Answer:
H₂O Gained Electron.
H₂O is <span>Bronsted-Lowry Base.
Explanation:
Due to amphoteric nature of water it can act as acid when reacted with strong base, also it can act as base when reacted with strong acid.
In given statement water is treated with strong acid hence it is acting as Bronsted-Lowery Base, as it accepting H</span>⁺. So those species which accepts proton are called as Bronsted-Lowry Base and those which donated proton are called as <span>Bronsted-Lowry Acid.
</span>
H₂O + H⁺ → H₃O⁺
Answer:
C. The samples contain the same substances in a fixed proportion.
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Answer:
% composition of glucose = 85 %
Explanation:
Given Data
moles of glucose (C₆H₁₂O₆) = 1.3 mole
Total mass of mixture = 276 g
percent composition of glucose (C₆H₁₂O₆) = ?
Solution
Number of moles are given so we find out the mass of C₆H₁₂O₆
<em>mass = moles × molar mass</em>
Molar mass of glucose = 180.156 g/mol
mass = 1.3 mol × 180.156 g/mol
mass = 234.2 g
Now we find out the percent composition of glucose
% composition = (mass of compound ÷ mass of mixture) × 100
% composition = ( 234.2 g ÷ 276 g) × 100
% composition = 0.85 × 100
% composition = 85 %
Answer:
Explanation:
H2CO3 = H+ + HCO3- H2CO3 = 0.530 M Ka =4.4 X10^-7
Ka =[H+][A-]/[HA] [H+] = [A-]
0.530 M x 4.4 X10^-7= [H+}^2
2.32 X10^-7 = [H+}^2
23.2 X10^-8 = [H+}^2
4.83 X 10^-4 = [H+]
pH = - log 4.83 X 10^-4
pH = -(.68-4)
pH =-(-3.32)
pH= 3.32
