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3 years ago

( Endocytosis / Exocytosis ) is the movement of substances out of a cell by vesicular transport.

Chemistry

1 answer:

tiny-mole [99]3 years ago

4 0

Answer:

Exocytosis

Explanation:

Some molecules are simply too big to move via a transport protein or the plasma membrane. To carry these macromolecules in or out of the cell, cells employ two more active transport pathways. Macromolecules or big particles are transported across the plasma membrane via Vesicles transport or other cytoplasmic structures. They are of two types, Endocytosis and Exocytosis

From the given information, Exocytosis is the right answer.

It is the process of vesicles combining with the plasma membrane thereby releasing their contents to the exterior of the cell. When a cell creates components for export, such as proteins, or when it gets rid of a waste product or a toxin, exocytosis occurs. Exocytosis is the process by which newly generated membrane proteins and membrane lipids are transported on top of the plasma membrane.

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Perform the following calculations to the correct number of significant figures.

SSSSS [86.1K]

Answer: 460.624

Explanation:
1. Multiply the numbers
(24.5260 x 2.56) + 397.84
= (62.784) + 397.84

2. Add the numbers
(62.784) + 397.84
= 460.624

4 0

2 years ago

Mendeleev arranged the elements by what physical property? atomic number atomic mass boiling point density

m_a_m_a [10]

Mendeleev arranged the elements according to their atomic masses

3 0

3 years ago

Read 2 more answers

The rock in a lead ore deposit contains 89 % PbS by mass. How many kilograms of the rock must be processed to obtain 1.5 kg of P

Zolol [24]

Answer:

Approximately 1.9 kilograms of this rock.

Explanation:

Relative atomic mass data from a modern periodic table:

Pb: 207.2;
S: 32.06.

To answer this question, start by finding the mass of Pb in each kilogram of this rock.

89% of the rock is $\rm PbS$ . There will be 890 grams of $\rm PbS$ in one kilogram of this rock.

Formula mass of $\rm PbS$ :

$M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}$ .

How many moles of $\rm PbS$ formula units in that 890 grams of $\rm PbS$ ?

$\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol$ .

There's one mole of $\rm Pb$ in each mole of $\rm PbS$ . There are thus $\rm 3.71980\; mol$ of $\rm Pb$ in one kilogram of this rock.

What will be the mass of that $\rm 3.71980\; mol$ of $\rm Pb$ ?

$m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg$ .

In other words, the $\rm PbS$ in 1 kilogram of this rock contains $\rm 0.770743\; kg$ of lead $\rm Pb$ .

How many kilograms of the rock will contain enough $\rm PbS$ to provide 1.5 kilogram of $\rm Pb$ ?

$\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg$ .

5 0

3 years ago

λmax for the π → π* transition in ethylene is 170 nm. Is the HOMO-LUMO energy difference in ethylene greater than or less than t

-Dominant- [34]

Answer:

the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene

Explanation:

The λmax is the wavelength of maximum absorption. We could use it to calculate the HOMO-LUMO energy difference as follows:

For ethylene

E= hc/λ= 6.63×10^-34×3×10^8/170×10^-9= 1.17×10^-18J

For cis,trans−1,3−cyclooctadiene

E= hc/λ=6.63×10^-34×3×10^8/230×10^-9=8.6×10^-19J

Therefore, the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene

8 0

3 years ago

Question 2 of 50

wolverine [178]

The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.

Further Explanation:

To solve this problem, follow the steps below:

Write the balanced chemical equation for the given reaction.
Convert the mass of calcium carbonate into moles.
Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

$CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}$

STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.

$mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}$

STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.

For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,

$mol \ CaO \ = 0.2498 \ mol$

STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.

$mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g$

Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.

Therefore,

$\boxed {mass \ CaO \ = 14 \ g}$

Learn More

Learn more about stoichiometry brainly.com/question/12979299
Learn more about mole conversion brainly.com/question/12972204
Learn more about limiting reactants brainly.com/question/12979491

Keywords: thermal decomposition, stoichiometry

5 0

3 years ago