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My name is Ann [436]
3 years ago
8

After reading “The Crisis, No. 1,” what do you think Paine means by the line, "These are the times that try men’s souls"? Cite e

vidence from the text to support your analysis.
Chemistry
2 answers:
lidiya [134]3 years ago
6 0

Answer:

America was going through a difficult period when Paine wrote "The Crisis, No. 1." Some soldiers were abandoning the fight for freedom, and therefore, Paine tried to convince them through these lines: "These are the times that try men's souls. The summer soldier and the sunshine patriot will, in this crisis, shrink from the service of their country; but he that stands by it now, deserves the love and thanks of man and woman."

Paine agreed that the time was full of troubles and problems, and it was testing the hard work of the people. However, he appealed to them to not abandon the war. He argued that hope was not yet lost, and victory was within reach as long as Americans continued their struggle.

Explanation: answer on plato/edmentum

posledela3 years ago
3 0
Paine is referring to the difficulties facing the colonists to stand up against the British Empire against imposed taxes and foreign rule. He says "<span>Tyranny, like hell, is not easily conquered; yet we have this consolation with us, that the harder the conflict, the more glorious the triumph". This emphasises his point that the fight is against tyranny, although the battle may be difficult.</span>
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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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2 years ago
How many carbon atoms are there in a diamond (pure carbon) with a mass of 55 mg ?
kari74 [83]
1 mol of Carbon = 12 grams.
x mol of Carbon = 55 grams

12*x = 1 * 55
x = 55/12
x = 4.583333 mols of carbon

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1/4.5833333 = 6.02 * 10^23/x
x = 4.58333* 6.02*10^23
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We are given with
M1 = 1.00 M

M2 = 0.300 M
V2 = 2.00 L

We are asked to get V1

Using material balance
M1 V1 = M2 V2
Substituting the given values
1.00 V1 = 0.300 M (2.00 L)
V1 = 0.600 L or 600 mL
THe volume needed is 600 mL<span />
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Goshia [24]

OPTION C

<h2>POTENTIAL ENERGY </h2>

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