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Stells [14]
3 years ago
12

Why must we do the a lot of quantity urine​

Chemistry
1 answer:
serg [7]3 years ago
7 0

Answer:

because

ExplanatioN:

<em>BeCaUsE</em>

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Answer: 64

Explanation:

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What are the rules for writing chemical formulas
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1: the symbol of the metal should be written first.
2: the valency of the elements or radicals should be interchanged.
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In which group on the periodic table does the atom pictures belong?
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A sample of argon has a volume of 1.2 L at STP. If the temperature is increased to 21 c and the pressure is lowered to 0.80 atm,
Nuetrik [128]

Answer:

The new volume is 1.62 L

Explanation:

Boyle's law says:

"The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure." It is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T}=k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. So this law indicates that the quotient between pressure and temperature is constant.

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T}=k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law.

\frac{P*V}{T}=k

Having an initial state 1 and a final state 2 it is possible to say that:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

Standard temperature and pressure (STP) indicate pressure conditions P = 1 atm and temperature T = 0 ° C = 273 ° K. Then:

  • P1= 1 atm
  • V1= 1.2 L
  • T1= 273 °K
  • P2= 0.80 atm
  • V2= ?
  • T2= 21°C= 294 °K

Replacing:

\frac{1 atm* 1.2 L}{273K} =\frac{0.8 atm*V2}{294K}

Solving:

V2=\frac{1 atm*1.2 L}{273 K} *\frac{294 K}{0.8 atm}

V2= 1.62 L

<u><em>The new volume is 1.62 L</em></u>

<u><em></em></u>

8 0
3 years ago
How many grams of RbOH are present in 35.0 mL of a 5.50 M RbOH solution?​
Ganezh [65]
I think 5.50 M x 35.0 mL x molar mass of RbOH = mass (g)
8 0
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