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3 years ago

Golgi body function

Chemistry

1 answer:

olasank [31]3 years ago

5 0

Explanation:

GOLGI BODY - transportation of lipids and steroids.

LYSOSOME - destruction of malfunctioning or worn out cell organelles.

MITOCHONDRIA - oxides glucose to give energy to the cell in a form of cellular respiration.

VACUOLE - FOOD VACUOLE for storage of food while WATER VACUOLE eliminates excess water by a process called OSMOREGULATION.

CELL WALL - provides rigidity for the plant cell.

CHLOROPLAST - pigmented organelle of the plant which functions to manufacture food by a phenomenon known as PHOTOSYNTHESIS.

RIBOSOME - synthesizes protein and enzymes.

ENDOPLASMIC RETICULUM - The ROUGH ENDOPLASMIC RETICULUM has RIBOSOMES attached to it while the SMOOTH ENDOPLASMIC RETICULUM has nothing attached, they transport proteins and enzymes

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The equilibrium constant, Kp, for the reaction H2(g) + I2(g) → 2HI(g) is 55.2 at 425°C. A rigid cylinder at that temperature con

Fiesta28 [93]

Answer: No, the reverse reaction must proceed to establish equilibrium.

Explanation:

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{(p_{HI})^2}{p_{H_2}\times p_{I_2}}$

We are given:

$p_{HI}=1.055atm\\p_{H_2}=0.127atm\\p_{I_2}=0.134atm$

Putting values in above equation, we get:

$Q_p=\frac{(1.055)^2}{0.127\times 0.134}=65.41$

We are given:

$K_p=55.2$

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium.

As, $Q_p>K_p$ , the reaction will be favoring reactant side or the reaction must proceed in the reverse direction.

Hence, no, the reverse reaction must proceed to establish equilibrium.

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