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postnew [5]
3 years ago
11

How many liters of oxygen gas (O2) are needed to produce 100 kJ of energy at STP? 2C6H6(I) + 1502(g) —> 12CO2(g)+6H2O(g)+3909

.9 kJ
Chemistry
1 answer:
makkiz [27]3 years ago
7 0

Answer: 8.59 L of oxygen gas are needed to produce 100 kJ of energy at STP

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atmosphere respectively.  

1 mole of every gas occupy volume at STP = 22.4 L

The balanced chemical reaction is:

2C_6H_6(I)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(g)

3909.9 kJ of of energy is produced by  = 15\times 22.4=336L

100 kJ of oxygen gas are needed to produce = \frac{336}{3909.9}\times 100=8.59L

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Answer:

Wind energy

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Wind energy is energy obtained from air moving at high velocity. This energy is harvested using windmills which convert mechanical energy to electrical energy.

Wind is inexpensive because it occurs naturally. However, a large expanse of land is required in order to mount sufficient number of windmills that will generate enough electrical energy for practical purposes.

This method of electricity generation is safe and does not lead to any environmental hazard unlike the burning of fossil fuels, use of nuclear energy or loss of habitat due to hydroelectric power generation.

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6 0
2 years ago
The pH of an acid solution is 5.82. Calculate the Ka for the monoprotic acid. The initial acid concentration is 0.010 M.
blagie [28]

Answer:

The answer is

<h3>Ka = 2.29 \times  {10}^{ - 14} moldm^{ - 3}</h3>

Explanation:

The Ka of an acid when given the pH and concentration can be found by

<h3>pH =  -  \frac{1}{2}  log(Ka)  -  \frac{1}{2}  log(c)</h3>

where

c is the concentration of the acid

From the question

pH = 5.82

c = 0.010 M

Substitute the values into the above formula and solve for Ka

We have

<h3>5.82 =   - \frac{1}{2}  log(Ka)  -  \frac{1}{2}  log(0.010)</h3><h3 /><h3>-  \frac{1}{2}  log(Ka)  = 5.82 + 1</h3><h3 /><h3>-  \frac{1}{2}  log(Ka)  = 6.82</h3>

Multiply through by - 2

<h3>log(Ka)  =  - 13.64</h3>

Find antilog of both sides

We have the final answer as

<h3>Ka = 2.29 \times  {10}^{ - 14} moldm^{ - 3}</h3>

Hope this helps you

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3 years ago
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I think the answer would be b
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