The volume of a pond being studied for the effects of acid rain is 35 kiloliters (kL). There are 1,000 liters (L) in 1 kL and 1 mc032-1.jpg 106 microliters (mL) in 1 L. What is the volume of this pond in microliters?

Answer:

The specific heat capacity of the metal is 1.143 J/g°C

Explanation:

A typical excersise of calorimetry:

Q = m . C . ΔT

ΔT = Final T° - Initial T°

m = mass

In this case, the heat released by the metal is gained by the water to rise its temperature.

Qmetal = Qwater

(We consider that metal was at the same T° of water)

22.5 g  . C . (65°C - 25.55°C) = 25 g .  4.184 J/g°C . (35.25°C - 25.55°C)

22.5 g  . C . 39.45°C = 25g . 4.184 J/g°C . 9.7°C

887.625 g.°C . C = 1014.62J

C = 1014.62J / 887.625 g.°C

C = 1.143 J/g°C

Given:

Distance of solvent front = 68 mm

Distance of unknown = 48 mm

To determine:

The rf value

Explanation:

The retention factor or the rf value is given by the ratio of distance traveled by the unknown to the distance traveled by the solvent front

RF = distance by unknown/distance by solvent

RF = 48/68 = 0.706

Ans: the RF value is 0.706

Final answer:

The Rf (Retention factor) value can be calculated using the given distances traveled by the solvent and the substance. The Rf value in this case is approximately 0.71.

Explanation:

The Rf value or Retention factor value in chromatography can be calculated with the given parameters of solvent front and the distance the substance travelled from the original spot. The formula to calculate the Rf value is: Rf = distance traveled by the substance / distance traveled by the solvent. So, in this case, it would be: Rf = 48mm / 68mm which is approximately 0.71.

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Here's your asnwer.

50

Answer:

posible mente es una equivocacion porque dijo que estaban refriados y les dió una medicina para la tos

Answer:the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.

Explanation:The given chemical reaction is:

Pb^4+(aq) + 2 Ce^3+(aq) -> Pb^2+(aq) + 2 Ce^4+(aq)

The standard cell potential (E°cell) for this electrochemical reaction is 0.06 V.

The standard cell potential for a galvanic cell can be calculated using the Nernst equation:

E°cell = E°cathode - E°anode

In this reaction, Pb^4+(aq) is being reduced to Pb^2+(aq), so it is the reduction half-reaction, and Ce^3+(aq) is being oxidized to Ce^4+(aq), so it is the oxidation half-reaction.

The standard reduction potentials (E°) for the half-reactions are as follows:

For the reduction half-reaction:

Pb^4+(aq) + 2 e^- -> Pb^2+(aq) E°red = x (we'll solve for x)

For the oxidation half-reaction:

2 Ce^3+(aq) -> 2 Ce^4+(aq) + 2 e^- E°red = 1.44 V (This value is usually given)

Now, plug these values into the Nernst equation:

E°cell = E°cathode - E°anode

0.06 V = x - 1.44 V

Now, solve for x:

x = 0.06 V + 1.44 V

x = 1.50 V

So, the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.