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lys-0071 [83]
3 years ago
7

NEED HELP ASAP

Chemistry
2 answers:
ira [324]3 years ago
8 0

Answer:

i think its the 3 thired one sorry if rong but good luck

Explanation:

pychu [463]3 years ago
7 0

Answer:

The sample is made up of a single element. t or f

Explanation:

Forgot about the t or f!   Its true.

Dam bro.....

1. T

2. F

3. T

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A sample of a compound contains ​0.300 mol​ ​of carbon​ and ​1.20 mol​ ​of hydrogen​. What is the ​empirical formula of the comp
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What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances
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The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope                    mass amu        Relative abundance

1                                77.9                     14.4

2                               81.9                     14.3

3                               85.9                      71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9  

% abundance of isotope 1 = 14.4% = \frac{14.4}{100}=0.144

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = \frac{14.3}{100}=0.143

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = \frac{71.3}{100}=0.713

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[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]

A=84.2amu

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

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