Rate = k * [A]^2 * [B]^1
<span>Use the data from any trial to calculate k. </span>
<span>k = (rate)/([A]^2 * [B]^1) </span>
<span>E.g., for Trial 1, we have </span>
<span>rate = 3.0×10−3 M/s </span>
<span>[A] = 0.50 M </span>
<span>[B] = 0.010 M </span>
<span>Plug those numbers in and crank out the answer. </span>
<span>Now with the calculated value of k, calculate the initial rate for [A] = 0.50 M and [B] = 0.075 M </span>
<span>rate = k * [A]^2 * [B]^1 </span>
<span>k = calculated value </span>
<span>[A] = 0.50 M </span>
<span>[B] = 0.075 M</span>
The rate of product formation by an enzyme-catalyzed reaction would be increased by temperature (until it reaches optimim temperature)
Answer:
a) The formal charge on N is 0 in both species.
Explanation:
The formal charge is calculated using the formular;
FC = V-N-B/2
Where;
V = Number of Valence electrons
N = number of nonbonding valence electrons
B = total number of electrons shared in bonds
In NO2-;
The formal charge of N is given as;
FC = 5 - 2 - 6/2
FC = 0
In HNO2
The formal charge of N is given as;
FC = 0
The correct option is;
a) The formal charge on N is 0 in both species.
Electrolysis of water<span> is the </span><span>decomposition reaction, because from one molecule (water) two molecules (hydrogen and oxygen) are produced. Water is separeted into two molecules:
</span>Reaction of reduction at cathode: 2H⁺(aq) + 2e⁻<span> → H</span>₂(g<span>).
</span><span><span>Reaction of oxidation at anode: 2H</span></span>₂<span><span>O(l) → O</span></span>₂<span><span>(g) + 4H</span></span>⁺(<span><span>aq) + 4e</span></span>⁻.<span><span>
</span></span>
An ion is an atom that has lost or gained an electron so the relation ship you would see would be a negative or a positive one depending on if it gained or lost an electron.<span />
