Answer:
(a) A strong acid
Explanation:
We have given the pH of the solution is 2.46
pH=2.46
So the concentration of 
solution having H+ concentration more than 
is acidic
Since in the given solution, H+ concentration is 0.00346 M which is more than 10^{-7}[/tex] so this is an acidic solution
Note-The concentration of 
decide the behavior of the solution that is, it is acidic or basic
Answer:
a. The original temperature of the gas is 2743K.
b. 20atm.
Explanation:
a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:
T₁n₁ = T₂n₂
<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>
<em />
<em>Replacing with values of the problem:</em>
T₁n₁ = T₂n₂
X*7.1g = (X+300)*6.4g
7.1X = 6.4X + 1920
0.7X = 1920
X = 2743K
<h3>The original temperature of the gas is 2743K</h3><h3 />
b. Using general gas law:
PV = nRT
<em>Where P is pressure (Our unknown)</em>
<em>V is volume = 2.24L</em>
<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>
R is gas constant = 0.082atmL/molK
And T is absolute temperature (2743K)
P*2.24L = 0.20mol*0.082atmL/molK*2743K
<h3>P = 20atm</h3>
<em />
<span>Displaced volume :
</span>Final volume - <span>Initial volume
</span>13.45 mL - 12.00 mL => 1.45 mL
Mass = 4.50 g
Therefore:
density = mass / volume
D = 4.50 / 1.45
<span>D = 3.103 g/mL </span>
Answer:
The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm
Explanation:
Using the ideal gas law
PV=nRT
if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore
Inicial state ) P₁V=n₁RT
Final state ) P₂V=n₂RT
dividing both equations
P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁
now we have to determine P₁ and n₂ /n₁.
For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)
p ar₁ = P₁ * x ar₁ → P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm
n₁ = n ar₁ + n N₁ = n ar₁ + n ar₁ = 2 n ar₁
n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁
n₂ /n₁ = 3/2
therefore
P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2 = 2.25 atm
P₂= 2.25 atm
