The given reaction:
<span>ch3ch2cooh (aq) ↔ ch3ch2coo- (aq) + h+ (aq)
is called a reversible reaction.
This means that, the reaction does not reach an end point.
In this type of reactions, reactants react together to form products, while products combine together to form reactants.
So, the reaction proceeds in both direction forming both reactants and products.</span>
Proton number = the atomic number (which is the smaller number
neutron number = the mass number (the bigger number) - the atomic number
number of electrons = the atomic number - the charge (it depends on the element but group 1 is +1 group 2 is +2 group 3 is +3 group five is -3 group six is -2 group seven is -1
i would solve the whole thing but its unclear hope this helps tho
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
Answer:
Empirical CHO
molecular C4H4O4
Explanation:
From the question, we know that it contains 41.39% C , 3.47% H and the rest oxygen. To get the % composition of the oxygen, we simply add the carbon and hydrogen together and subtract from 100%.
This means : O = 100 - 41.39 - 3.47 = 55.14%
Next is to divide the percentage compositions by their atomic masses.
C = 41.39/12 = 3.45
O = 55.14/16 = 3.45
H = 3.47/1 = 3.47
Now we divide by the smallest value which is 3.45. We can deduce that this will definitely give us an answer of 1 all through as the values are very similar.
Hence the empirical formula of Maleic acid is CHO
Now we go on to deduce the molecular formula.
To do this we need the molar mass. I.e the amount in grammes per one mole of the compound.
Now we can see that 0.378mole = 43.8g
Then 1 mole = xg
x = (43.8*1)/0.378 = 115.87 = apprx 116
[CHO]n = 116
(12 + 1 + 16]n = 116
29n = 116
n = 116/29 = 4
The molecular formula is thus C4H4O4
The new pressure : P₂ = 1038.39 mmHg
<h3>Further explanation</h3>
Given
1.5 L container at STP
Heated to 100 °C
Required
The new pressure
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure).
So P₁ = 1 atm = 760 mmHg
T₁ = 273 K
T₂ = 100 °C+273 = 373 K
Gay Lussac's Law
When the volume is not changed, the gas pressure is proportional to its absolute temperature
Input the value :
P₂=(P₁.T₂)/T₁
P₂=(760 x 373)/273
P₂ = 1038.39 mmHg
