The volume occupied by the N₂ gas change : <u>5 L</u>
<h3>Further explanation </h3>There are several gas equations in various processes:
PV = NkT
N =number of gas particles
n = number of moles
R = gas constant (8,31.10 ^ 3 J / kmole K
k = Boltzmann constant (1,38.10 ^ -23)
n = = N / No
n = m / m
n = mole
No = Avogadro number (6.02.10 ^ 23)
m = mass
M = relative molecular mass
In the same temperature and pressure, in the same volume conditions, the gas contains the same number of molecules
So it applies: the ratio of gas volume will be equal to the ratio of gas moles
<h3> V1: V2 = n1: n2 </h3>
At a fixed temperature, the gas volume is inversely proportional to the pressure applied
<h3> p1.V1 = p2.V2 </h3>
When the gas pressure is kept constant, the gas volume is proportional to the temperature
<h3> V1 / T1 = V2 / T2 </h3>
When the volume is not changed, the gas pressure in the tube is proportional to its absolute temperature
<h3> P1 / T1 = P2 / T2 </h3>
Combined with Boyle's law and Gay Lussac's law
P1.V1 / T1 = P2.V2 / T2
P1 = initial gas pressure (N / m2 or Pa)
V1 = initial gas volume (m3)
P2 = gas end pressure
V2 = the final volume of gas
T1 = initial gas temperature (K)
T2 = gas end temperature
From the attached picture, it shows that if the stopvalve is opened then the two gas systems / containers will be interconnected (almost the same as the related vessel system). So that the total volume filled by the two gases (O₂ and N₂) becomes:
total volume = O₂ container volume + N₂ container volume
total volume = 3 L + 2 L
total volume = 5 L
<h3>Learn more </h3>
a description of Charles’s law
Charles's law
State Boyle's, Charles's, and Gay-Lussac's laws
Keywords: containers, valve, O₂, N₂, volume, gas
Answer:
The correct answer is "1.0100".
Explanation:
Let the volume of mixture be 100 ml.
then,
The volume of DMSO will be 10 mL as well as that of water will be 90 mL.
DMSO will be:
=
=
The total mass of mixture will be:
=
=
Density of mixture will be:
=
=
=
hence,
Specific gravity of mixture will be:
=
=
=
Answer:- 0.138 M
Solution:- The buffer pH is calculated using Handerson equation:
acts as a weak acid and
as a base which is pretty conjugate base of the weak acid we have.
The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:
Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.
So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.
Let's plug in the values in the equation:
Taking antilog:
On cross multiply:
[base] = 1.38(0.10)
[base] = 0.138
So, the concentration of the base that is required to make the buffer is 0.138M.
The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.
The volume of water is .
Since, 1 gal= 3785.41 mL
Thus,
Density of water is 1 g/mL thus, mass of water will be .
Since, 1 grams of chlorine → grams of water.
1 g of water → g of chlorine and,
of water →86.6 g of chlorine
Since, the solution is 9% chlorine by mass, the volume of solution will be:
Thus, volume of chlorine solution is 9.62\times 10^{2} mL.