Answer:
Hence,
1) removed
2) drained through the stopcock
3) get eye level with
4) slow the draining
5) switch to a new flask
Explanation:
After mixing the solutions in a separatory funnel, the stopper should be removed and the liquid should be drained through the stopcock, and the layers allowed to separate. When you get close to the interface between the layers, get eye level with the funnel and turn over to slow the draining heat up until the first layer is collected. Switch to a new flask get eye level with it to collect the second layer.
Answer:
D & E
Explanation:
I think this is dealing with latent heat and D & E would be the range where you will find solid and liquid phases in equilibrium, cuz it starts as gas at from A to B, B to C is gas and liquid equilibrium, C to D is liquid, D to E solid and liquid, and then E to F is solid.
Answer:
[K2SO4] = 4,75x10⁻⁷M ; [K⁺] = 9.50x10⁻⁷M ; [SO4⁻²] = 4,75x10⁻⁷M
SO4⁻²: 0.045ppm ; K⁺: 0.037ppm
[SO4⁻²] = 4,70x10⁻⁷ F
Explanation:
Determine the equation
K2SO4 → 2K⁺ + SO4⁻²
Each mole of potassium sulfate generates two moles of potassium cation and one mole of sulfate anion
Molar mass K2SO4: 174.26 g/m
Moles of K2SO4: grams / molar mass
2.07x10⁻⁴g / 174.26 g/m = 1.18x10⁻⁶ moles
Molarity: Moles of solute in 1 L of solution
1.18x10⁻⁶ moles / 2.5 L = 4,75x10⁻⁷M (K2SO4)
K⁺ : 4,75x10⁻⁷M . 2 = 9.50x10⁻⁷M
SO4⁻²: 4,75x10⁻⁷ M
1 mol of K2SO4 has 2 moles of K and 1 mol of SO4
1.18x10⁻⁶ moles of K2SO4 has 1.18x10⁻⁶ moles of SO4 and 2.37x10⁻⁶ moles of K.
1.18x10⁻⁶ moles of SO4⁻² are 1.13x10⁻⁴ grams (moles. molar mass)
2.37x10⁻⁶ moles of K are 9.26x10⁻⁵ grams (moles. molar mass)
These grams are in 2.5 L of water, so we need μg/mL to get ppm
2.5 L = 2500 mL
1.13x10⁻⁴ grams SO4⁻² are 113.35 μg (1 μg = 1x10⁶ g)
9.26x10⁻⁵ grams K⁺ are 92.6 μg (1 μg = 1x10⁶ g)
113.35 μg /2500 mL = 0.045ppm
92.6 μg /2500 mL = 0.037ppm
Formal concentration of SO4⁻² :
Formality = Number of formula weight of solute / Volume of solution (L)
(1.13x10⁻⁴ grams / 96.06 g ) / 2.5 L = 4,70x10⁻⁷ F
The correct option is B.
A voltaic cell is an electrochemical cell that uses a chemical reaction to produce electrical energy. The important parts of a voltaic cell: The anode is an electrode where oxidation occurs. The cathode is an electrode where reduction occurs.
A galvanic cell is an electrochemical cell that converts the free energy of a chemical process into electrical energy. A photogalvanic cell is one that generates species photochemically which react resulting in an electrical current through an external circuit.
Learn more about energy conversion here brainly.com/question/961052
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Answer:
B) chemical energy to electrical energy
Explanation:
Answer:
newtons 3rd law of motion
Explanation:
Answer:
Unsweetened tea
Explanation:
A concentrated solution is a solution that has more of the solute in it than the solvent (water).
A diluted solution is a solution that has more solvent (water) than the solute.
From the above we can say that the sweet tea contains more of the tea than water i.e it is concentrated hence, the taste is sweet. On the other hand, we can say that the unsweetened tea contains more water than the tea i.e it is diluted hence, the unsweetened taste.
a. protons.
b. electrons.
C. neutrons.
d. isotopes.
Answer:
Atomic number is protons
Explanation:
Protons = positive charge