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Katyanochek1 [597]
3 years ago
13

Please Help.....................................................................................................................

................

Chemistry
1 answer:
S_A_V [24]3 years ago
5 0
Mixtures:
physically combined
components retain properties

compounds:
chemically combined
components change properties

both:
made of elements
variable proportions 
fixed properties
2 or more compounds

hope this helps 
comment me if you have any questions 


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5 0
3 years ago
Air containing 0.06% carbon dioxide is pumped into a room whose volume is 12,000 ft3. The air is pumped in at a rate of 3,000 ft
castortr0y [4]

Answer:

A(t)=1.8+34.2*e^{-t}

Explanation:

The concentration of CO2 in the room will be the amount of CO2 in the room at time t, divided by the volume of the room.

Let A(t) be the amount of CO2 in the room, in ft3 CO2.

The air entering the room is 3000 ft3/min with 0.06% concentrarion of CO2. That can be expressed as (3000*0.06/100)=1.8 ft3 CO2/min.

The mixture leaves at 3000 ft3/min but with concentration A(t)/V. We can express the amount of CO2 leaving the room at any time is A(t).

We can write this as a differential equation

dA/dt=v_i-v_o=1.8-A

We can rearrange and integrate

dA/dt=v_i-v_o=1.8-A\\\\dA/(A-1.8)=-dt\\\\\int(dA/(A-1.8) = -\int dt\\\\ln(A-1.8)=-t+C\\\\A-1.8=e^{-t}* e^{C}=C*e^{-t}\\\\A=1.8+C*e^{-t}

We also know that A(0) = 12000*(0.3/100)=36 ft3 CO2.

A(0)=1.8+C*e^{-0}\\36=1.8+C*1\\C=34.2

Then we have the amount A(t) as

A(t)=1.8+34.2*e^{-t}

5 0
3 years ago
A race car travels a circular track at an average rate of 135 mi/hr. The radius of the track is 0.450 miles. What is the centrip
MaRussiya [10]
Thank you for posting your question here. I hope my answer will help. below are the choices that can be found elsewhere:

a. 300 mi/hr2 
<span>b. 61 mi/hr2 </span>
<span>c. 8,201 mi/hr2 </span>
<span>d. 40,500 mi/hr
</span>
Below is the solution:

<span>135 * 135 / 0.450 = 40,500 mi/hr2 </span>
4 0
3 years ago
Read 2 more answers
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