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Zolol [24]
3 years ago
9

How many moles of O2 are needed to react with 2.35 mol of C2H2?

Chemistry
2 answers:
adelina 88 [10]3 years ago
7 0

Answer : The number of moles of O_2 needed are 5.88 moles.

Solution : Given,

Moles of C_2H_2 = 2.35 mole

The balanced chemical reaction will be,

2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 moles of C_2H_2 react with 5 moles of O_2

So, 2.35 moles of C_2H_2 react with \frac{2.35}{2}\times 5=5.88 moles of O_2

Therefore, the number of moles of O_2 needed are 5.88 moles.

Mrrafil [7]3 years ago
5 0
2 C2H2 + 5 O2 --> 4 CO2 + 2 H2O

2.35 mol C2H2 - x mol O2
2 mol C2H2 - 5 mol O2

x =  \frac{2.35 \times 5}{2}  = 5.875 \: mol
answer: 5.875 mol
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The number of glyceraldehyde-3-phosphate molecules that would be produced from 24 turns of the calvin cycle would be
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The Calvin cycle, also called the light-independent or carbon fixation reactions, is the second stage of photosynthesis where water, and carbon dioxide (CO2) from air, are converted into organic compounds (i.e. sugars) using the energy from short-lived electronically excited carriers (ATP and NADPH) for the reactions. These organic compounds can then be used by the producing organism (i.e. plants) and the animals that feed on it. 

One product of the Calvin cycle is the glyceraldehyde-3-phosphate (G3P), which is later on used in the production of glucose and in the regeneration of <span>Ribulose 1,5-bisphosphate (RuBP), which is an organic compound</span> essential to the reactions in the cycle. 

One turn of Calvin Cycle produces 2 G3P molecules, each comprising of 3 carbons. This gives a total of 6 carbons. Five (5) of these carbons will be used to regenerate RuBP and only 1 will be available to form a surplus G3P later on. This surplus G3P will be used for the production of glucose (a 6-carbon sugar). 

Thus, 3 turns of the carbon cycle will produce 1 surplus G3P. There are 8 sets of 3-turns in 24 cycles, therefore, 

                       1 net G3P molecule * 8 sets of 3-turns  = 8 G3P molecules

Therefore, there are 8 net or surplus G3P molecules produced for 24 cycles of the Calvin Cycle. The total G3P molecules produced, including the ones that participated in the regeneration of RuBP would be 48 G3Ps. 

For every 3 turns, 6 G3P molecules are produced, 5 of which will be used in the regeneration of RuBP and 1 will be the net or surplus, to be used for the production of glucose. The 48 G3Ps then come from the calculation, 

                       6 total G3P molecules * 8 sets of 3-turns  = 48 G3P molecules


The figure below shows the products of the cycle after 3 turns (Source: https://ka-perseus-images.s3.amazonaws.com/2f4bdc8f8275834d3f5ef434d93bf16b991b2357.png). 

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2 years ago
A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing point depression o
Ludmilka [50]

Answer:

0.297 °C

Step-by-step explanation:

The formula for the <em>freezing point depression </em>ΔT_f is

ΔT_f = iK_f·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For glucose,

       glucose(s) ⟶ glucose(aq)

1 mole glucose ⟶ 1 mol particles     i = 1

Data:

Mass of glucose = 10.20 g

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                 ΔT_f = 1.86 °C·kg·mol⁻¹

Calculations:

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n = 10.20 g × (1 mol/180.16 g)

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(b) <em>Kilograms of water </em>

m = 355 g × (1 kg/1000 g)

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(c) <em>Molal concentration </em>

b = moles of solute/kilograms of solvent

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(d) <em>Freezing point depression </em>

ΔT_f = 1 × 1.86 × 0.1595

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