Answer:
2CH2Cl2(g) Doublearrow CH4(g) + CCl4(g)
0.205 moles of CH2Cl2 is introduced. Let by the time an equilibrium is reached x moles each of CH4 and CCl4 are formed => remaining moles of CH2Cl2 are 0.205-x
i.e at equilibrium the concentration on CH2Cl2 is (0.205-2x) mol/L, CH4 is x mol/L, CCl4 is x mol/L
Now the equilibrium constant equation : K = [CH4][CCl4]/[CH2Cl2]^2 ([.] - stands for concentration of the term inside the bracket)
10.5 = x*x/(0.205-2x)^2
=> 10.5(4x^2-0.82x+0.042) = x^2
=>42x^2-8.61x+0.441=x^2
=>41x^2-8.61x+0.441 = 0
This is a Quadratic in x, solving for the roots, we get x = 0.0886 , x = 0.121
The second solution for x will lead 0.205-2x to become negative, so is an infeasible solution.
Therefore equilibrium concentrations of the products and reactants correspond to x=0.0886 and they are , [CH2Cl2] = 0.205-2*0.0886 =0.0278 mol/L , [CH4] = 0.0886 mol/L , [CCl4] = 0.0886 mol/L
In oxygen- starved muscle cells in anaerobic respiration lactic acid is formed while in anaerobic respiration by yeast cells it produces ethanol and carbon dioxide.
Explanation:
Respiration occurring in absence of oxygen in muscles leads to accumulation of lactic acid and very less amount of energy. This is fermentation process as the end product formed is lactic acid.
The process of fermentation takes place in yeast bacteria and oxygen starved cells of muscles.
The yeast can ferment simple sugars like glucose and fructose into ethyl alcohol and carbon dioxide.
In general human respiration is aerobic but during strenuous exercise, oxygen does not meet the demand of working out muscle, so they respire anaerobically and produce lactic acid and 2 ATP i.e very less compared to 38 molecules of ATP in aerobic respiration.
Ether<span> has two Carbons (single) bonded to an Oxygen atom. An </span>Ester<span>has a double Oxygen bond with a Carbon atom</span>
Answer: 0.3794 moles of Iodine gas are produced.
Explanation:
Volume of iodine gas produced at STP =8.5 L
At STP, the 1 mol of gas occupies volume = 22.4 L
So, 8.5 L of volume will be occupied by:
0.3794 moles of Iodine gas are produced.
Answer:
THE NEW VOLUME OF THE OXYGEN GAS AT 28 PSI FROM 72.5 PSI IS 0.078 L.
Explanation:
Initial volume of the oxygen in the container = 30.0 mL = 30 / 000 L = 0.03 L
Initial pressure of the oxygen = 72.5 psi = 1 psi = 6890 pascal
Final pressure = 28 psi
Final volume = unknown
First convert the mL to L and since both pressures are in similar unit that is psi; there is no need converting them to pascal or other standard unit of pressure. They cancel each other out.
This question follows Boyle's equation of gas laws and mathematically it is written as:
P1 V1 = P2 V2
Re-arranging by making P2 the subject of the formula, we have:
V2 = P1 V1 / P2
V2 = 72,5 * 0.03 / 28
V2 = 2.175 /28
V2 = 0.0776 L
The new volume of the oxygen gas at a change in pressure from 72.5 psi to 28 psi is 0.078 L.
