Taking into account the ideal gas law, the pressure is 2.52 atm.
An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas are related by a simple formula called the ideal gas law. This equation relates the three variables if the amount of substance, number of moles n, remains constant. The universal constant of ideal gases R has the same value for all gaseous substances. The numerical value of R will depend on the units in which the other properties are worked.
P×V = n×R×T
In this case, you know:
- P=?
- V= 500 L
- n= 52.1 moles
- R= 0.082
- T= 22 C= 295 K (being 0 C=273 K)
Replacing in the ideal gas law:
P×500 L = 52.1 moles ×0.082 ×295 K
Solving:
P= (52.1 moles ×0.082 ×295 K)÷ 500 L
<u><em>P= 2.52 atm</em></u>
Finally, the pressure is 2.52 atm.
Learn more about ideal gas law:
Energy released from changing the phase of a
substance from the liquid phase to solid phase can be calculated by using the
latent heat of fusion. The heat of fusion of iron is calculated as follows:
<span>Energy = mL(f)</span>
<span>9840 cal (4184 J / 1 cal) = .200 kg x L(f)</span>
L(f) = 2.06 x10^8 J/kg
Answer:
25.0 g of sodium carbonate are present in 220 ml of the solution.
Explanation:
Hi there!
I have found the complete question on the web:
<em>Calculate the volume in milliliters of a 1.07 M sodium carbonate solution that contains 25.0g of sodium carbonate Na2CO3. Be sure your answer has the correct number of significant digits.</em>
First, let's find how many moles of sodium carbonate have a mass of 25.0 g.
The molar mass of Na₂CO₃ is 106 g/mol.
So, if 106 g of sodium carbonate is equal to 1 mol, then 25.0 g will be:
25.0 g · (1 mol / 106 g) = 0.236 mol Na₂CO₃
The solution contains 1.07 mol sodium carbonate in 1 liter.
So, if 1.07 mol sodium carbonate is present in 1 l solution, then, 0.236 mol will be present in:
0.236 mol · (1000 ml / 1.07 mo) = 221 ml (220 ml without any intermediate rounding).
25.0 g of sodium carbonate are present in 220 ml of the solution.