All categories

3 years ago

The leaves of plants contain specialized parts that absorb energy from sunlight. Which feature would help a plant

Chemistry

2 answers:

denis-greek [22]3 years ago

8 0

Answer:

larger overall leaf size

Explanation:

Ber [7]3 years ago

6 0

Chlorophyll pigment found in chloroplast of leaves absorbs sunlight.

The plants has a technique in which they lean towards sunlight which helps them better to compete for sunlight.

Explanation:

The leaves of the plant consist of chloroplasts in which light absorbing pigment chlorophyll is found. The chloroplast consist of thylakoid where light reaction of photosynthesis takes place.

The presence of chlorophyll a in large amount would help plant better use the sunlight this amounts to increase number of chloroplast.

Leaning towards the sunlight is feature which helps them using more sunlight.

You might be interested in

A freight train rushed down the track going south at 60 mph. This is an example of ____.

dalvyx [7]

This Question Would Be "b"

4 0

3 years ago

Anthony made a pitfall trap by burying a small glass in the ground so that the top of the glass was exactly level with the groun

torisob [31]

Answer:

Spiders and ground beetles.

Why is Anthony so mean? Don't hurt the beetles!

4 0

3 years ago

A body-centered cubic unit cell has an edge length of 588 pm. What is the radius of the atoms composing the unit cell?

blagie [28]

Answer:

hi follow me it well help

5 0

3 years ago

Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

5 0

3 years ago

Write the simple equation that describes the action of the enzyme catalase.

Nataly [62]

Here we have to write a simple equation which describes the action of the enzyme catalase.

The equation is: The concentration of the complex [ES] = $\frac{[E]0}{1+\frac{Km}{[S]} }$

Let us consider an enzyme catalyses reaction E + S ⇄ ES → E + P

Where E, S, ES and P are enzyme, substrate, complex and product respectively.

The concentration of the complex [ES] = $\frac{[E]0}{1+\frac{Km}{[S]} }$ , where $K_{m}$ is the Michaelis constant.

[E]₀ and [S] is the initial concentration of enzyme and concentration of substrate respectively.

5 0

3 years ago