The ER takes up a lot of space in some cells<span>. The endoplasmic reticulum may be “rough” or “smooth.” ER that has no attached ribosomes is called smooth endoplasmic reticulum. </span>
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
This
can be solved using Dalton's Law of Partial pressures. This law states that the
total pressure exerted by a gas mixture is equal to the sum of the partial
pressure of each gas in the mixture as if it exist alone in a container. In
order to solve, we need the partial pressures of the gases given. Calculations
are as follows:<span>
P = P1 + P2 + P3
<span>P = (0.08206 atm.L/mol.K)( 298.15
K)/1.50) x (0.158 mole + 0.09 mol + 0.044 mol) = <span>4.76
atm</span></span></span>
How many protons are in am element
Answer:
Mass = 47.04 g
Volume = 23.94 L
Solution:
The equation for given reaction is as follow,
BaCO₃ + 2 HNO₃ → Ba(NO₃)₂ + CO₂ + H₂O
According to this equation,
197.34 g (1 mole) BaCO₃ produces = 44 g (1 mole) of CO₂
So,
211 g of BaCO₃ will produce = X g of CO₂
Solving for X,
X = (211 g × 44 g) ÷ 197.34 g
X = 47.04 g of CO₂
As we know,
44 g (1 mole) CO₂ at STP occupies = 22.4 L volume
So,
47.04 g of CO₂ will occupy = X L of Volume
Solving for X,
X = (47.04 g × 22.4 L) ÷ 44 g
X = 23.94 L Volume
