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mel-nik [20]
2 years ago
11

You are given two fatlike solid substances and determine that sample a has a higher melting point than sample

Chemistry
1 answer:
Alchen [17]2 years ago
5 0
Given that <span>sample a has a higher melting point than sample
b. Therefore, sample a is a longer chain of a </span><span>fatlike solid substance. It could also be that the bonds present in sample a is much stronger which will require more energy to break. Hope this answers the question.</span>
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A mixture of0.161 moles of C is reacted with 0.117 moles of O2 in a sealed, 10.0 L-vessel at 500.0 K, producing a mixture of CO
labwork [276]

Answer:

number of moles of CO2 is 0.054

number of moles of CO is 0.107

number of moles of O2 remaining is 0.01 mole

mole fraction of CO is 0.63

Explanation:

Firstly, we write the equation of reaction;

3C(s) +2O2(g) → CO2(g) +2CO(g)

Now, we proceed.

From the written equation, we can deduce that

3 mol C = 2 mol O2 = 1 mol CO2 = 2 mol CO

No of mol of C reacted = 0.161 mol

limiting reactant according to the question is Carbon

a. no of mol of CO2 formed = 0.161*1/3 = 0.054 moles ( no of moles of CO2 formed is one-third of no of moles of carbon reacted. This is obtainable from their mole ratio 1:3)

b. no of mol of CO formed = 0.161*2/3 = 0.107 mol

c. no of mol of O2 remaining = 0.117 - (0.151*2/3) = 0.117-0.107 = 0.01 mole

d. mole fraction of CO = no of mol of CO/Total number of moles

= 0.107/(0.107+0.054+0.01)

= 0.625730994152 which is approximately 0.63

5 0
2 years ago
Alex dragged a log across the yard in 30 the log weighted 400n she did 900 j of work how much power did I she have
In-s [12.5K]

The power used by Alex to drag the log across the yard is determined as 2,656 W.

<h3>Mass of the log</h3>

The mass of the log is calculated as follows;

W = mg

m = W/g

m = (400)/9.8

m = 40.82 kg

<h3>Velocity of the log</h3>

K.E = ¹/₂mv²

v² = 2K.E/m

v² = (2 x 900)/(40.82)

v² = 44.096

v = 6.64 m/s

<h3>Power used by Alex</h3>

P = Fv

P = 400 x 6.64

P = 2,656 W

Learn more about power here: brainly.com/question/13881533

#SPJ1

5 0
1 year ago
Convert 63.3 kg to grams
Zepler [3.9K]

Answer:

63300

Explanation:

63300

5 0
2 years ago
Read 2 more answers
A certain atom has a number if four. How many protons does it have
user100 [1]
4 protons are there in the atom
3 0
2 years ago
Read 2 more answers
What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?
Marta_Voda [28]

Answer : The excess reactant in the combustion of methane in opem atmosphere is O_{2} molecule.

Solution : Given,

Mass of methane = 23 g

Molar mass of methane = 16.04 g/mole

The Net balanced chemical reaction for combustion of methane is,

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

First we have to calculate the moles of methane.

\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}} = \frac{23g}{16.04g/mole} = 1.434 moles

From the above chemical reaction, we conclude that

1 mole of methane react with the 2 moles of oxygen

and 1.434 moles of methane react to give \frac{2moles\times 1.434moles}{1moles} moles of oxygen

The Moles of oxygen = 2.868 moles

Now we conclude that the moles of oxygen are more than the moles of methane.

Therefore, the excess reactant in the combustion of methane in open atmosphere is O_{2} molecule.


6 0
3 years ago
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