<u>Answer:</u> The volume of hydrogen gas required for the given amount of ethylene gas is 113 L
<u>Explanation:</u>
At STP:
1 mole of a gas occupies 22.4 L of volume
We are given:
Volume of ethylene = 113 L
For the given chemical equation:
By Stoichiometry of the reaction:
of ethylene reacts with of hydrogen gas
So, 113 L of ethylene gas will react with = 
of hydrogen gas
Hence, the volume of hydrogen gas required for the given amount of ethylene gas is 113 L
Answer:
Following are the responses to the given question:
Explanation:
In the given question the OH could be generated as the H-bond with both the N (nitrogen), which is used to place its N as the partially positive value(+). It is used to simplifies the addition for either its AA with the imine C. In the above H-bond, the w N is not required as the possibility to use the OCH3
<span>When pKas of polyprotic intermediates have a difference of 2 or more you just average them using the equation: pH = (pKa2 + pKa3) / 2 </span>
<span>pKa2 = -log(Ka2) ; pKa3 = -log(Ka3) </span>
<span>so, for this problem, REGARDLESS OF THE CONCENTRATION GIVEN, the answer is: </span>
<span>pH = (7.2076+12.3767) / 2 </span>
<span>pH = 9.79</span>
Answer:
b: Neither atoms will have a full valence shell
Explanation:
I think sooo
Fat: 1 gram = 9 calories Protein: 1 gram = 4 calories Carbohydrates: 1 gram = 4 calories Alcohol: 1 gram = 7 calories"
So, (12 g carbohydrates)*4 Cal=48 Cal; (9 g fat) *9 Cal=81 Cal; (9 g protein)*4 Cal=36 Cal
48 Cal + 81 Cal + 36 Cal = 165 Cal in one glass of milk
