A sample of fluorine gas occupies 810 milliliters at 270 K and 1.00 atm. What volume does the gas occupy when the pressure is doubled, and the temperature increases to 400 K

The answer to your question is Volume = 600 ml

Explanation:

Data

Volume 1 = 810 ml

Temperature 1 = 270°K

Pressure 1 = 1 atm

Volume 2 = ?

Pressure = 2 atm

Temperature 2 = 400°K

Process

1.- To solve this problem, use the Combine gas law.

               P1V1 / T1 = P2V2 / T2

- Solve for V2

                V2 = P1V1T2 / T1P2

2.- Substitution

                 V2 = (1)(810)(400) / (270)(2)

3.- Simplification

                 V2 = 324000 / 540

4.- Result

                 V2 = 600 ml

600mL

Explanation:

The following were obtained from the question:

V1 = 810mL

P1 = 1atm

T1 = 270K

V2 =?

P2 = 2atm(since the pressure is doubled)

T2 = 400K

The new volume can be obtained by doing the following :

P1V1/T1 = P2V2/T2

1 x 810/270 = 2 x V2 /400

Cross multiply to express in linear form

270 x 2 x V2 = 810 x 400

540 x V2 = 324000

Divide both side by 540

V2 = 324000/540

V2 = 600mL

The new volume of the gas is 600mL

a plasma is a warm ionized gasoline consisting of approximately equal numbers of definitely charged ions and negatively charged electrons. the characteristics of plasmas are notably unique from the ones of ordinary neutral gases in order that plasmas are considered a awesome "fourth kingdom of remember."

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"