Answer:
130ml of HCl(36%) in 4.90L solution => pH = 1.50
Explanation:
Need 4.90L of HCl(aq) solution with pH = 1.5.
Given pH = 1.5 => [H⁺] = 10⁻¹·⁵M = 0.032M in H⁺
[HCl(36%)] ≅ 12M in HCl
(M·V)concentrate = (M·V)diluted
12M·V(conc) = 0.032M·4.91L
=> V(conc) needed = [(0.032)(4.91)/12]Liters = 0.0130Liters or 130 ml.
Mixing Caution => Add 131 ml of HCl(36%) into a small quantity of water (~500ml) then dilute to the mark.
Explanation:
The given data is as follows.
(NaCl) = 
(H-O=C-ONO) = 
(HCl) = 
Conductivity of monobasic acid is 
Concentration = 0.01 
Therefore, molar conductivity (
) of monobasic acid is calculated as follows.

= 
= 
= 
Also,
= 
= 
= 
Relation between degree of dissociation and molar conductivity is as follows.

= 
= 0.1254
Whereas relation between acid dissociation constant and degree of dissociation is as follows.
K = 
Putting the values into the above formula we get the following.
K = 
= 
= 
= 
Hence, the acid dissociation constant is
.
Also, relation between
and
is as follows.

= 
= 3.7454
Therefore, value of
is 3.7454.
Answer:
Error of parallax it usually occurs when the cylinder is above or below the eye level,and thus resulting in differences In reading the millimeters
Answer:
Mole Fraction (H₂O) = 0.6303
Mole Fraction (C₂H₅OH) = 0.3697
Explanation:
(Step 1)
Calculate the mole value of each substance using their molar masses.
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
200.0 g H₂O 1 mole
--------------------- x ------------------ = 11.10 moles H₂O
18.014 g
Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol
Molar Mass (C₂H₅OH): 46.068 g/mol
300.0 g C₂H₅OH 1 mole
---------------------------- x -------------------- = 6.512 moles C₂H₅OH
46.068 g
(Step 2)
Using the mole fraction ratio, calculate the mole fraction of each substance.
moles solute
Mole Fraction = ------------------------------------------------
moles solute + moles solvent
11.10 moles H₂O
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH
Mole Fraction (H₂O) = 0.6303
6.512 moles C₂H₅OH
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH
Mole Fraction (C₂H₅OH) = 0.3697
The chemical reaction is expressed as:
2H2 + O2 = 2H2O
To determine the amount of oxygen used in the reaction, we use the amount of water produced and the relation of the substances in the reaction we do as follows:
209 g H2O ( 1 mol / 18.02 g ) ( 1 mol O2 / 2 mol H2O ) ( 32 g / 1 mol ) = 185.57 g O2