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nirvana33 [79]
3 years ago
8

Blank moles of carbon dioxide are required to make 7.2 moles of glucose. A plant using 618 grams of carbon dioxide and plenty of

water can make
Chemistry
1 answer:
Virty [35]3 years ago
7 0

Answer:

43.2 moles of carbon dioxide are required and 421g of glucose could be produced

Explanation:

Based on the reaction:

6CO2 + 6H2O → C6H12O6 + 6O2

1 mole of glucose, C6H12O6, requires 6 moles of carbon dioxide. 7.2moles of glucose requires:

7.2mol C6H12O6 * (6mol CO2 / 1mol C6H12O6) =

<h3>43.2 moles of carbon dioxide are required</h3><h3 />

618g of CO2 -Molar mass: 44.01g/mol- are:

618g * (1mol / 44.01g) = 14.04moles CO2

Moles C6H12O6:

14.04moles CO2 * (1mol C6H12O6 / 6mol CO2) = 2.34moles C6H12O6

Mass glucose -Molar mass: 180.156g/mol-

2.34moles C6H12O6 * (180.156g / mol) =

<h3>421g of glucose could be produced</h3>
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A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o
klasskru [66]
Hello!

<span>We have the following statement data:
</span>
Data:
P_{Total} = 800 mmHg
P\% N_{2} = 60\%
P\% O_{2} = 40\%
P_{partial} = ? (mmHg)

<span>As the percentage is the mole fraction multiplied by 100:

</span>P =  X_{ O_{2} }*100

<span>The mole fraction will be the percentage divided by 100, thus:
</span><span>What is the partial pressure of oxygen in this mixture? 
</span>
X_{ O_{2} }  =  \frac{P}{100}
X_{ O_{2}} =  \frac{40}{100}
\boxed{X_{ O_{2}} = 0.4}


<span>To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:
</span>
In relation to O_{2} :

\frac{P O_{2} }{P_{total}} = X_O_{2}
\frac{P O_{2} }{800} = 0.4
P_O_{2} = 0.4*800
\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark
<span>
Answer:
</span><span>b. 320.0 mm hg </span>
7 0
3 years ago
What is the volume in (a) liters and (b) cubic yards of a room that is 10. meters wide by 15 meters long and 8.0 ft high?
deff fn [24]

Answer:

V = 364500 L, 476.748 yard³

Explanation:

Given that,

The dimensions of a room are 10 meters wide by 15 meters long and 8.0 ft high.

l = 10 m, b = 15 m, h = 8 ft = 2.43 m

The volume of the room is :

V = lbh

So,

V = 10×15×2.43

V = 364.5 m³

As 1 m³ = 1000 L

364.5 m³ = 364500 L

Also, 1 m³ = 1.30795 yard³

364.5 m³ = 476.748 yard³

Hence, this is the required solution.

7 0
2 years ago
How many moles of Ar gas are<br> present in a container with a<br> volume of 78.4 L at STP?
rusak2 [61]
<h3><u>Answer:</u></h3>

<u>1 mole of a gas at STP occupies 22.4 L volume </u>

<u>Now the volume is given =78.4 therefore,</u>

<u>No. of moles of gas = 78.4 ÷ 22.4 = 3.5 moles</u>

<u>I hope it helps you~</u>

7 0
1 year ago
Which of the following is altered by a catalyst?
Vika [28.1K]
<span>In chemistry, a catalyst can speed up the reaction (or make it initiate easier) by altering the activation energy, lowering it enough to allow the reactants to react more easily. Some negative catalysts or inhibitors can do the same by increasing the activation energy.
</span>
4 0
2 years ago
How many moles of O2 are required to generate 18 moles of H2O in the given reaction? 2C8H18 + 25O2 16CO2 + 18H2O
Veronika [31]
There's a slight error in your equation. I think you were trying to present it like this:

2C8H18 + 25O2 -> 16CO2 + 18H2O

Mole Ratio
O2 : H20
25 : 18
? moles : 18 moles
(18/18)×25 : 18 moles

25 moles : 18 moles

Final answer would be 25 moles of O2. :)

If you have any doubts that you want to clarify with me, please ask me! :)
I will do my utmost best to help you.
7 0
3 years ago
Read 2 more answers
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