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4 years ago

Select all that apply. what intermolecular forces stabilize a lipid bilayer?

1 answer:

4 0

Lipid Bilayer is composed of two simillar characterterized lipid layers.
The intermolecular forces that stabilize a lipid layer are:
-Hydrogen bonds
-Ion-dipole interactions
-Dispersion forces
-Salt links

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The ____________________ package and distribute materials to other parts of the cell.

Alex777 [14]

The Golgi, that’s the answer:)

7 0

3 years ago

Read 2 more answers

4. Consider the following statement made in Chapter 4 of the text: "... consider that if the

AveGali [126]

The claim: "If the nucleus were the size of a grape, the electrons would be one mile away on average" is reasonably accurate because the ratios between the nucleus's sizes and the distances (between electrons and nucleus) for the two given examples are in the same order of magnitude.

To know if the claim is accurate we need to calculate the ratio of the size of the nucleus (the same as a grape) and the distance between the electrons and the nucleus for example 1 (r₁):

$r_{1} = \frac{s_{1}}{d_{1}}$ (1)

and to compare it with the ratio of the size and the distance given in example 2 (r₂):

$r_{2} = \frac{s_{2}}{d_{2}}$ (2)

Where:

s₁: is the size of the nucleus (like the size of a grape)

d₁: is the distance between electrons and nucleus of example 1 = 1 mile

s₂: is the average diameter of the nucleus = 10⁻¹³ cm

d₂: is the average distance between electrons and nucleus of example 2 = 10⁻⁸ cm

Assuming that the diameter of a grape is 3 cm (in a spherical way), the ratio of the first example is (eq 1):

$r_{1} = \frac{3 cm}{1 mi*\frac{160934 cm}{1 mi}} = 1.86 \cdot 10^{-5}$

Now, the ratio of the second example is (eq 2):

$r_{2} = \frac{10^{-13} cm}{10^{-8} cm} = 1.00 \cdot 10^{-5}$

Since r₁ and r₂ are in the same order of magnitude (10⁻⁵), we can conclude that the given claim is reasonably accurate.

You can learn more about the nucleus of an atom here: brainly.com/question/10658589?referrer=searchResults

I hope it helps you!

3 0

3 years ago

1. If there are 100 navy beans, 27 pinto beans and 173 blackeyed peas in a container, what is the percent abundance in the conta

kompoz [17]

The answers to the two questions are:

1. The percent abundance in the container which has 100 navy, 27 pinto, and 173 black-eyed peas beans is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The weighted average score for the scores of 85, 75, 96 obtained from the evaluations of exams (20%), labs (75%), and homework (96%) is 84.1.

1. The percent abundance by type of bean is given by:

$\% = \frac{n}{n_{t}} \times 100$ (1)

Where:

n: is the number of each type of beans

$n_{t}$ : is the total number of beans

The total number of beans can be calculated by adding the number of all the types of beans:

$n_{t} = n_{n} + n_{p} + n_{b}$ (2)

Where:

$n_{n}$ : is the number of navy beans = 100

$n_{p}$ : is the number of pinto beans = 27

$n_{b}$ : is the number of black-eyed peas beans = 173

Hence, the total number of beans is (eq 2):

$n_{t} = 100 + 27 + 173 = 300$

Now, the percent abundance by type of bean is (eq 1):

Navy beans

$\%_{n} = \frac{100}{300} \times 100 = 33.3 \%$

Pinto beans

$\%_{p} = \frac{27}{300} \times 100 = 9.0 \%$

Black-eyed peas beans

$\%_{b} = \frac{173}{300} \times 100 = 57.7 \%$

Hence, the percent abundance by type of bean is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The average score (S) can be calculated as follows:

$S = e*\%_{e} + l*\%_{l} + h*\%_{h}$ (3)

Where:

e: is the score for exams = 85

l: is the score for lab reports = 75

h: is the score for homework = 96

$%_{e}$ $\%_{e}$ : is the percent for exams = 70.0%

$\%_{l}$ : is the percent for lab reports = 20.0%

$\%_{h}$ : is the percent for homework = 10.0%

Then, the average score is:

$S = 85*0.70 + 75*0.20 + 96*0.10 = 84.1$

We can see that if the score for each evaluation is 100, after multiplying every evaluation for its respective percent, the final average score would be 100.

Therefore, the weighted average score will be 84.1.

Find more about percents here: