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Anestetic [448]
3 years ago
15

How many moles of C6H1206 are consumed if 6 moles of O2 are consumed?​

Chemistry
2 answers:
Butoxors [25]3 years ago
6 0

Answer:

1 mole

Explanation:

To answer the following question we must use the formula for aerobic respiration :

C6H12O6 + 6O2 -> 6CO2 + 6H2O

Looking at this equation, you can see molar ratio between C₆H₁₂O₆ and O₂ is one to six. Meaning that if six moles of oxygen are devoured, one mole of glucose will be eaten.

DerKrebs [107]3 years ago
5 0
The molar ratio between glucose and oxygen is 1:6. Therefore, we can see that if 6 moles of oxygen are consumed, ONE mole of glucose will be consumed
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olganol [36]

Answer:

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Explanation:

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2 years ago
Interphase Milosis: Prophese Mitosis: Melaphese Cytokinesis Anaphose<br>drag and drop​
Alinara [238K]
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5 0
3 years ago
How many kilograms of potassium iodide (ki) are needed to make 1.25 l of a 4.41 m ki solution?
Lapatulllka [165]
C = n/V
n = C×V
n = 4,41M × 1,25L
n = 5,5125 mol

mKI: 39+127 = 166 g/mol

1 mol --------- 166g
5,5125 mol --- X
X = 166×5,5125 = 915,075g KI

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7 0
3 years ago
Question 9
Evgesh-ka [11]

Answer:

0.382 atm

Explanation:

In order to find the pressure, you need to know the moles of carbon dioxide (CO₂) gas. This can be found by multiplying the mass (g) by the molar mass (g/mol) of CO₂. It is important to arrange the conversion in a way that allows for the cancellation of units.

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

15 grams CO₂               1 mole
----------------------  x  ------------------------  =  0.341 moles CO₂
                                44.007 grams

To find the pressure, you need to use the Ideal Gas Law equation.

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas Constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

After you convert Celsius to Kelvin, you can plug the given and calculated values into the equation and simplify to find the pressure.

P = ? atm                              R = 0.08206 atm*L/mol*K

V = 20 L                               T = 0 °C + 273.15 = 273.15 K

n = 0.341 moles

PV = nRT

P(20 L) = (0.341 moles)(0.08206 atm*L/mol*K)(273.15 K)

P(20 L) = 7.64016

P = 0.382 atm

7 0
2 years ago
Gas law problem: A gas has a volume of 800 mL at -25℃ and 600 kPa. What would the volume of the gas be at 227℃ and 1000 kPa?
Sati [7]
<h3><u>Answer;</u></h3>

= 930.23 mL

<h3><u>Explanation</u>;</h3>

Using the combined gas law;

P1V1/T1 = P2V2/T2

Where; P1 = 600 kPa, V1 = 800 mL, and T1 = -25 +273 = 258 K, and

V2= ?, P2 = 1000 kPa, and T2 = 227 +273 = 500 K

Thus;

V2 = P1V1T2/T1P2

     = (600 ×800 ×500) / (258 × 1000)

     = 930.23 mL

5 0
3 years ago
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