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Alchen [17]
3 years ago
12

PLEASE HELP•Your body has organ systems that perform the functions you need to live. Where the body has specialized organs withi

n specialized systems, the cell has organelles. Each organelle has a specialized function within the cell.
Question: Choose one organelle discussed in this lesson and compare it's function to an organ or organ system in the human body.
Chemistry
2 answers:
jek_recluse [69]3 years ago
7 0

An organelle could be the nucleus of a cell.

The nucleus of a cell would be comparable to the nervous system in the body, which is considered an organ system.

Hope this helps!

Alona [7]3 years ago
6 0

it would be digestive system or somthing like that thanks for the free points bro


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A family takes a summer vacation to Florida. They drive for 8 hours before stopping at a hotel for the night. They
andrezito [222]

Average speed of the trip = 52 km/hr

<h3>Further explanation  </h3>

Distance is the length traveled by an object within a certain of time .

Average speed = total distance/total amount of  time ,

Can be formulated :

\tt S(speed)=\dfrac{D}{T}

Total distance travelled : 1560 km

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7 0
3 years ago
Which were Lenin's actions before and during the Russian Revolution? Check all that apply.
Stolb23 [73]

Answer:

He supported the ideology of Marxism.

He opposed the tsar and was exiled.

He led the Bolsheviks.

Explanation:

Lenin was Russian Revolution political theorist. He served as first founding head of government of Soviet Russia. He supported the ideology of Marxism and opposed Tsar.

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3 years ago
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7 0
3 years ago
In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.
Serhud [2]

Answer:

Explanation:

Chemical equation:

C₁₀H₈O + COCl₂  → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?

Given data:

Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 250000 g/ 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

                     C₁₀H₈O       :         C₁₁H₇O₂Cl

                        1               :                1

                       1734.1         :             1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 1734.1 mol × 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg×358091.65 g/ 1000 g  = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given data:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 100 g/ 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles  = mass/ molar mass

Number of moles =  100 g/ 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

                     C₁₀H₈O        :         C₁₁H₇O₂Cl

                        1                :                1

                       0.694        :              0.694

                    COCl₂          :             C₁₁H₇O₂Cl

                        1                :                1

                       1.0              :              1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of  C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass =  0.694 mol × 206.5 g/mol

Mass = 143.3 g

Theoretical yield  =  143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given data:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula :

Percent yield = actual yield / theoretical yield × 100

Now we will put the values in formula.

Percent yield = 118 g/ 143.3 g × 100

Percent yield = 0.82 × 100

Percent yield = 82%

5 0
2 years ago
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