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MariettaO [177]

3 years ago

12

How are cellular respiration and photosynthesis different? (4 points)

1 answer:

avanturin [10]3 years ago

3 0

Answer b, cellular and photosynthesis are reversible

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Is it possible to replace our dependency on fossil fuels with alternative renewable energy's?

77julia77 [94]

Answer:

yes, coal for example produces energy by burning but there are several alternatives such as water or solar which are 2 growing energy sources.

Explanation:

hope this helps!

5 0

2 years ago

How to set up the rate expressions for the following mechanism?

Anestetic [448]

Answer:

Explanation:

From the given information:

A → B k₁

B → A k₂

B + C → D k₃

The rate law = $\dfrac{d[D]}{dt}=k_3[B][C] --- (1)$

$\dfrac{d[B]}{dt}=k[A] -k_2[B] -k_3[B][C]$

Using steady-state approximation;

$\dfrac{d[B]}{dt}=0$

$k_1[A]-k_2[B]-k_3[B][C] = 0$

$[B] = \dfrac{k_1[A]}{k_2+k_3[C]}$

From equation (1), we have:

$\mathbf{\dfrac{d[D]}{dt}= \dfrac{k_3k_1[A][C]}{k_2+k_3[C]}}$

when the pressure is high;

k₂ << k₃[C]

$\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_3[C]}= k_1A \ \ \text{first order}$

k₂ >> k₃[C]

$\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_2}= \dfrac{k_1k_3}{k_2}[A][C] \ \ \text{second order}$

3 0

3 years ago

What term is used for the electrons in the outermost shell

son4ous [18]

Answer:

valence electrons

Explanation:

4 0

2 years ago

Read 2 more answers

What are 3 permeable materials

gizmo_the_mogwai [7]

Permeable materials for what

4 0

4 years ago

Read 2 more answers

Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.

STALIN [3.7K]

Answer: Solution A : $[H_3O^+]=0.300\times 10^{-7}M$

Solution B : $[OH^-]=0.107\times 10^{-5}M$

Solution C : $[OH^-]=0.177\times 10^{-10}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

$pH=-\log[H_3O^+]$

$pOH=-log[OH^-}$

$pH+pOH=14$

$[H_3O^+][OH^-]=10^{-14}$

a. Solution A: $[OH^-]=3.33\times 10^{-7}M$

$[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M$

b. Solution B : $[H_3O^+]=9.33\times 10^{-9}M$

$[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M$

c. Solution C : $[H_3O^+]=5.65\times 10^{-4}M$

$[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M$

7 0

3 years ago