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Leya [2.2K]
2 years ago
15

Molecules that are too large to flow across the membrane need help from what structures?

Chemistry
2 answers:
Pachacha [2.7K]2 years ago
6 0
Some molecules, such as carbon dioxide and oxygen, can diffuse across the plasma membrane directly, but others need help to cross its hydrophobic core. In facilitated diffusion, molecules diffuse across the plasma membrane with assistance from membrane proteins, such as channels and carriers.
Troyanec [42]2 years ago
5 0

Answer:

Some molecules, such as carbon dioxide and oxygen, can diffuse across the plasma membrane directly, but others need help to cross its hydrophobic core. In facilitated diffusion, molecules diffuse across the plasma membrane with assistance from membrane proteins, such as <u>channels</u> and <u>carriers</u>.

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Which particle is emitted when an atom of 85Kr spontaneously decays?
stiv31 [10]
A Beta particles is emitted when an atom of 85Kr spontaneously decays.
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2 years ago
4. Al2O3 (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H20(1) Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.
slava [35]

Answer:

Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over

Explanation:

For the reaction:

Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)

10,0g of Al₂O₃ are:

10,0g ₓ\frac{1mol}{102g} = <em>0,0980 moles</em>

And 10,0g of HCl are:

10,0 gₓ\frac{1mol}{36,5g} = <em>0,274 moles</em>

<em />

For a total reaction of 0,274 moles of HCl you need:

0,274×\frac{1molesAl_{2}O_3}{6 mole HCl} = <em>0,0457 moles of Al₂O₃</em>

Thus, limiting reactant is HCl

The grams produced of AlCl₃ are:

0,274 moles HCl ×\frac{2 moles AlCl_{3}}{6 moles HCl} × 133\frac{g}{mol} = <em>12,1 g of AlCl₃</em>

<em />

The moles of Al₂O₃ that don't react are:

0,0980 moles - 0,0457 moles =<em> </em>0,0523 moles

And its mass is:

0,0523 molesₓ\frac{102g}{1mol} = <em>5,33 g of Al₂O₃ </em>

<em />

I hope it helps!

7 0
3 years ago
Complete the following radioactive decay problem.<br> 238<br> Np →<br> 93<br> 0<br> e +<br> - 1
JulijaS [17]

Answer:

238 94Pu

Explanation:

From the question given, we noticed that the element is undergoing beta minus decay. This means that the daughter nuclei obtained will have the same mass number as the parent element and the the atomic number of the daughter nuclei will increase by 1.

Please see attached photo on how to arrive at the answer.

5 0
3 years ago
What element does this model represent?
AleksandrR [38]

Answer:

$\tex$ Option 2: Aluminum (Al)

Explanation:

When we are given Bohr models, we will be given a circle with rings surrounding it. The circle in the center of the model represents the nucleus, which contains the neutrons and the protons. The rings will have spherical structures that are attached to them in an orderly fashion - these model the electrons of an atom.

  • Protons are positively-charged subatomic particles that also identify the atom's chemical identity and atomic number. Using the number of protons, we are able to identify the element.
  • The neutrons are the neutrally-charged subatomic particles that give an atom its weight. When you look at a traditional periodic table, you'll see that the square that houses an element has its symbol, atomic number, and atomic mass. The atomic mass is equivalent to the sum of the protons and neutrons.
  • Electrons are negatively-charged subatomic particles that give an atom its overall charge. In order for an atom to be stable and neutral, the electrons <u>must</u> equal the protons. Otherwise, we have an unstable atom called an ion with either a positive or a negative charge. This is dependent on whether an atom has gained or lost electrons.

When we reference the model, we will see that there are 13 "p" and 14 "n" within the green circle. The "n" refers to <em>neutrons </em>and the "p" refers to <em>protons</em>. We can also count the red spheres and make quick observations about these: there are 2 red spheres on the innermost ring - for simplicity reasons, we will title this ring as r = 1. There are 8 red spheres on the middle ring - this ring will be titled r = 2. Finally, we can see that there are 2 more electrons in the outermost ring - this ring will be titled r = 3.

Now, because we have 13 protons, we know that the protons are equivalent to the atomic number.

  1. If we check the periodic table, we will see that Silicon (Si) has an atomic number of 14. This doesn't match the number of protons, so we can rule out that a silicon atom is not the element shown.
  2. When we use the same process and check aluminum, we discover that Aluminum (Al) has an atomic number of 13. Since the number of protons and the atomic number are equal, we can conclude that this is the element.
  3. If we check Helium (He), we see that it has an atomic number of 2, so this is definitely not our element in question.

Now that we have concluded that Aluminum is our element, we can check this to be sure. If we use the formula m = n + p (where m is the atomic mass, n is the neutrons, and p is the protons), we can check to be sure we have selected the right element.

The given mass of aluminum on the periodic table is 26.982 atomic mass units. We round to the nearest integer when it comes to this, so we round 26.982 up to 27 even.

Now that we have determined the atomic mass and we are given the number of both protons and neutrons, we can act as if we weren't given the amount of neutrons and only the mass and amount of protons. We can then use the equation to solve for the amount of neutrons and check that the selection we made is correct.

<u>Steps</u>

  1. Substitute 27 for <em>m</em> and 13 for <em>p</em>.
  2. Subtract 13 from both sides to isolate the <em>n</em> and place the constants on the same side of the equation.
  3. Combine like terms by taking the appropriate operations (in this case, this is subtracting 13 from 27).
  4. Finally, because we are solving for <em>n</em>, you can reverse the equation (place the constant on the right and place the variable on the left; i.e., 72 = x  →  x = 72).

27 = n + 13\\\\27 - 13 = n\\\\14 = n\\\\n = 14

The work we just performed will allow us to confirm that because we solved for <u>14 neutrons</u>, Aluminum (Al) is indeed the element represented by the model.

Hope this helps! :)

5 0
2 years ago
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