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Sergio039 [100]
3 years ago
9

A sample of fluorine contains isotopes with different masses; F-19 and F-18. In the sample 94% of the

Chemistry
1 answer:
zalisa [80]3 years ago
7 0

The relative mass of the sample : 18.94

<h3>Further explanation </h3>

The elements in nature have several types of isotopes

Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

Atomic mass is the average atomic mass of all its isotopes

Mass atom X = mass isotope 1 . % + mass isotope 2.%

F-19 = 94% of the sample

F-18 = 100%-94%=6% of the sample

The relative mass :

\tt = 0.94\times 19+0.06\times 18\\\\=18.94

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This plateau is where a phase change occurs. Likely, this cooling curve observes the changing of a liquid into a solid (so that the temperature is quantifiable). As something cools, it’s losing energy to the surroundings; when a phase change occurs, like liquid to solid, energy isn’t lost directly from the atom, but instead the energy maintained by free motion of the atom is used. This lack of motion reduces it into a lattice (all while maintaining a constant internal energy), finalizing the phase change, after which energy is lost directly from the atom again.
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What is the atomic mass of a carbon isotope that has 6 protons and 7 neutrons?
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The atomic mass of a carbon isotope that has 6 protons and 7 neutrons is<u> </u><u>13</u>

Explanation:

Remember that whilst the atomic number represents the number of protons in an atom, the mass number represents the summation of protons and neutrons particles in the atomic nuclei. Therefore, in this case, the carbon will have a mass number of;

6 +  7 = 13

Isotopes of an element usually have the same atomic number but different mass numbers -because they have slightly different numbers of neutrons. An example is isotopes of Carbon; C-14 and C-12

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Which condition must be met in order for an equation to be balanced? The elements in the reactants are the same as the elements
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Calculate the freezing point of a solution made from 220g of octane (C Hua), molar mass = 114,0 gmol dissolved in 1480 g of benz
stiv31 [10]

Answer: Freezing point of a solution will be -1.16^0C

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(5.50-T_f)^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

K_f = freezing point constant = 5.12^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

(5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}

(5.50-T_f)^0C=6.68

T_f=-1.16^0C

Thus the freezing point of a solution will be -1.16^0C

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