There are 16electrons in 02
The delta H of -484 kJ is the heat given off when 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O. You don't have anywhere near that much reactants, only 1/4 as much
<span>actual delta H = 0.34 moles H2 x (-484 kJ / 2 moles H2) = 823 kJ </span>
<span>delta E = delta H - PdeltaV = 823 kJ - 0.41 kJ = 822 kJ</span>
Given that the question gives us concentration (M) and volume, we can use these so get moles. remember that Molarity (M)= mol/ Liters, so if we want the moles, then
moles= M x L
moles= 1.13 x 5.60= 6.33 moles KI
now to get the mass in grams, we need the molar mass of potassium iodide (KI) which can be determined using the periodic table and add the masses of each atom.
molar mass KI= 39.1 + 127= 166.1 g/mol
6.33 mol (166 g/ 1 mol)= 1050 grams KI
Answer:
Exocytosis
Explanation:
Some molecules are simply too big to move via a transport protein or the plasma membrane. To carry these macromolecules in or out of the cell, cells employ two more active transport pathways. Macromolecules or big particles are transported across the plasma membrane via Vesicles transport or other cytoplasmic structures. They are of two types, Endocytosis and Exocytosis
From the given information, Exocytosis is the right answer.
It is the process of vesicles combining with the plasma membrane thereby releasing their contents to the exterior of the cell. When a cell creates components for export, such as proteins, or when it gets rid of a waste product or a toxin, exocytosis occurs. Exocytosis is the process by which newly generated membrane proteins and membrane lipids are transported on top of the plasma membrane.
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
