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Luden [163]
3 years ago
7

When a substance is oxidized it is called a(n): select one:

Chemistry
1 answer:
BlackZzzverrR [31]3 years ago
4 0
When a substance is oxidized, its oxidation number increases. The reduction in oxidation number is due to lose of electron(s) to another substance in the reactant side. Substance which gets oxidize must reduce others. Thus it is a reducing agent.
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Which element requieres the least amount of energy to remove the outermost electron
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Answer: Sodium

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HELP ASAP!! What happens to the value of K when pressure increases??
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The value turns into P

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An aluminum oxide component must not fail when a tensile stress of 16.5 mpa is applied. determine the maximum allowable surface
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90 per square surface of the inner difference between both.

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2 years ago
In Vinegar Analysis lab, at the end point of a titration, there is still a drop of NaOH solution hanging on the tip of burrette.
podryga [215]

Answer:

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7 0
3 years ago
A 0.773 mol sample of Xe(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the
soldi70 [24.7K]

Answer : The value of q\text{ and }\Delta U is 193.8 J and 193.8 J respectively.

Explanation : Given,

Moles of sample = 0.773 mol

Change in temperature = 20.1 K

First we have to calculate the heat absorbed by the system.

Formula used :

q=n\times c_v\times \Delta T

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.773 mol

\Delta T = Change in temperature = 20.1 K

c_v = heat capacity at constant volume of Xe (mono-atomic molecule) = \frac{3}{2}R

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

q=0.773mol\times \frac{3}{2}\times 8.314J/mol.K\times 20.1K

q=193.8J

Now we have to calculate the change in internal energy of the system.

\Delta U=q+w

As we know that, work done is zero at constant volume. So,

\Delta U=q=193.8J

Therefore, the value of q\text{ and }\Delta U is 193.8 J and 193.8 J respectively.

4 0
3 years ago
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