Explanation:
temperature changes affect seawater density as water cools its density increases. As water cools H2O molecules pack more closely together because the molecules are vibrating less at low temperatures and take up less volume. The same number of water molecules in smaller volume results higher density
Answer:
The answer to your question is 308.1 g of KmNO₄
Explanation:
Data
Volume = 6 L
Concentration = 0.325 M
Molar mass of Potassium permanganate = 39 + 55 + (16 x 4) = 158 g
Process
1.- Calculate the moles of KMnO₄
Formula
Molarity = moles/volume
moles = molarity x volume
moles = 0.325 x 6
moles = 1.95
2.- Calculate the grams of KMnO₄
158 g KMnO₄ ------------------ 1 mol
x ------------------ 1.95
x = (1.95 x 158)/1
x = 308.1 g
Answer:
Time (months) is the independent variable
Growth (measured) is the dependent variable.
Explanation:
Let y = growth
Let x = time
The result will yield a relation of the form
y = y(x), or y is a function of x.
x, the time can be controlled. The experimenter can decide how many months to conduct the experiment. It is the independent variable.
y, the growth is a response to the amount of time allowed for growth. It depends on x, therefore y is the dependent variable.
Given:
0.060 mol of NiC2O4
Ksp = 4 x 10⁻¹⁰
1.0 L of solution
Kf of Ni(NH3)6 2⁺ = 1.2 x 10⁹
<span>NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- </span>
<span>NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp </span>
<span>Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf </span>
Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48
K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶<span>
</span>0.48 = (0.060)² / [NH3]⁶<span> ... (dissolved C2O4 2- = 0.060M)
</span><span>[NH3]</span>⁶<span> = (0.060)</span>²<span> / 0.48 = </span>0.0036 / 0.48 = 0.0075
NH3 = ⁶√0.0075
NH3 = 0.44 M