Answer : The temperature will be, 392.462 K
Explanation :
According to the Arrhenius equation,

or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at
= 
= rate constant at
= 
= activation energy for the reaction = 66.41 kJ/mole = 66410 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature = 293 K
= final temperature = ?
Now put all the given values in this formula, we get:
![\log (\frac{3K_1}{K_1})=\frac{66410J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{293K}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B3K_1%7D%7BK_1%7D%29%3D%5Cfrac%7B66410J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B293K%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)

Therefore, the temperature will be, 392.462 K
The pressure of the gas will increase with the temperature as each molecule moves around. think of hot air balloons the way they inflate that big balloon is a flame underneath the gas which increases temperature and pressure.
Answer:
a) 40 %
b) 
c) 
Explanation:
For a) we will have to calculate the <u>molar mass</u> of
, so the first step is to find the <u>atomic mass</u> of each atom and multiply by the <u>amount of atoms</u> in the molecule.
C => 12*(6) = 72
H => 1*(12) = 12
O => 6*(16) = 96
Molar mass = 180 g/mol
Then we can calculate the percentage by mass:

For b) we have to start with the <u>reaction of glucose</u>:

Then we have to convert the grams of glucose to moles, the moles of glucose to moles of carbon dioxide and finally the moles of carbon dioxide to grams. To do this we have to take into account the<u> following conversion ratios</u>:
-) 180 g of glucose = 1 mol glucose
-) 1 mol glucose = 6 mol carbon dioxide
-) 1 mol carbon dioxide = 44 g carbon dioxide

For C, we have to start with the conversion from grams of glucose to moles, the moles of glucose to moles of oxygen and finally the moles of oxygen to molecules. To do this we have to take into account the <u>following conversion ratios</u>:
-) 180 g of glucose = 1 mol glucose
-) 1 mol glucose = 6 mol oxygen
-) 1 mol oxygen = 6.023x10^23 molecules of O2

Answer:
17.6% is the percentage error
Explanation:
The percentage error is used to determine accuracy of a measurement (That is, how closely is the measure to the real or theoretical value). The equation is:
|Measure - Real| / Real * 100
In the problem, the measure was 22.7cm³ and real value is 19.3cm³. Solving for percentage error:
|22.7cm³ - 19.3cm³| / 19.3cm³ × 100
<h3>17.6% is the percentage error</h3>