1. rancidification fixation of water is CHEMICAL CHANGE
2. Tearing of paper is PHYSICAL CHANGE
3. Rusting if iron is CHEMICAL CHANGE
4. Electrolysis of water is CHEMICAL CHANGE
<u>Given:</u>
Concentration of Ba(OH)2 = 0.348 M
<u>To determine:</u>
pOH of the above solution
<u>Explanation:</u>
Based on the stoichiometry-
1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion
Therefore, concentration of OH- ion = 2*0.348 = 0.696 M
pOH = -log[OH-] = - log[0.696] = 0.157
Ans: pOH of 0.348M Ba(OH)2 is 0.157
Answer;
FeSO4 (aq) + H2 (g)
Explanation;
The reaction between iron and dilute sulfuric acid gives a salt and hydrogen gas. (iron (ii) sulfate and hydrogen gas).
The equation for the reaction is;
Fe(s) + H2SO4 (aq) = FeSO4(aq) +H2(g)
Answer:
3 × 10¯¹⁰
Explanation:
9×10² ÷ 3×10¹²
The above expression can be simplified as follow:
9×10² ÷ 3×10¹²
Recall:
9 = 3²
9×10² ÷ 3×10¹² = 3²×10² ÷ 3×10¹²
Recall:
a^m ÷ a^n = a^(m – n)
3²×10² ÷ 3×10¹² = 3^(2 – 1) × 10^(2 – 12)
= 3¹ × 10¯¹⁰
Recall:
a¹ = a
3¹ × 10¯¹⁰ = 3 × 10¯¹⁰
Answer:
Number of moles is 3.0moles
Explanation:
Volume (v) = 67.2L
Pressure at STP = 1atm
Temperature at STP = 273.15K
R = 0.082J/mol.K
Number of moles (n) = ?
From ideal gas equation,
PV = nRT
P = pressure
V = volume of the gas
n = number of moles
R = ideal gas constant
T = temperature of the gas
PV = nRT
solving for n,
n = PV / RT
n = (1.0 * 67.2) / (0.082 * 273.15)
n = 67.2 / 22.398
n = 3.0 moles
The number of moles present is 3.0
