Answer:
10−8 M.
Explanation:
In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × 10−8 M.
The theoretical yield of I2 in the reaction would be 0.23 g
<h3>Theoretical yield</h3>
This refers to the stoichiometric yield of a reaction.
From the equation of the reaction:
Ca(IO3)2 + 10 KI + 12 HCl → 6 I2 + CaCl2 + 10 KCl + 6 H2O
The mole ratio of Ca(IO3)2 and I2 is 1: 6
Mole of 15.00 mL, 0.0100 M Ca(IO3)2 = 15/1000 x 0.0100
= 0.00015 mole
Equivalent mole of I2 = 0.00015 x 6
= 0.009 mole
mass of 0.0009 I2 = 0.0009 x 253.809
= 0.23 g
More on stoichiometric calculations can be found here: brainly.com/question/6907332
Answer:
1.022ppm is the unknown concentration of the metal
Explanation:
Based on Lambert-Beer law, the increasing in signal of a detector is directly proportional to its concentration.
The unknown concentration (X) produces a signal of 0.255
99mL * X + 1mL * 100ppm / 100mL produces a signal of 0.502
0.99X + 1ppm produce 0.502, thus, X is:
0.255 * (0.99X + 1 / 0.502) =
X = 0.503X + 0.508
0.497X = 0.508
X =
1.022ppm is the unknown concentration of the metal
Answer:
Q sln = 75.165 J
Explanation:
a constant pressure calorimeter:
∴ m sln = m Ba(OH)2 + m HCl
∴ molar mass Ba(OH)2 = 171.34 g/mol
∴ mol Ba(OH)2 = (0.06 L)(0.3 mol/L) = 0.018 mol
⇒ mass Ba(OH)2 = (0.018 mol)(171.34 g/mol) = 3.084 g
∴ molar mass HCl = 36.46 g/mol
∴ mol HCl = (0.06 L)(0.60 mol/L) = 0.036 mol
⇒ mass HCl = (0.036 mol)(36.46 g/mol) = 1.313 g
⇒ m sln = 3.084 g + 1.313 g = 4.3966 g
specific heat (C):
∴ C sln = C H2O = 4.18 J/g°C
∴ ΔT = 26.83°C - 22.74°C = 4.09°C
heat absorbed (Q):
⇒ Q sln = (4.3966 g)(4.18 J/g°C)(4.09°C)
⇒ Q sln = 75.165 J
