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anyanavicka [17]
3 years ago
11

An ideal gas in a cylindrical container of radius r and height h is kept at constant pressure p. The bottom of the container is

maintained at temperature T0 while the top at temperature T1. Assuming a linear temperature distribution along the cylinder, calculate the total mass of gas within the container.
Chemistry
1 answer:
Juli2301 [7.4K]3 years ago
8 0

Answer:

m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}  

Explanation:

The gas ideal law is  

PV= nRT (equation 1)

Where:

P = pressure  

R = gas constant  

T = temperature  

n= moles of substance  

V = volume  

Working with equation 1 we can get  

n =\frac{PV}{RT}

The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.

\frac{m}{mw} =\frac{PV}{RT}  or  

m =\frac{P*V*mw}{R*T}   (equation 2)

The cylindrical container has a constant pressure p  

The volume is the volume of a cylinder this is

V =(pi)*r^{2}*h

Where:

r = radius  

h = height  

(pi) = number pi (3.1415)

This cylinder has a radius, r and height, h so the volume is  V =(pi)*r^{2}*h

Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:  

T =\frac{T_{1} + T_{O}}{2}  

Replacing these values in the equation 2 we get:

m =\frac{P*V*mw}{R*T}   (equation 2)

m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}    

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Consider the following reaction where Kc = 1.80×10-2 at 698 K:2 HI (g) H2 (g) + I2 (g)A reaction mixture was found to contain 0.
Iteru [2.4K]

Answer:

Answer for the given statements: (1) T , (2) F , (3) T , (4) F , (5) F

Explanation:

At the given interval, concentration of HI = \frac{0.304}{1}M=0.304M

Concentration of H_{2} = \frac{5.07\times 10^{-2}}{1}M=5.07\times 10^{-2}M

Concentration of I_{2} = \frac{4.57\times 10^{-2}}{1}M=4.57\times 10^{-2}M

Reaction quotient,Q_{c} , for this reaction = \frac{[H_{2}][I_{2}]}{[HI]^{2}}

species inside third bracket represents concentrations at the given interval.

So, Q_{c}=\frac{(5.07\times 10^{-2})\times (4.57\times 10^{-2})}{(0.304)^{2}}=2.51\times 10^{-2}

So, the reaction is not at equilibrium.

As Q_{c}> K_{c} therefore reaction must run in reverse direction to reduce Q_{c} and make it equal to K_{c}. That means HI(g) must be produced and H_{2} must be consumed.

3 0
2 years ago
For the reaction ? Fe+? H2o ⇀↽? Fe3o4+? H2 , a maximum of how many grams of fe3o4 could be formed from 354 g of fe and 839 g of
Evgesh-ka [11]

The given reaction is:

3Fe + 4H2O → Fe3O4 + 4H2

Given:

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Mass of H2O = 839 g

Calculation:

Step 1 : Find the limiting reagent

Molar mass of Fe = 56 g/mol

Molar mass of H2O = 18 g/mol

# moles of Fe = mass of Fe/molar mass Fe  = 354/56 = 6.321 moles

# moles of H2O = mass of h2O/molar mass of H2O = 839/18 = 46.611 moles

Since moles of Fe is less than H2O;  Fe is the limiting reagent.

Step 2: Calculate moles of Fe3O4 formed

As per reaction stoichiometry:

3 moles of Fe form 1 mole of Fe3O4

Therefore, 6.321 moles of Fe = 6.321 * 1/ 3 = 2.107 moles of Fe3O4

Step 4: calculate the mass of Fe3O4 formed

Molar mass of Fe3O4 = 232 g/mol

# moles = 2.107 moles

Mass of Fe3O4 = moles * molar mass

= 2.107 moles * 232 g/mol = 488.8 g (489 g approx)

 


7 0
3 years ago
Read 2 more answers
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Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

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0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

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Answer:

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Explanation:

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