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dmitriy555 [2]
3 years ago
13

A generic element, G, is composed of two isotopes, 132G and 128G. 132G has a natural abundance of 90% and an isotopic mass of 13

1.90 amu, and 128G has a natural abundance of 10% and an isotopic mass of 127.90 amu. What is the average atomic mass of this element?
Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
8 0

Answer: The average atomic mass of this element is 131.5 amu

Explanation:

Mass of isotope 1 = 131.90 amu

% abundance of isotope 1 = 90% = \frac{90}{100}=0.9

Mass of isotope 2 = 127.90 amu

% abundance of isotope 2 = 10% = \frac{10}{100}=0.1

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(131.90\times 0.9)+(127.90\times 0.1)]=131.5amu

The average atomic mass of this element is 131.5 amu

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When humans burn fossil fuels, most of the carbon quickly enters the_______
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Atmosphere.

Explanation:

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All the atoms have negatively charged electrons which are counter balanced by the presence of equal number of protons in the nucleus of an atom and thus, the element is neutral in nature and possess no charge.

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3 years ago
A. Identify the steps of the scientific method?
DedPeter [7]

Answer:

1. The steps of the scientific method consist of seven things: Ask a question, get the research needed for a experiment, (What are you trying to find out? What materials do I need to complete the experiment? What are the procedures needed to complete the experiment?) create a hypothesis, (your educated guess on what you think the results of the experiment will be) conduct a experiment (complete multiple different trials preferably three) to test your hypothesis, make observations and record data during the experiment, draw a conclusion, (Was your hypothesis correct? Yes or no and explain why it is wrong or right) present your findings.

2. A hypothesis is a educated guess that the scientist believe will be the answer at the end of the experiment which is why they conduct a experiment in the first place to find viable data that will support their hypothesis.

3.  Predictions consist of theories that will test the hypothesis (educated guess) because they're the reasoning to why you believe your guess is correct.

4. A control group consists of variables that do not go through change during a experiment, things that remain the same.

5. Data can be presented in many different ways in the form of graphs, charts, or a research paper. You find the data by completing multiple different trials in a experiment, to make sure you have valid results to write down in your data research.

6. Remember that your hypothesis is your educated guess at the beginning of the experiment, what YOU thought was going to happen during the experiment and if the data you received during the experiment supports your hypothesis.

7. Your procedures, perhaps you measured a variable wrong, perhaps you used to much, or to little of a variable, all depends on your experiment.

8. Having people review and test your data by conducting their own experiments will not only make sure that your data is valid but it insures you even more of your conclusion, making it more accepted by other scientists.

Hope this helps.

4 0
3 years ago
An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder
seraphim [82]

Answer:

\Delta H=-11897J

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

\Delta H=\Delta U+V\Delta P

Whereas the change in the internal energy is computed by:

\Delta U=nCv\Delta T

So we compute the initial and final temperatures for one mole of the ideal gas:

T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K  }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K  }{n}

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J

Then, the volume-pressure product in Joules:

V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J

Finally, the change in the enthalpy for the process:

\Delta H=-7140J-4757J\\\\\Delta H=-11897J

Best regards.

7 0
3 years ago
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