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3 years ago

Out of 13 students how many saw themselves in the horoscope?

1 answer:

5 0

Answer: Astrology is not astronomy! ... Astronomers and other scientists know that stars many light years* away have no effect on the ... Imagine a straight line drawn from Earth through the Sun and out into space way ... The 13 constellations in the zodiac. ... on the phases of the Moon), each month got a slice of the zodiac all to itself.

Explanation: Hope this helps

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____________ heat is the amount of heat required to change the temperature of 1 gram of a substance by 1°C, and it is related to

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Answer:

specific heat.

Explanation:

Definition:

The amount of heat required to raise the temperature of one gram of substance by one degree is called specific heat.

Formula:

Q = m. c. ΔT

Q = amount of heat required

m = mass of substance

c = specific heat of substance

ΔT = change in temperature

The substance with greater value of specific heat require more heat to raise the temperature while the substance with lower value will raise its temperature very quickly by absorbing smaller heat.

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A student does an experiment to determine the molar solubility of lead(II) bromide. She constructs a voltaic cell at 298 K consi

attashe74 [19]

Answer:

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

$E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}$

E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.

At equilibrium, E = 0, therefore:

$E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\$

[red] = 0.010 M

The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.

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