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lisabon 2012 [21]
3 years ago
10

How many moles of H atoms are there in 2 moles of CH3CH2CH2CH3?

Chemistry
1 answer:
Advocard [28]3 years ago
3 0

AnSweR: 10

ExplanAtion: My random idiot friend just told me

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Vlad1618 [11]

Answer:

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Explanation:

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3 years ago
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What is the change in temperature when 75 grams of water releases -2657 J of energy? The specific heat of water is 4.18 J/g°C
Naily [24]
I think the answer for this is 4702.5 J/g*k Depending on if it is water as a solid liquid or gas. I used water as a liquid when I solved it. J=(75g)(4.18 J/g*k)(15K)
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3 years ago
how will you seperate dye from black ink? draw a neatly labelled same?? fast I will mark brainly 100%​
Nonamiya [84]

Answer:

put it through a distillery

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3 years ago
Calculate the concentration in mol/L, M, of an aqueous sugar solution with a concentration of 23.5% (w/w) and density of 1.005 g
Temka [501]

Answer:

The concentration in mol/L is 0.683M

Explanation:

23.5% (w/w)

This data means that in 100 g of solution, we have 23.5 grams of solute.

From this point, we can calculate the moles of sugar.

Moles = Mass / Molar mass

Moles = 23.5 g /342.30 g/m

Moles = 0.068 moles

Density data make us know, the volume of our solution.

solution δ = solution mass / solution volume

δ = 1.005 g/mL = 100 g /  solution volume

solution volume = 100g / 1.005 g/ml

solution volume = 99.5 mL

In conclusion, 0.068 moles are in 99.5 mL

Molarity (M) is mol/L

Let's convert 99.5 mL in L

99.5 mL / 1000 = 0.0995 L

0.068 m / 0.0995L = 0.683

<em>If we convert moles in mmoles, we can also get Molarity (M)</em>

<em>mmoles / mL = M</em>

<em>0.068 moles . 1000 = 68 mmoles</em>

<em>68 mmmoles / 99.5mL = 0.683</em>

8 0
3 years ago
A 1.25g sample of copper (cCu=0.386Jg∘C) is initially at a temperature of 25.0∘C. If the sample absorbs 87.4J of heat, what is i
professor190 [17]

Answer: Final temperature = 206∘C

Explanation:

Heat Energy is given as  

q= mCΔT

ehere

q= Heat energy = 87.4J

m= mass=1.25g

C=specific heat c= 0.386Jg∘C) ,

ΔT =  Change in temperate of which the final temperature= 25.0∘C

 q= mCΔT

ΔT = q/mC

ΔT = 87.4/ 1.25 X 0.386=181.14∘C

But,

T final- T initial = ΔT

T final = T initial + ΔT

T final = 25.0∘C +181.14∘C=206.14∘C rounded to 206∘C

4 0
3 years ago
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