm H₂O (g)

Total pressure = 3.6+1.2= 4.8 atm

which gives

3.36 atm CO

0.96 atm H₂O (g)

0.24 atm H₂ (g)

That is

CO+H₂O→CO(g)+H₂ (g)

therefore the mixture contained

0.24 atm CO₂ and the total pressure =

3.36+0.96+0.24+0.24 = 4.8 atm

when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂

At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857

adding 1.8 atm CO gives 4.46 atm hence we have

 (0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857

which gives x = 0.031 atm or x = -0.6183 atm

Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm

7 0
3 years ago
What is the percent of Cr in Cr2O3
IgorLugansk [536]
The percentage of Chromium in Chromium Oxide is calculated as follow,

Step 1:
           Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
     2(51.99) + 3(16) = 103.98 + 48 = 151.98 u

Step 2:
          Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
                   = \frac{103.98}{151.98} × 100
                   
                   = 68.41 %
So, the %age composition of chromium in chromium oxide is 68.41 %.
5 0
3 years ago
Read 2 more answers
what is the molar mass of a gaseous flouride of sulfur containing 70.4% F and having a density of approximately 4.5g/L at 20 deg
zlopas [31]

Answer:

The molar mass is 180.2 g/mol

Explanation:

<u>Step 1:</u> Data given

% of F = 70.4 %

Density = 4.5 g/L

Temperature = 20 °C

Pressure = 1 atm

<u>Step 2:</u> Calculate the number of moles

PV = nRT

 ⇒ with P = the pressure = 1.00 atm

⇒ with V = the volume = Assume this is 1L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature : 20°C = 293 Kelvin

1 atm*1L= n(0.08206 L-atm/mol-K)*(293 K)

n = 0.04159 moles

<u>Step 3</u>: Calculate molar mass

Molar mass = Mass / moles

4.5 grams / 0.04159 moles = 108.2 g/mol

<u>Step 4:</u> Calculate moles of F

Moles = Mass / molar mass

Moles F = 70.4 g / 19 g/mol

Moles F =  3.70 moles

Moles S = 29.6g / 32.07 g/mol

Moles S = 0.923 moles S

<u>Step 5:</u> Divide by the smallest amount of moles

F = 3.70 / 0.923 = 4

S = 0.923 / 0.923 = 1

The empirical formula is SF4

The molar mass of SF4 = 32.07 + 4*19 = 108.07 g/mol

This means the empirical formula is the same as the molecular formula SF4

The molar mass is 180.2 g/mol

4 0
3 years ago
Which is the larger atom As or P
sashaice [31]

Answer:

I think that As is larger

6 0
3 years ago
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