how many atmospheres of pressure would there be if you started at 5.75 atm and changed the volume from 5 l to 1 l ?

Answers

Answer 1

The pressure would be 28.75 atm if the volume is changed from 5 L to 1 L, starting from an initial pressure of 5.75 atm.

To solve this problem, we can use the combined gas law equation, which relates the pressure, volume, and temperature of a gas:

P1V1/T1 = P2V2/T2

where P1 and V1 are the initial pressure and volume, T1 is the initial temperature, P2 and V2 are the final pressure and volume, and T2 is the final temperature. Since the temperature is constant in this problem, we can simplify the equation to:

P1V1 = P2V2

Substituting the given values, we get:

5.75 atm × 5 L = P2 × 1 L

Solving for P2, we get:

P2 = (5.75 atm × 5 L) / 1 L = 28.75 atm.

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Related Questions

a 1.25 g sample of co2 is contained in a 750. ml flask at 22.5 c. what is the pressure of the gas, in atm?

Answers

The pressure of gas is 1.05 atm when a 1.25 g sample of CO₂ is contained in a 750ml flask at 22.5°C.

Molecular weight of CO₂ is 1.25g ,Volume of CO₂ is 750ml,Temperature of CO₂ is 22.5°C and the gas constant is 0.08206 L atm/mol K.

Using the ideal gas law equation the pressure is found to be 1.05 atm.

To calculate the pressure of the gas, we can use the ideal gas law equation: [tex]PV=nRT[/tex]
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we need to convert the volume to liters by dividing by 1000: 750 ml = 0.75 L.
Next, we need to calculate the number of moles of CO₂ present in the flask. We can use the molecular weight of CO₂ to convert from grams to moles:

[tex]1.25 * (1 /44.01 ) = 0.0284 mol[/tex]
Now we can plug in the values into the ideal gas law equation:

[tex]PV=nRT[/tex]
[tex]P * 0.75 L = 0.0284 mol * 0.08206 L*atm/mol*K * (22.5 + 273.15) K[/tex]
Simplifying and solving for P, we get:
[tex]P = (0.0284 * 0.08206 * 295.65) / 0.75 = 1.05 atm[/tex]
Therefore, the pressure of the gas in the flask is 1.05 atm.

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What is the concentration (in molality) of an aqueous solution of NaCl made by adding
4.56 g of NaCl to enough water to give 20.0 mL of solution. Assume the density of the
solution is 1.03 g/mL

Answers

Answer:

data given

mass of NaCl 4.56

dissolved volume 20ml(0.02l)

density of solution 1.03g/ml

Required molality

Explanation:

molarity=m/mr×v

where

m is mass

mr molar mass

v is volume

now,

molarity=4.56/58.5×0.02

molarity =3.9

: .molarity is 3.9mol/dm^3

According to molal concentration, the concentration (in molality) of an aqueous solution of NaCl is 0.0047 mole/kg.

What is molal concentration?

Molal concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molal concentration is moles/kg.

The molal concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molal concentration is calculated by the formula, molal concentration=mass/ molar mass ×1/mass of solvent in kg.

In terms of moles, it's formula is given as molal concentration= number of moles /mass of solvent in kg.

Substitution in formula gives the answer but first mass of solution is determined which is density×volume= 1.03×20=20.6 g , mass of solvent= 20.6-4.56=16.05, thus molal concentration=4.56/58.5×1/16.05=0.0047 moles/kg.

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