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KonstantinChe [14]
3 years ago
15

Help me. Dmskdjjddjjcjcfjfjgjghhffufuf

Chemistry
1 answer:
DENIUS [597]3 years ago
5 0

The answer Is D due to the fact the car is the heaviest one on there and would require the greates amount of force to move

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Which of these is caused by a chemical change?
IRISSAK [1]
I think that is A is the answer.
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Where is most of the mass of an atom? •
yaroslaw [1]
B. inside the nucleus
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What is a solution?
Marizza181 [45]

A solution is a homogeneous type of mixture of two or more substances. A solution has two parts: a solute and a solvent.

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One atom of an element possesses 80 protons, 80 electrons, and 121 neutrons. Given this information, what is the atomic mass of
IRINA_888 [86]

Answer:

I think the answer is 4) 41

Explanation:

APE= atomic number, proton and the electrons are the same number

MAN= mass = atomic number - neutrons

121 - 80 = 41

i haven't done this in a while so hope this helps :)

3 0
2 years ago
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
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