First, let's write down the balanced chemical reaction between the given reactants:
NO₂ + NO → N₂O + O₂
The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:
Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)
For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1
<u>Answer:</u> The element represented by M is Strontium.
<u>Explanation:</u>
Let us consider the molar mass of metal be 'x'.
The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol
It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:
The metal atom having molar mass as 87.62/mol is Strontium.
Hence, the element represented by M is Strontium.
To determine the Keq, we need the chemical reaction in the system. In this case it would be:
CO + 2H2 = CH3OH
The Keq is the ration of the amount of the product and the reactant. We use the ICE table for this. We do as follows:
CO H2 CH3OH
I .42 .42 0
C -0.13 -2(0.13) 0.13
-----------------------------------------------
E = .29 0.16 0.13
Therefore,
Keq = [CH3OH] / [CO2] [H2]^2 = 0.13 / 0.29 (0.16^2)
Keq = 17.51
Answer: the HCO3- to act as a base and remove excess H by the formation of H2CO3
Explanation:
H2CO3 in an aqueous solution is a buffer. This means the reaction is the following:
H2CO3 ------ HCO3- + H+
Then, the HCO3- that was formed acts as a base (absorbing a proton) like this
HCO3- + H+ ------- H2CO3
If there was an increase in H+, there would be an increase in the second reaction in an effort to neutralize that acid, thus making the H2CO3 more concentrated
To know the acidity of a
solution, we calculate the pH value. The formula for pH is given as:
<span>pH = - log [H+] where H+ must be in Molar</span>
We are given that H+ = 3.25 × 10-2 M
Therefore the pH is:
pH = - log [3.25 × 10-2]
pH = 1.488
Since pH is way below 7, therefore the solution
is acidic.
To find for the OH- concentration, we must
remember that the product of H+ and OH- is equivalent to 10^-14. Therefore,
[H+]*[OH-] = 10^-14 <span>
</span>[OH-] = 10^-14 / [H+]
[OH-] = 10^-14 / 3.25 × 10-2
[OH-] = 3.08 × 10-13 M
Answers:
Acidic
[OH-] = 3.08 <span>× 10-13 M</span>
