Answer:
1.76
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.</em>
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Step 1: Calculate the molarity of HI(aq)
M = mass of solute / molar mass of solute × liters of solution
M = 0.660 g / 127.91 g/mol × 0.300 L
M = 0.0172 M
Step 2: Write the acid dissociation reaction
HI(aq) ⇄ H⁺(aq) + I⁻(aq)
HI is a strong acid, so [H⁺] = 0.0172 M
Step 3: Calculate the pH
pH = -log [H⁺]
pH = -log 0.0172
pH = 1.76
Answer:
m = 180 g
Explanation:
Given data:
Energy absorbed = 108 J
Mas of gold = ?
Initial temperature = 25°C
Final temperature = 29.7 °C
Specific heat capacity of gold = 0.128 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT =29.7 °C - 25°C
ΔT = 4.7 °C
108 J = m ×0.128 J/g.°C ×4.7 °C
108 J = m ×0.60 J/g
m = 108 J/0.60 J/g
m = 180 g
Answer:
Fe2(SO4)3 + 3BaCl2 → 2FeCl3 + 3BaSO4
%Mass
Ar C = 12 g/mol, Mr C₄H₁₀ = 58 g/mol, Ar H = 1 g/mol
or
