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nordsb [41]
3 years ago
11

How many liters of C3H6O are present in a sample weighing 25.6 grams?

Chemistry
1 answer:
lawyer [7]3 years ago
4 0

To Find :

Number of moles of C₃H₆O present in a sample weighing 25.6 grams.

Solution :

Molecular mass of C₃H₆O is :

M = (6×12) + (6×1) + (16×1) grams

M = 94 grams/mol

We know, number of moles of 25.6 grams of C₃H₆O is :

n = \dfrac{Given \ Mass \ Of \ C_3H_6O }{Molar\ Mass \ Of \ C_3H_6O }\\\\n = \dfrac{25.6}{94}\ mole\\\\n = 0.27 \ mole

Hence, this is the required solution.

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7) What is the hydrogen concentration of a solution with a pH of 2.1? *
lesya [120]

Answer:

Hydrogen concentration = 7.94×10^-3 M

Explanation:

from potenz Hydrogen ( pH ) definition

pH = -log[H+]

2.1 = -log[H+]

2.1/-log = -log[H+]/-log

10^-2.1 = [H+]

[H+] = 7.94×10^-3M

8 0
3 years ago
Here is a multiple choice I need help on:
Ivenika [448]
I'm pretty sure it's B because carbon atoms are in all living organisms. They can also be bonded in different varations.
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3 years ago
Pablo's experiments confirm the relationship between pressure and the number of gas molecules. as the pressure in a system incre
horrorfan [7]
The correct answer is:  [C]:  
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          "<span>pressure and the number of gas molecules are directly related."
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<u>Note</u>:  The conclusion was: "</span> as the pressure in a system increases, the number of gas molecules increases" — over the course of many trials.
          This means that the "pressure" and the "number of gas molecules" are directly related.

Furthermore, this conclusion is consistent with the "ideal gas law" equation:

     " PV = nRT " ; 
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in which:

    "P = Pressure" ; 
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        All other factors held equal, when "n" (the "number of gas molecules") 
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6 0
3 years ago
Read 2 more answers
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

7 0
3 years ago
What happens to the temperature of the remaining liquid when some of the liquid evaporates
Dovator [93]
Turns into vapor. not all of the molecules are liquid have the same energy
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