NaOH was titrated to the phenolphthalein endpoint with a known concentration of HCl. The student conducting the experiment place
1 answer:
Answer:
The concentration of NaOH solution is 3,65M
Explanation:
The neutralization of NaOH with HCl is:
NaOH + HCl → NaCl + H₂O
You are titrating 12,67 mL HCl 2,74M with your NaOH solution, the moles of HCl in the beaker are:
12,67 mL ≡ 0,01267L× = <em>0,0347 moles HCl </em>≡ 0,0347 moles NaOH
The volume that you require for total neutralization of 12,67 mL HCl 2,74M is:
13,65 mL - 4,50 mL = <em>9,50 mL </em>≡ 0,0095L
Thus, the concentration of NaOH solution is:
= <em>3,65M</em>
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Answer:
A. CaO
Explanation:
Reactant are the species that take part in and also undergoes the reaction to form species which are known as products.
Symbolic representation of the chemical reaction with the help of the formulas and symbols is known as chemical equation. In the chemical equation, reactants are written at left-hand side as well as products are expressed or written on right-hand side. The coefficients which are written before the formulas or the symbols are known as stoichiometric numbers.
Thus, in the reaction shown below:-
<u>Thus, The product of the reaction is:- CaO</u>
Answer:
The solution to the given problem is done below.
Explanation:
a)
i) =( 0.002 μg / L )( 1mg / 1000 μg )( 1L / kg )( 1000 mil / 1 billion) = 0.002 ppb
ii) =( 0.002 μg / L )( 1mg / 1000 μg )( 1L / kg )( 1,000,000 mil / 1 trillion) = 2 ppt
iii) =( 0.002 μg / L )( 1 mole / 540g ) = 3.7 x μM.
b)
i) =( 0.002 μg / g ) = 0.002 ppm
ii) In solids, ppb = μg/kg
=( 0.002 μg / g )( 1000 mil/ 1 billion) = 2 ppb