The second one.
Remember, temperature is the AVERAGE kinetic energy. So, looking for highest average kinetic energy means looking for highest temperature, which is 55.
Note: the amount of the substance does not determine the average kinetic energy.
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 />
Answer:
Here's what I get
Explanation:
(g) Titration curves
I can't draw two curves on the same graph, but I can draw two separate curves for you.
The graph in part (d) had an equivalence point at 20 mL.
In the second titration, the NaOH was twice as concentrated, so the volume to equivalence point would be half as much — 10 mL.
The two titration curves are below.
(h) Evidence of reaction
HCl and NaOH are both colourless.
They don't evolve a gas or form a precipitate when they react.
The student probably noticed that the Erlenmeyer flask warmed up — a sign of a chemical change.
The balanced chemical reaction is written as:
4Al + 3O2 = 2Al2O3
To determine the mass of oxygen gas that would react with the given amount of aluminum metal, we use the initial amount and relate this amount to the ratio of the substances from the chemical reaction. We do as follows:
moles Al = 16.4 g ( 1 mol / 26.98 g ) = 0.61 mol Al
moles O2 = 0.61 mol Al ( 3 mol O2 / 4 mol Al ) = 0.46 mol O2
mass O2 = 0.46 mol O2 ( 32.0 g / mol ) = 14.59 g O2
Therefore, to completely react 16.4 grams of aluminum metal we need a minimum of 14.59 grams of oxygen gas.