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3 years ago

A plant that has a high amount of orange pigment carotene would have leaves what color?

Chemistry

1 answer:

coldgirl [10]3 years ago

7 0

The orange pigment carotene absorbs the orange wavelength of light very poorly, which makes the leaves appear orange.

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C. Another radioisotope of americium exists which has an atomic mass of 242.

ollegr [7]

Answer:

1 g

Explanation:

The half-life of Am-242 (16 h) is the time it takes for half of it to disappear.

We can make a table of the mass left after each half-life.

$\begin{array}{cccc}\textbf{No. of} & & \textbf{Percent} & \textbf{Mass}\\\textbf{Half-lives} & \textbf{Time/h} & \textbf{Remaining} & \textbf{Remaining/g}\\0& 0 & 100 & 8\\1 & 16 &50 & 4\\2 & 32 & 25 & 2\\3 & 48 & 12.5 & 1\\4 & 64 & 6.25 & 0.5\\\end{array}$

The mass remaining after 48 h is 1 g.

7 0

3 years ago

What is the predominant intermolecular force in the liquid state of each of these compounds: ammonia (NH3), methane (CH4), and n

MAVERICK [17]

Answer:

The predominant intermolecular force in the liquid state of each of these compounds:

ammonia (NH3)

methane (CH4)

and nitrogen trifluoride (NF3)

Explanation:

The types of intermolecular forces:

1.Hydrogen bonding: It is a weak electrostatic force of attraction that exists between the hydrogen atom and a highly electronegative atom like N,O,F.

2.Dipole-dipole interactions: They exist between the oppositely charged dipoles in a polar covalent molecule.

3. London dispersion forces exist between all the atoms and molecules.

NH3 ammonia consists of intermolecular H-bonding.

Methane has London dispersion forces.

Because both carbon and hydrogen has almost similar electronegativity values.

NF3 has dipole-dipole interactions due to the electronegativity variations between nitrogen and fluorine.

3 0

3 years ago

The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal

spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0

3 years ago

Which statement about enzymes is true

artcher [175]

The lower the activation energy required for a chemical reaction

3 0

3 years ago

There are four different starting molecules that one might use to synthesize the illustrated alkyl halide as the major product u

Mkey [24]

Answer:

Explanation:

An electrophilic addition reaction occurs when an electrophile attacks a substrate, with the end result being the inclusion of one or many comparatively straightforward molecules along with multiple bonds.

In the given question, the hydrogen bromide provides the electrophile while the bromide is the nucleophile. The mechanism proceeds with the attack of the electrophile on the carbon, followed by deprotonation. This process is continued with a formation of carbocation and the bromide(nucleophile) finally bonds to the carbocation to form a stable product.

The first diagram showcases the possible various starting molecules for the synthesis while the second diagram illustrates their mechanism.

6 0

3 years ago