Molarity of a solution if 124.86 g of rbf are dissolved into a solution of water that has a final volume of 2.00L is 0.59.
<h3>What is molarity?</h3>
Molarity is used for dilute aqueous solutions held at a constant temperature. In general, the difference between molarity and molality for aqueous solutions near room temperature is very small and it won't really matter whether you use a molar or molal concentration.
MOLARITY = no of moles of solute/volume of soln in litres
No of moles of rbf = 124.6/104.46
= 1.19
Volume of soln = 2
Molarity=1.19/2 = 0.59
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For the first question, its revise the hypothesis and develop a new experiment to test it
For the second question, the fertilizer independent variable
I think it is the third one.
Answer: 15062.4 Joules
Explanation:
The quantity of heat energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = ?
Mass of food = 200.0g
C = 4.184 j/g°C
Φ = (Final temperature - Initial temperature)
= 83.0°C - 65.0°C = 18°C
Then, Q = MCΦ
Q = 200.0g x 4.184 j/g°C x 18°C
Q = 15062.4 J
Thus, 15062.4 joules of heat energy was contained in the food.
Answer:
3.62x10⁻⁷ = Kb
Explanation:
The acid equilibrium of a weak acid, HX, is:
HX + H₂O ⇄ X⁻ + H₃O⁺
Where Ka = [X⁻] [H₃O⁺] / [HX]
And basic equilibrium of the conjugate base, is:
X⁻ + H₂O ⇄ OH⁻ + HX
Where Kb = [OH⁻] [HX] / [X⁻]
To convert Ka to Kb we must use water equilibrium:
2H₂O ⇄ H₃O⁺ + OH⁻
Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]
Thus, we can obtain:
Kw = Ka*Kb
Solving for Kb:
Kw / Ka = Kb
1x10⁻¹⁴ / 2.76x10⁻⁸ =
3.62x10⁻⁷ = Kb
