Answer:
150ml
Explanation:
For this question,
NaOH completely dissociates. It is a strong base
HCl also completely dissociates. It is a strong acid
So we have this equation
m1v1 = m2v2 ----> equation 1
M2 = 2m
V1= ??
M2 = 6m
V2 = 50m
When we input these into equation 1, we have:
2m x v1 = 6m x 50ml
V1 = 6m x 50ml/2
V1 = 300/2
V1 = 150ml
Therefore NaOH that is required to neutralize the solution of hydrochloric acid is 150ml.
Thank you
Answer:
2.78 x 10²³
Explanation:
1 mole contains 6.02 x 10²³ hydrogen atoms => 0.46 mole contains 0.46(6.02 x 10²³) hydrogen atoms or 2.78 x 10²³ atoms.
Caution => When to use H vs H₂ => This problem is specific for 'hydrogen atoms' but some may simply say hydrogen. In such cases use H₂ or 'molecular hydrogen' is the focus. it's a matter of semantics, H vs H₂.
Answer:
Explanation:
Oxyacids are acid containing oxygen; they are also known as acid-alcohol or acid-phenol. As said earlier, the strength of these acids increases with increases in the polarity of these compounds. So, what makes the polarity is as a result of the electronegative substituents attached to it. Halogen family possesses the highest electronegativity in the periodic table, and electronegativity decreases down the group.
The ranking of the oxyacids in order of decreasing acid strength from strongest to weakest acid is:
HClO3 > . HClO2 > HClO > HBrO
Solution :
Given :
Amount of anserine solution = 0.200 M
pH value is = 7.20
Preparation of 0.04 M solution of anserine from the 0.2 M solution.
0.2 M x 
= 0.04 M x 1000 ml
= 200 ml
So the 200 ml of 0.2 M anserine solution is required to prepare0.04 M of anserine.
0.1 M x = 0.04 x 1000 ml
= 400 ml
Therefore, 400 ml of HCl is needed.
Answer:
Explanation:
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
ionisation constant = 1.36 x 10⁻⁴ .
molecular weight of lactic acid = 90 g
moles of acid used = 20 / 90
= .2222
it is dissolved in one litre so molar concentration of lactic acid formed
C = .2222M
Let n be the fraction of moles ionised
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionisation constant Ka
Ka = nC x nC / C - nC
= n²C ( neglecting n in the denominator )
n² x .2222 = 1.36 x 10⁻⁴
n = 2.47 x 10⁻²
nC = 2.47 x 10⁻² x .2222
= 5.5 x 10⁻³
So concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per litre .
