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IceJOKER [234]
3 years ago
14

Which cell organelle is like the construction site, where proteins are built? Which cell organelle packages and transports the p

rotein to the food processing plant?
O vacuole, endoplasmic reticulum
O cell membrane, lysosome
Onbosomes, golgi bodies
O nucleus, mitochondria
Chemistry
1 answer:
german3 years ago
7 0

Explanation:

The endoplasmic reticulum consists of a network of a tube-like passageway through which proteins from the ribosomes are able to be moved within a cell as the road system allows for movement throughout the city.

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The ears work with the brain and other body parts through a series of steps that help
o-na [289]

Answer: B: Your ears send a message through the nerves to the brain.

5 0
2 years ago
Ethanol (CH3CH2OH), the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuri
Fed [463]

Answer:

a) 88.48%

b) 0.05625 mol

Explanation:

2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g)         Reaction 1

CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g)                        Reaction 2

a) CH₃CH₂OH = 46.0684 g/mol

   CH₃CH₂OCH₂CH₃ = 74.12 g/mol

1 mol CH₃CH₂OH ______  46.0684 g

x                            ______   50.0 g

x = 1.085 mol  CH₃CH₂OH

1 mol  CH₃CH₂OCH₂CH₃ ______  74.12 g g

y                           ______   35.9 g

y = 0.48 mol   CH₃CH₂OCH₂CH₃

100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃

w                _____  0.48 mol CH₃CH₂OCH₂CH₃

w = 88.48%

b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.

4 0
3 years ago
Combustion of 25.0 g of a hydrocarbon produces 86.5 g of co2. what is the empirical formula of the compound?
Step2247 [10]
Combustion is a reaction between a combustible substance and oxygen, to ultimately produce carbon dioxide and water. Reaction between carbon and oxygen would give,

                               C     +     O2      ------>  CO2

Here, we have 86.5 grams of carbon dioxide, CO2, which is a product of combustion. Dividing this mass by the molar mass of CO2, which is 44 grams, we can determine the number of moles of CO2. 

                          <u>     86.5 g CO       </u>   = 1.966 moles CO2
                            44 g CO2/ mole

Considering that CO2 is composed of 1 mole of carbon and 2 moles of oxygen, and that with complete combustion, 1 mole of carbon reacts to produces 1 mole of CO2, we can then determine the mass of the carbon in the hydrocarbon fuel. 

        1.966 moles CO2   x   <u>   1 mole C   </u>    x   <u>   </u><u>12 g C   </u>  = 23.59 g C
                                             1 mole CO2          1 mole C

We were given 25.0 grams of the fuel hydrocarbon. A hydrocarbon is a substance consisting of carbon and hydrogen. To determine the mass of the hydrogen in the fuel, we simply subtract 23.59 grams from 25.0 grams. 


            25.0 g - 23.59 g = 1.41 grams Hydrogen 

To know the number of moles of hydrogen, we divide the mass of the hydrogen in the fuel by the molar mass of hydrogen, which is 1.01 g/mole. Thus, we have 1.396 mole hydrogen. 

To determine the empirical formula, we divide the number of moles carbon by the number of moles hydrogen, and find a factor that would give whole number ratios for the carbon and hydrogen in the fuel, 

Carbon:     <u>  1.966 mol   </u>   = 1.408   x   5 (factor)     = 7
                    1.396 mol

Hydrogen:  <u>  1.396 mol   </u>    = 1.00   x    5 (factor)    = 5
                     1.396 mol

Thus, the empirical formula is C7H5

       
4 0
4 years ago
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mostly water

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