The molecule that could diffuse across the plasma membrane is methane (CH4).
<h3>What is diffusion?</h3>
Diffusion is the movement of fluids or substances from regions of high concentration toward regions of lower concentration.
The plasma membrane is the semipermeable membrane that surrounds the cytoplasm of a cell. The semipermeability means that it allows some molecules through but blocks other substances.
The semipermeable plasma membrane readily allows the passage of small hydrophobic and polar molecules.
Therefore, the molecule that could diffuse across the plasma membrane is methane (CH4).
Learn more about semipermeability at: brainly.com/question/1652796
#SPJ1
Answer:
52 da
Step-by-step explanation:
Whenever a question asks you, "How long to reach a certain concentration?" or something similar, you must use the appropriate integrated rate law expression.
The i<em>ntegrated rate law for a first-order reaction </em>is
ln([A₀]/[A] ) = kt
Data:
[A]₀ = 750 mg
[A] = 68 mg
t_ ½ = 15 da
Step 1. Calculate the value of the rate constant.
t_½ = ln2/k Multiply each side by k
kt_½ = ln2 Divide each side by t_½
k = ln2/t_½
= ln2/15
= 0.0462 da⁻¹
Step 2. Calculate the time
ln(750/68) = 0.0462t
ln11.0 = 0.0462t
2.40 = 0.0462t Divide each side by 0.0462
t = 52 da
I would personally convert the 12 mg to g so I could see what I was working with. So 12 mg to grams is 0.012 g...
so 1 tablet is 0.012g. the patient needs 0.024 g.
so 0.024g/0.012g = 2 tablets or 0.012g X 2 is 0.024 g
hope this helps :)
Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
