Answer:
The characteristics of cathode rays do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube.
Answer:
4.00 is the pH of the mixture
Explanation:
The ethyl amine reacts with HNO3 as follows:
C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻
To solve this question we need to find the moles of ethyl amine and the moles of HNO3:
<em>Moles C2H5NH2:</em>
0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine
<em>Moles HNO3:</em>
0.201L * (0.025mol/L) = 0.005025 moles HNO3
That means HNO3 is in excess. The moles in excess are:
0.005025 moles HNO3 - 0.00500 moles ethyl amine =
2.5x10⁻⁵ moles HNO₃
In 50 + 201mL = 251mL = 0.251L:
2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]
As pH = -log [H+]
pH = -log 9.96x10⁻⁵M
pH = 4.00 is the pH of the mixture
This is a missing part of your question:
The equilibrium system between sulfur dioxide gas, oxygen gas, and sulfur trioxide gas is given.
So you need the equilibrium balanced equation of SO2, O2, SO3 reaction:
First, we will start with the original equation which is not balanced yet (to understand how we get it):
SO2 + O2 ↔ SO3
Here the number of O atom is not equal at the to sides
So we will start to balance our equation by make the number of O atom equal each other on both sides:
So we will start to put 2SO3 instead of SO3
and put 2SO2 instead of SO2 to balance also the S atom on both sides
So we will get this:
2SO2(g) + O2(g) ↔ 2SO3(g) (This is our equilibrium balanced equation)
know we have a number of O atom equals on each side = 6
and the sulfur equals on each side = 2
8.509 kilocalories I think
