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IRINA_888 [86]
3 years ago
10

Please help, due soon!

Chemistry
1 answer:
algol133 years ago
5 0

Answer:

three types of stimuli consists of

Mechanical,

thermal,

and chemical.

Explanation:

I really hope this helps have an amazing day, dont forget YOU ARE WORTH IT

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Margaret [11]
Periodic properties
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If there are 3 cats but they each die of different diseases one of heart fail, second of Brian tumor and 3rd of experiment gone
Snowcat [4.5K]

Answer:

none

Explanation:

It is given that there are 3 cats.

Each of the cats die with different diseases. One of cats died due to heart failure, the second cats died because of brain tumor and the third cat died due to an experiment gone wrong.

So there are none of the cats left out. All the cats died due to different reasons.

7 0
3 years ago
What is the electron configuration silver, Ag? exeception​
Setler79 [48]

Answer

2,8,18,18,1

Explanation:

7 0
3 years ago
0.005 molar Ca(OH)2 solution is used in the titration.calculate the pH oh solution
Rus_ich [418]

Ca(OH)₂: strong base

pOH = a . M

a = valence ( amount of OH⁻)

M = concentration

Ca(OH)₂ ⇒ Ca²⁺ + 2OH⁻ (2 valence)

so:

pOH = 2 x 0.005

pOH = 0.01

pH = 14 - 0.01 = 13.99

6 0
3 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
4 years ago
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