This family (ethane, propane, butane, etc) of materials is likely to have following set of properties.
- The alkanes are non- polar solvents.
- The alkanes are immiscible in water but freely miscible in other non-polar solvent .
- The alkanes are consisting of weak dipole dipole bonds can not breaks the strong hydrogen bond.
- The alkanes having only carbon (C) and hydrogen (H) atom which is bonded by a single bonds only.
- The alkanes posses weak force of attraction that is weak van der waals force of attraction.
The ethane, propane, butane, belong to alkanes family.The alkanes are also considers as saturated hudrocarbons. Ethane is found in gaseous stae Ethane is the second alkane followed by propane followed by butane.
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Nitrogen, for example, is a gas that liquefies at about −200° C and freezes around −210° C, whereas bismuth is a solid melting at 271° C and boiling at about 1,560° C.
The original mass of krypton 81 that is present in the ice is 6.70 grams.
<h3>How do we calculate original mass?</h3>
Original mass of any substance will be calculated as below for the decomposition reaction is:
N = N₀(1/2)ⁿ, where
N = remaining mass of krypton-81 = 1.675g
N₀ = original mass of krypton-81 = ?
n will be calculated as:
n = T/t, where
T = total time period = 458,000 years
t = half life time = 229,000 years
n = 458,000/229,000 = 2
Now putting all these values on the above equation, we get
N₀ = 1.675 / (1/2)²
N₀ = 6.70 g
Hence required mass is 6.70 g.
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Independent variables don't change
Dependent variables change as a result of the alteration made during the experiment
Answer:
12.34 amu
Explanation:
Let the 1st isotope be A
Let the 2nd isotope be B
Let the 3rd isotope be C
From the question given above, the following data were obtained:
1st Isotope (A):
Mass of A = 12.32 amu
Abundance (A%) = 19.5%
2nd isotope (B):
Mass of B = 13.08 amu
Abundance (B%) = 26.23%
3rd isotope (C):
Mass of C = 11.99 amu
Abundance (C%) = 54.27%
Atomic mass of X =?
The atomic mass of the element X can be obtained as follow:
Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]
= [(12.32 × 19.5)/100] + [(13.08 × 26.23)/100] + [(11.99 × 54.27)/100]
= 2.402 + 3.431 + 6.507
= 12.34 amu
Thus, the atomic mass of the element X is 12.34 amu
