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3 years ago

IF YOU CAN HELP ME THAT WOULD BE GREAT

1 answer:

4 0

Answer:
I. Changing the pressure:
Increasing the pressure: the amount of H₂S(g) will increase.
Decreasing the pressure: the amount of H₂S(g) will decrease.
II. Changing the temperature:
Increasing the temperature: the amount of H₂S(g) will decrease.
Decreasing the temperature: the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
Increasing the H₂ concentration: the amount of H₂S(g) will increase.
Decreasing the H₂ concentration: the amount of H₂S(g) will decrease.
Explanation:
Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
I. Changing the pressure:
When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
For the reaction: CH₄(g) + 2H₂S(g) ⇄ CS₂(g) + 4H₂(g),
The reactants side (left) has 3.0 moles of gases and the products side (right) has 5.0 moles of gases.
Increasing the pressure: will shift the reaction to the side with lower moles of gas (left side), amount of H₂S(g) will increase.
Decreasing the pressure: will shift the reaction to the side with lower moles of gas (right side), amount of H₂S(g) will decrease.
II. Changing the temperature
The reaction is endothermic since the sign of ΔH is positive.
So the reaction can be represented as:
CH₄(g) + 2H₂S(g) + heat ⇄ CS₂(g) + 4H₂(g).
Increasing the temperature:
The T is a part of the reactants, increasing the T increases the amount of the reactants. So, the reaction will be shifted to the right to suppress the effect of increasing T and the amount of H₂S(g) will decrease.
Decreasing the temperature:
The T is a part of the reactants, increasing the T decreases the amount of the reactants. So, the reaction will be shifted to the left to suppress the effect of decreasing T and the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
H₂ is a part of the products.
Increasing the H₂ concentration:
H₂ is a part of the products, increasing H₂ increases the amount of the products. So, the reaction will be shifted to the left to suppress the effect of increasing H₂ and the amount of H₂S(g) will increase.
Decreasing the H₂ concentration:
H₂ is a part of the products, decreasing H₂ decreases the amount of the products. So, the reaction will be shifted to the right to suppress the effect of decreasing H₂ and the amount of H₂S(g) will decrease.

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Explanation:

The work is to calculate the molar mass of the solute (adrenaline) from the elevation of the boling point and compare with the mass of the structural formula. If they both are reasonably equal then you conclude that the results are in agreement, else they are not in agreement.

Since, you did not include the structural formula, I can explain the whole procedure to calculate the molar mass from the boiling point elevation, and then you can compare with the mass of the structural formula that only you know.

Determination of the molar mass from the boiling point elevation.

1) With the elevation of the boiling point, which is a colligative property, you can find the molality of the solution, using the formula:

ΔTb = i * Kb * m

Where:

> ΔTb is the increase of the boiling point of ths solvent, i.e. CCl4.

> i van't Hoff constant = 1 (because the solute is non ionic)

> m is the molality of the solution

2) Clearing m you get:

m = ΔTb / Kb

Kb is a datum that you must find in a table of internet (since the statement does not include it).

I found Kb = 5.02 °C/m

Then, m = 0.49°C / (5.02°C/m) = 0.09760956 m

3) With m and the mass of solvent you find the number of moles of solute using the formula:

m = number of moles of solute / kg of solvent

You have the mass of the solvent = 36.0 g = 0.0360 kg, so you can solve for the number of moles of solute:

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4) With the nuimber of moles and the mass you find the molar mass:

molar mass = mass in grams / number of moles = 0.64 g / 0.0035139 moles = 182 g/mol <------------- this is the important result

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3 years ago

consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of

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Answer:

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

$\frac{44}{56}$

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14.4 g of CaO must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

Needed = 11.31 g

Available CO2 = 13.8 g (given)

So CO2 is in excess , hence CaO is the limiting reagent and product will produce from 14.4 g of CaO

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO will produce 1 mole pf CaCO3

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1 g of CaO produce =

$\frac{100}{56}$

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= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

Percent Yield =

$\frac{Actual\ yield}{Theoritical\ yield}\times 100$

$\frac{19.4}{25.71}\times 100$

= 75.44 %

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