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Marysya12 [62]
3 years ago
6

Which is a factor on determining the average atomic mass of an element

Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

Relative abundance of each isotopes of an element.

Explanation:

Consider the following example:

Uranium is used in nuclear reactors and is a rare element on earth. Uranium has three common isotopes. If the abundance of 234U is 0.01% the abundance of 235U is 0.71% and the abundance of 238U is 99.28% what is the average atomic mass of uranium

Abundance of U²³⁴ = 0.01%

Abundance of U²³⁵ = 0.17%

Abundance of U²³⁸ = 99.28%

Average atomic mass = ?

Solution:

Average atomic mass of uranium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) +(abundance of 3rd isotope × its atomic mass) / 100

Average atomic mass of uranium= (234×0.01)+(235×0.71)+(238×99.28)/100

Average atomic mass of uranium=  2.34 + 166.85 + 23628.64 / 100

Average atomic mass of uranium= 23797.83 / 100

Average atomic mass of uranium= 237.98 amu.

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0.257 L

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The Environmental Protection Agency has determined that safe drinking
dmitriy555 [2]

Answer:

b) \bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}

The confidence interval for this case is given (6.21, 6.59)

So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter

c) Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.

Explanation:

We assume that part a is test the claim. And we can conduct the following hypothesis test:

Null hypothesis: \mu =7

Alternative hypothesis \mu \neq 7

The statistic is to check this hypothesi is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

We know the following info from the problem:

\bar X = 6.4 , s=0.5, n =30

Replacing we got:

t = \frac{6.4-7}{\frac{0.5}{\sqrt{30}}}= -6.573

And the p value would be:

p_v= 2*P(Z

Since the p value is very low compared to the significance assumed of 0.05 we have enough evidence to reject the null hypothesis that the true mean is equal to 7 moles/liter

Part b

The confidence interval is given by:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}

The confidence interval for this case is given (6.21, 6.59)

So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter

Part c

Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.

6 0
3 years ago
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