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3 years ago

What happens when molecules diffuse across the cell membrane?

Chemistry

2 answers:

mafiozo [28]3 years ago

7 0

Answer:

The correct answer is from areas of high concentration to low concentration.

Explanation:

A concentration gradient exists for these molecules, so they have the potential to diffuse into (or out of) the cell by moving down it.

Hope this Helps!

antiseptic1488 [7]3 years ago

5 0

U give birth to a crusty child :)

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A 0.08541 g sample of gas occupies 10.0-ml at 288.5 k and 1.10 atm. upon further analysis, the compound is found to be 13.068% c

topjm [15]

C2Br2 First, we need to determine how many moles of the gas we have. For that, we'll use the Ideal Gas Law which is PV = nRT where P = pressure (1.10 atm = 111458 Pa) V = volume (10.0 ml = 0.0000100 m^3) n = number of moles R = Ideal gas constant (8.3144598 (m^3 Pa)/(K mol) ) T = Absolute temperature Solving for n, we get PV/(RT) = n Now substituting our known values into the formula. (111458 Pa * 0.0000100 m^3) / (288.5 K * 8.3144598 (m^3 Pa)/(K mol)) = (1.11458/2398.721652) mol = 0.000464656 mol Now let's calculate the empirical formula for this compound. Atomic weight carbon = 12.0107 Atomic weight bromine = 79.904 Relative moles carbon = 13.068 / 12.0107 = 1.08802984 Relative moles bromine = 86.932 / 79.904 = 1.087955547 So the relative number of atoms of the two elements is 1.08802984 : 1.087955547 After dividing all numbers by the smallest, the ratio becomes 1.000068287 : 1 Which is close enough to 1:1 for me to consider the empirical formula to be CBr Now calculate the molar mass of CBr 12.0107 + 79.904 = 91.9147 Finally, let's determine if the compound is actually CBr, or something like C2Br2, or some other multiple. Using the molar mass of CBr, multiply by the number of moles and see if the result matches the mass of the gas. So 91.9147 g/mol * 0.000464656 mol = 0.042708701 g 0.0427087 g is a lot smaller than 0.08541 g. So the compound isn't exactly CBr. Let's divide them to see what the factor is. 0.08541 / 0.0427087 = 1.99982673 1.99982673 is close enough to 2 to within the number of significant digits we have for me to claim that the formula for the unknown gas isn't CBr, but instead is C2Br2.

3 0

3 years ago

Read 2 more answers

Fe(no3)3 + na0h > fe(oh)3 + nano3 is balanced or unbalanced?

VashaNatasha [74]

Answer:

Unbalanced

Explanation:

$\text{Fe}(\text{NO}_3)}_3+ \text{NaOH} \rightarrow \text{Fe(OH)}_3 + \text{NaNO}_3\\\\\text{Balanced reaction:}\\\\\text{Fe}(\text{NO}_3)}_3+ 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaNO}_3$

5 0

2 years ago

1. Draw the Dot Structure for Na!

stepladder [879]

Answer:

Na ·

Explanation:

Hello!

In this case, since the Lewis dot structure of any element illustrates the number of valence electrons at the outer shell; we can set up the electron configuration of sodium to obtain:

$Na^1^1\rightarrow 1s^2,2s^2,2p^6,3s^1$

We can see there is only one valence electron, for that reason the Lewis dot structure is:

Na ·

Whereas the dot represents the valence electron.

Best regards!

4 0

3 years ago

A sample of chromium oxide is 76.5hromium by weight. what is the simplest formula of the oxide?

otez555 [7]

CrO and Cr₂O₃ make up the simplest chromium oxide formula.

What name does Cr₂O₃ use?

Chromium oxide (Cr₂O₃)sometimes referred to as chromium sesquioxide or chromic oxide, is a compound in which chromium is oxidized to a +3 state. Sodium dichromate is calcined with either carbon or sulfur to produce it.

Eskolaite, a mineral that bears the name of the Finnish geologist Pentti Eskola, is a kind of chromium oxide green that may be found in nature. The metallic glassy green surface of this unusual material has an unsettling moss-like look that may be used to conceal oneself in the environment.

Studies on humans have conclusively shown that chromium (VI) breathed is a potential carcinogen, increasing the likelihood of developing lung cancer. According to animal studies, chromium (VI) exposure by inhalation can result in lung cancers.

Learn more about chromium here:

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6 0

1 year ago

Group the following electron configurations in pairs that would represent similar chemical properties at their atoms;

marishachu [46]

Answer: Pairs are: (a) and (d), (b) and (f), (c) and (e)

Explanation:

In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

Elements arranged in a group show similar chemical properties because of the presence of same number of valence electrons.

Valence electrons are defined as the electrons which are present in the outermost shell of an atom. Outermost shell has the highest value of 'n' that is principal quantum number.

For the given options:

For a:

The given electronic configuration is: $1s^22s^22p^63s^2$

The number of valence electrons in the given configuration are 2

For b:

The given electronic configuration is: $1s^22s^22p^63s^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

For c:

The given electronic configuration is: $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For d:

The given electronic configuration is: $1s^22s^2$

The number of valence electrons in the given configuration are 2

For e:

The given electronic configuration is: $1s^22s^22p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For f:

The given electronic configuration is: $1s^22s^22p^63s^23p^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

Electronic configuration of (a) and (d) will form a pair, (b) and (f) will form a pair, (c) and (e) will form a pair and will have similar chemical properties.

Hence, the pairs are: (a) and (d), (b) and (f), (c) and (e)

4 0

3 years ago