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ehidna [41]
3 years ago
15

Thermochemistry

Chemistry
1 answer:
max2010maxim [7]3 years ago
4 0
THe Answer is 105 J

hope this helps and please brainliest, I was first :)
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Nh4cl = 117.165 gram ( solution is attached )

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What is the [OH-] of solution with pH of 3.4?
arsen [322]

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(A) 4×0.0001M

Explanation:

[OH-] = 10^-3.4 which is equivalent to 4×0.0001M

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An ion of oxygen- 16 contains 8 protons and has a 2- charge. How many electrons does it have?
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i would say 10, so the anser is A.

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2 years ago
A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

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

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
3 years ago
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