Answer:
a measure of concentration equal to the gram equivalent weight per liter of solution.
Explanation:
Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.
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Answer: Many fronts cause weather events such as rain, thunderstorms, gusty winds and A weather front is a transition zone between two different air masses at the Lifted warm air ahead of the front produces cumulus or cumulonimbus clouds
Explanation:
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Answer is: the hydronium ion concentratio is 1.71×10⁻⁷ mol/dm³ and pH<6.76.
The Kw (the ionization constant of water) at 40°C is 2.94×10⁻¹⁴ mol²/dm⁶ or 2.94×10⁻¹⁴ M².
Kw = [H₃O⁺] · [OH⁻].
[H₃O⁺] = [OH⁻] = x.
Kw = x².
x = √Kw.
x = √2.94×10⁻¹⁴ M².
x = [H₃O⁺] = 1.71×10⁻⁷ M; concentration of hydronium ion.
pH = -log[H₃O⁺].
pH = -log(1.71×10⁻⁷ M).
pH = 6.76.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity an aqueous solution.
Answer:
P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)
Explanation:
Given:
P₁ = 90 atm P₂ = ?
V₁ = 18 Liters(L) L₂ = 12 Liters(L)
=> decrease volume => increase pressure
=> volume ratio that will increase 90 atm is (18L/12L)
T₁ = 272 Kelvin(K) T₂ = 274 Kelvin(K)
=> increase temperature => increase pressure
=> temperature ratio that will increase 90 atm is (274K/272K)
n₁ = moles = constant n₂ = n₁ = constant
P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)
By rule of sig. figs., the final answer should be rounded to an accuracy equal to the 'measured' data value having the least number of sig. figs. This means P₂ ≅ 100 atm based on the given value of P₁ = 90 atm.
Answer:
14 mol e⁻
Explanation:
Step 1: Write the balanced half-reaction for the reduction of permanganate to manganese
8 H⁺(aq) + 7 e⁻ + MnO₄⁻(aq) ⇒ Mn(s) + 4 H₂O(l)
Step 2: Calculate the moles corresponding to 110 g of manganese
The molar mass of Mn is 55 g/mol.
110 g × 1 mol/55 g = 2 mol
Step 3: Calculate the number of moles of electrons needed to produce 2 moles of Mn
According to the half-reaction, 7 moles of electrons are required to produce 1 mole of Mn.
2 mol Mn × 7 mol e⁻/1 mol Mn = 14 mol e⁻
