I can't really answer this question without knowing the given picture. Since the pictures are not shown,I can only give the CORRECT representation of the compound. Since it contains C, H and O atoms only, this is an organic compound. Since there are two atoms of O, we can conclude that it has a carboxylic group. Rearranging the elements, it must be written as CH₃COOH which is names as acetic acid. The structural formula is shown in the attached picture.
Answer:
Excess Reagent = oxygen
Explanation:
Limiting reagent: The substance that is totally consumed when the reaction is completed.
Excess reagent: The substance left after the limiting reagent is consumed completely
The balanced chemical equation for formation of water is as follow:
This means when 2 moles of hydrogen reacts with 1 mole of oxygen, 2 moles of water is produced.
Hence the ratio in which hydrogen and oxygen gas reacts is 2:1
Now if 2 mole hydrogen require 1 mole of oxygen ,then 4 mole hydrogen need 2 mole of oxygen.
or
Here 5 mole of oxygen is reacting but only 2 mole is required .
Oxygen is in excess.
Answer:
it will take 444.44 min to reach 0.25 M from 0.75 M
Explanation:
since the reaction constant is k= 6.00*10⁻³ L*mol⁻¹min⁻¹. For a reaction rate of the form
-dCa/dt=k*Caⁿ
doing a dimensional analysis
[dCa/dt]= mol/(L*min) = [k]*[Caⁿ] = L/(mol*min) * (mol/L)ⁿ
then only n=2 can comply with the dimensional analysis, therefore we get a the second order reaction . Thus
-dCa/dt=k*Ca²
-dCa/Ca² =k*dt
-∫dCa/Ca² =k*∫dt
(1/Ca₂ - 1/Ca₁)= k*Δt
Δt= 1/k*(1/Ca₂ - 1/Ca₁)
replacing values
Δt= 1/k*(1/Ca₂ - 1/Ca₁) = 1/(6.00*10⁻³ L*mol⁻¹min⁻¹)*(1/0.25 M - 1/0.75 M)= 444.44 min
Let us check each statement one by one
a) Sb has a lower ionization energy but a higher electronegativity than I. : As per values given : Definitely Sb has lower ionization energy however the electronegativity of Sb is lower than that of iodine
b) Sb has a higher ionization energy but a lower electronegativity than I. FAlse:
Sb has lower ionization energy than I
c) Sb has a lower ionization energy and a lower electronegativity than I. True
d) Sb has a higher ionization energy and a higher electronegativity than I. False
The question is incomplete, here is the complete question:
Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO₂H) and 0.230 mol of sodium formate (NaCO₂H) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is 1.77 × 10⁻⁴
a) 2.099
b) 10.463
c) 3.546
d) 2.307
e) 3.952
<u>Answer:</u> The pH of the solution is 3.546
<u>Explanation:</u>
We are given:
Moles of formic acid = 0.370 moles
Moles of sodium formate = 0.230 moles
Volume of solution = 1 L
To calculate the molarity of solution, we use the equation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
= negative logarithm of acid dissociation constant of formic acid = 3.75
pH = ?
Putting values in above equation, we get:
Hence, the pH of the solution is 3.546