What is a similarity between the temperate climate zone and polar climate zone? (2 points)

What are the three end products of the oxidation-reduction reactions involved in metabolism?

san4es73 [151]

Answer: The three end products of the oxidation-reduction reactions involved in metabolism are WATER, CARBON DIOXIDE and ATP.

Metabolism refers to the sum of all chemical reactions occurring in the body. It includes both catabolism and anabolism. Every metabolic reaction is characterized by oxidation-reduction reactions. The metabolic pathways include respiration, photosynthesis, etc. In all these reactions water is produced. In respiration CO2 is also a product. In general. in every metabolic reaction either an organic molecule is broken or synthesized producing CO2, water and energy.

7 0

3 years ago

What are intermediates in a reaction? Give an example.

Andrews [41]

In a reaction, molecular entity that is formed from the reactants. For example, consider this hypothetical stepwise reaction: A + B → C + D.

8 0

3 years ago

g An oxidized silicon (111) wafer has an initial field oxide thickness of d0. Wet oxidation at 950 °C is then used to grow a thi

77julia77 [94]

Answer:

Explanation:

From the information given:

oxidation of oxidized solution takes place at 950° C

For wet oxidation:

The linear and parabolic coefficient can be computed as:

$\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big]$

Using $D_o$ and $E_a$ values obtained from the graph:

Thus;

$\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr$

$B= 386 \ exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\ = 0.1719 \ \mu m^2/hr$

So, the initial time required to grow oxidation is expressed as:

$t_{ox} = \dfrac{x}{B/A}+ \dfrac{x^2}{B} - t_o (initial)$

$where; \\ \\ t_{ox} = 2 \ hrs;\\ \\ x = 0.5 \\ \\ B/A = 0.2535 \\ \\ B = 0.1719$

∴

$2= \dfrac{0.5}{0.2535}+ \dfrac{0.5^2}{0.1719} - t_o (initial)$

$2 = 3.4267 - t_o (initial) \\ \\ t_o(initial) = 3.4267 - 2 \\ \\ t_o(initial) = 1.4267 \ hr$

NOW;

$1.4267 = \dfrac{d_o}{0.2535} + \dfrac{d_o^2}{0.1719} \\ \\ 1.4267 = 3.9448 \ d_o + 5.8173 \ d_o^2 \\ \\ d_o^2 + 0.6781 \ d_o = 0.2453$

$d_o = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

$d_o = \dfrac{-(0.6781) \pm \sqrt{(0.6781)^2-4(1)(-0.245)}}{2(10)}$

$d_o = \dfrac{-(0.6781) \pm \sqrt{0.45981961+0.98}}{20}$

$d_o = \dfrac{-(0.6781) \pm \sqrt{1.43981961}}{20}$

$d_o = \dfrac{-(0.6781) + \sqrt{1.43981961}}{20} \ OR \ \dfrac{-(0.6781) - \sqrt{1.43981961}}{20}$

$d_o =0.02609 \ OR \ -0.0939$

Thus; since we will consider the positive sign, the initial thickness $d_o$ is ;

≅ 0.261 μm

3 0

3 years ago

If dissolving carbon dioxide in water to form carbonic acid is the forward reaction, what is the reverse reaction that balances

kenny6666 [7]

Answer:

$H_2CO_3\rightleftharpoons H_2O + CO_2$

Explanation:

When carbon dioxide dissolves and reacts with water, the water and the gaseous $CO_2$ reacts to form a dilute mixture solution of $H_2CO_3$ (carbonic acid ).

The reaction is $H_2O + CO_2\rightleftharpoons H_2CO_3$

This is a forward reaction. And the symbol $\rightleftharpoons$ shows that the reaction can be reversible. It means that the reaction can be carried in forward direction as well as in the backward direction.

The reaction attains chemical equilibrium until the reactants and the products no longer changes with time.

The carbonic acid can also dissociates into carbon dioxide and water in the backward direction.

$H_2CO_3\rightleftharpoons H_2O + CO_2$

6 0

4 years ago

In order to simulate epipelagic zone conditions in a model aquarium design, what light and temperature conditions would be requi

Radda [10]

C)High amount of sunlight, cold temperature

3 0

3 years ago

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