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k0ka [10]
3 years ago
8

Each side of a square is increasing at a rate of 7 cm/s. At what rate is the area of the square increasing when the area of the

square is 16 cm2?
Chemistry
1 answer:
zepelin [54]3 years ago
3 0

Ans1) Each side of a square is increasing at a rate of 2 cm/s. At what rate is the area of the square increasing when the area of the square is 81 cm 2?. 2) A cylindrical tank with radius 5 m is being filled with water at a rate of 2 m 3 /min. wer:

Explanation:

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What is the molecular formula for chloroform? none of these choices CH3C CHCI3 CH2C12 CCI4
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Answer:

CHCI₃

Explanation:

Chloroform, IUPAC name, trichloromethane, is organic compound with the molecular  formula CHCl₃. It is colorless and sweet-smelling liquid having high density which is produced on a large scale precursor of PTFE , and for various refrigerants .

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A gas exerts a pressure of 1.8 atm at a temperature of 60 degrees celsius. What is the new temperature when the pressure of the
Kamila [148]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final temperature of the gas is 740 K</em></u>

Explanation:

Given:

Initial pressure of the gas (P_{1}) = 1.8 atm

Final pressure of the gas (P_{2})  = 4 atm

Initial temperature of the gas (T_{1}) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas (T_{2})

We know;

From the ideal gas equation;

we know;

P  × V = n × R × T

So;

we can tell from the above equation;

 <u>   P ∝ T</u>

(i.e.)

      <em> </em>\frac{P}{T}<em> = constant</em>

        \frac{P_{1} }{P_{2} } = \frac{T_{1} }{T_{2} }

Where;

P_{1}  = initial pressure of a gas

P_{2} = final pressure of a gas

T_{1} = initial temperature of a gas

T_{2} = final temperature of a gas

        \frac{1.8}{4} = \frac{333}{T_{2} }

   T_{2} =\frac{333*4}{1.8}

    T_{2} = 740 K

<u><em>Therefore the final temperature of the gas is 740 K</em></u>

8 0
3 years ago
What is the molar mass of copper(II)nitrate
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an element has an isotope with a mass of 203.973 amu and and abundance of 1.40%. another isotope has a mass of 205.9745 amu with
ryzh [129]

Answer is: the average atomic mass 217.606 amu.

Ar₁= 203.973 amu; the average atomic mass of isotope.

Ar₂ = 205.9745 amu.

Ar₃ = 206.9745 amu.

Ar₄ = 207.9766 amu.

ω₁ = 1.40% = 0.014; mass percentage of isotope.

ω₂ = 24.10% = 0.241.

ω₃ = 22.10% = 0.221.

ω₄ = 57.40% = 0.574.

Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.  

Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.

Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.

Ar = 217.606 amu.

But abundance of isotopes is greater than 100%.

It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.

7 0
3 years ago
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