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GREYUIT [131]
3 years ago
15

A hypothetical element has two main isotopes with mass numbers of 62 and 65. If 66.00% of the isotopes have a mass number of 62

amu, what atomic weight should be listed on the periodic table for this element? Answer in units of amu.
Chemistry
1 answer:
Radda [10]3 years ago
7 0

Answer is 63.04amu.

Explanation: We are given in the question that an element has two main isotopes 1 and 2.

Mass Number of isotope 1 = 62 amu

Mass Number of 2 isotope = 65 amu

% abundance of isotope 1 = 66%

Fractional abundance of isotope 1 = 0.66

Total fractional abundance = 1

Fractional abundance of isotope 2 = (1-0.66)

                                                          = 0.34

For calculating he atomic mass of element, we use

\text{atomic mass of element}=\left [(\text{fractional abundance of isotope 1})\times (\text {atomic mass of isotope 1)} \right ]+\left [(\text{fractional abundance of isotope 2})\times (\text {atomic mass of isotope 2}) \right ]+...

We have only 2 isotopes for an element, so the formula stops at isotope 2 only.

By putting the values in the above equation, we get

\text{atomic mass of element}=(62amu\times 0.66)+(65amu}\times 0.34)

Atomic mass of element = 63.04amu

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3 0
2 years ago
What does lower case k stand for?
DaniilM [7]
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4 0
3 years ago
What process is being shown by water being given off from each bond site?
34kurt
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4 0
3 years ago
A student heats a sample of Copper (II) sulfate in a crucible and records the data shown in the table. What is the complete form
liberstina [14]

Explanation:

Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.

Mass of sample (CuSO4 * nH2O)

= 152.00g - 128.10g = 23.90g.

Mass of water loss during heating

= 152.00g - 147.60g = 4.40g.

Molar mass of H2O = 18g/mol

Moles of H2O in sample

= 4.40g / (18g/mol) = 0.244mol.

Mass of anhydrous sample (CuSO4)

= 23.90g - 4.40g = 19.50g

Molar mass of CuSO4 = 159.61g/mol

Moles of CuSO4 in sample

= 19.50g / (159.61g/mol) = 0.122mol.

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6 0
2 years ago
Read 2 more answers
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
cricket20 [7]

Answer:

21.2 gm

Explanation:

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

butane is the hydrocarbon C4H10  

in combustion, we react hydrocarbons with O2 to form CO2 and H2O

so

C4H10  + O2---------------->  CO2 + H2O

BALANCE

2C4H10 + 1302--------> 8CO2 + 10 H2O

the molar mass of CO2 is 12 + 16X2 = 44

64.1 gm of CO2 is

64.1/44 = 1.46 MOLES OF  CO2,

FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE  IT IS A

8:2 OR 4:1 RATIO.  THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2

SO

THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES

THE MOLAR MASS OF BUTANE IS 58.12

0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm

6 0
2 years ago
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