The question is incomplete, here is the complete question:
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1 M⁻¹s⁻¹

Suppose a vessel contains SO₃ at a concentration of 1.44 M. Calculate the concentration of SO₃ in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
<u>Answer:</u> The concentration of
in the vessel after 0.240 seconds is 0.24 M
<u>Explanation:</u>
For the given chemical equation:

The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken= 0.240 second
[A] = concentration of substance after time 't' = ?
= Initial concentration = 1.44 M
Putting values in above equation, we get:
![14.1=\frac{1}{0.240}\left (\frac{1}{[A]}-\frac{1}{1.44}\right)](https://tex.z-dn.net/?f=14.1%3D%5Cfrac%7B1%7D%7B0.240%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B1.44%7D%5Cright%29)
![[A]=0.245M](https://tex.z-dn.net/?f=%5BA%5D%3D0.245M)
Hence, the concentration of
in the vessel after 0.240 seconds is 0.24 M
Answer:

Explanation:
Since the temperature is held constant, we only need to focus on the volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula is:

Originally, the gas had a volume of 150 milliliters and a pressure of 3.0 atmospheres. We can substitute these values into the left side of the equation.

The original gas was compressed to a volume of 50 milliliters, but we don't know the volume.

Now, we need to solve for the new pressure (P₂). Multiply on the left side first.

Since we are solving for the pressure, we need to isolate the variable. It is being multiplied by 50 mL. The inverse of multiplication is division. Divide both sides by 50 mL.


The units of milliliters will cancel.


The new pressure is <u>9 atmospheres.</u>
Im assuming that the 2 is a sub of the Al and not the # for O and that u made a type of AL which is Al
its been a while but
Al2O3 + 6HCl = 2ALCl3 + 3H2O
Answer:
<u>Molar</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>diprotic</u><u> </u><u>acid</u><u> </u><u>is</u><u> </u><u>4</u><u>2</u><u>4</u><u> </u><u>grams</u>
Explanation:
[hint: <u>diprotic</u><u> </u><u>acid</u><u> </u><u>only</u><u> </u><u>contains</u><u> </u><u>2</u><u> </u><u>hydrogen</u><u> </u><u>protons</u><u>]</u>
Ionic equation:

first, we get moles of potassium hydroxide in 28.94 ml :

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.
Therefore, <u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>c</u><u>i</u><u>d</u><u> </u><u>t</u><u>h</u><u>a</u><u>t</u><u> </u><u>r</u><u>e</u><u>a</u><u>c</u><u>t</u><u>e</u><u>d</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>8</u><u>7</u><u>4</u><u>3</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u>.</u>

for the molar mass:
Answer:
Explanation:
T₁ = 100 + 273 = 373K
T₂ = 273 + 119 = 392 K
V₁ = initial volume
V₂ = Final volume
P₁ = P₁
P₂ = .85P₁
Using gas law equation


V₂ = 1.236 V₁
% increase in volume
= V₂-V₁ / V₁ x 100
= (1.236 V₁ - V₁ / V₁)x 100
= .236 x 100
= 23.6 % .