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Alik [6]
2 years ago
13

How many moles of O2 are produced when 0.500mol of KO2 reacts in this fashion?

Chemistry
1 answer:
Shtirlitz [24]2 years ago
6 0
I think you means the KO2 reacts with H2O. The equation of this reaction is 4KO2+2H2O->4KOH +3O2. The ratio of mole number of O2 and KO2 is 3:4. So the mole number of O2 produced is 0.500/4*3=0.375 mol.
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A major problem caused by humans is the contamination and depletion of _____ resources. A. solar B. water C. wind
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Water is the only one of these that would work by process of elimination.
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What is a common factor associated with most severe storms?
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A common factor is low pressure system. 
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How is the separation of dna like a zipper
Sedbober [7]

The two strands must be separated like the two sides of a zipper, by breaking the weak hydrogen bonds that link the paired bases.

<u>Explanation:</u>

  • A double helix structure formed by two polypeptide chains is separated like the two sides of a zipper. A zipper is formed by breaking the weak hydrogen bonds that link the paired bases. During replication, an enzyme "Helicase" travels down the DNA and splits the chain and it forms 2 separate strands.
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3 0
2 years ago
A 32.5 g piece of aluminum (which has a specific heat capacity of 0.921 J/g°C) is heated to 82.4°C and dropped into a calorimete
N76 [4]

Answer:

The mass of water = 219.1 grams

Explanation:

Step 1: Data given

Mass of aluminium = 32.5 grams

specific heat capacity aluminium = 0.921 J/g°C

Temperature = 82.4 °C

Temperature of water = 22.3 °C

The final temperature = 24.2 °C

Step 2: Calculate the mass of water

Heat lost = heat gained

Qlost = -Qgained

Qaluminium = -Qwater

Q = m*c*ΔT

m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)

⇒with m(aluminium) = the mass of aluminium = 32.5 grams

⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C

⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C =  -58.2 °C

⇒with m(water) = the mass of water = TO BE DETERMINED

⇒with c(water) = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C

32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9

-1742.1 = -7.95m

m = 219.1 grams

The mass of water = 219.1 grams

8 0
2 years ago
In this experiment, 0.170 g of caffeine is dissolved in 10.0 ml of water. the caffeine is extracted from the aqueous solution th
zmey [24]

solution:

Weight of caffeine is W = 0.170 gm.

Volume of water is V= 10 ml

Volume of methylene chloride which extracted caffeine is v= 5ml

No of portions n=3

Distribution co-efficient= 4.6

Total amount of caffeine that can be unextracted is given by

w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms

amount of caffeine un extracted is 0.125gms

amount of caffeine extracted=0.170-0.125

                                                       =0.045 gms


6 0
2 years ago
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