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Jlenok [28]
3 years ago
13

8 questions please please help i will make you barinliest 1. Photosynthesis is an important process that converts light energy i

nto ____________ energy.
A. Chemical
B.Nuclear
C.Water
D.Thermal

2. This cell would be found in which type of organism?

A.Animal
B.Bacteria
C.Virus
D.Plant

3. All Cells get their energy from glucose, but it must first be broken down in the mitochondria through a process called ______________________.
A. Photosynthesis
B. Cell Respiration
C.Cell Oxidation
D.ATP

4. Which of the following is an advantage for a single-celled organism like the Euglena?
A. It can become very large
B.It does not have to reproduce
C.It can reproduce quickly
D.It does not have to get energy

5.The items listed below were found in a science classroom. How would you describe these items?
A.Models
B.Experiments
C.Varaibles
C.Controls

6. Growth and repair in multicellular organisms are the result of
A.Excretion
B.Locomotion
C.Cell division
D.Decomposition

7. What is the result of cellular respiration?
A.Energy for cell processes is released.
B.Oxygen is released for photosynthesis.
C.Cells undergo decomposition
D.Nutrients are excreted to prevent the buildup of body fat.

8.Why is Photosynthesis important?

A.Makes organic molecules (glucose) out of inorganic materials (carbon dioxide and water).
B.It begins all food chains/webs. Thus all life is supported by this process.
C.It also makes oxygen gas which animals need.
D.All of the above

Chemistry
1 answer:
blagie [28]3 years ago
3 0

d.thermal

d.plant

a.photosynthesis

c.it can produce quickly

b.models

a.excretion

d

d

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Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/
Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of C_2H_4 and H_2 = 25.0 atm

Temperature of C_2H_4 and H_2 = 250^oC=273+250=523K

Volume of C_2H_4 = 1050 L per min

Volume of H_2 = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of C_2H_6 = 30 g/mole

First we have to calculate the moles of C_2H_4 and H_2 by using ideal gas equation.

For C_2H_4 :

PV=nRT\\\\n=\frac{PV}{RT}

n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}

n=611.34moles

For H_2 :

PV=nRT\\\\n=\frac{PV}{RT}

n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}

n=902.46moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

C_2H_4+H_2\rightarrow C_2H_6

From the balanced reaction we conclude that

As, 1 mole of C_2H_4 react with 1 mole of H_2

So, 611.34 mole of C_2H_4 react with 611.34 mole of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and C_2H_4 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of C_2H_6.

As, 1 mole of C_2H_4 react to give 1 mole of C_2H_6

As, 611.34 mole of C_2H_4 react to give 611.34 mole of C_2H_6

Now we have to calculate the mass of C_2H_6.

\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6

\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g

The theoretical yield of C_2H_6 = 18340.2 g

The actual yield of C_2H_6 = 14.0 kg = 14000 g      (1 kg = 1000 g)

Now we have to calculate the percent yield of C_2H_6

\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%

Therefore, the percent yield of the reaction is, 76.34 %

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4 years ago
We dissolve 93 grams of sodium sulfate (Na2SO4) in 20 grams of water. What is the percent by mass of sodium sulfate in this solu
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The empirical formula for carbohydrates is CH2O. The molecular weight for a particular carbohydrate is 90.09 grams. What is the
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We find the weight of the empirical formula:
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