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hjlf
3 years ago
11

How do isotopes affect the average atomic mass of an element?

Chemistry
1 answer:
kirill [66]3 years ago
7 0

Answer:

The versions of an element with different neutrons have different masses and are called isotopes. The average atomic mass for an element is calculated by summing the masses of the element's isotopes, each multiplied by its natural abundance on Earth.

Explanation:

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Non examples of chemical formulas
Bezzdna [24]

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Explanation:

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What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115 g of
Citrus2011 [14]

Answer:

freezing point   (°C) of the solution =  - 3.34° C

Explanation:

From the given information:

The freezing point (°C) of a solution can be prepared by using the formula:

\Delta T = iK_fm

where;

i = vant Hoff factor

the vant Hoff factor is the totality of the number of ions in the solution

Since there are 1 calcium ion and 2 nitrate ions present in Ca(NO3)2, the vant Hoff factor = 3  

K_f = 1.86 °C/m

m = molality of the solution and it can be determined by using the formula

molality = \dfrac{mole \ of \ solute }{kg \ of \ solvent }

which can now be re-written as :

molality = \dfrac{mole \ of \ Ca(NO_3)_2}{kg \ of \  water}

molality = \dfrac{\dfrac{mass \ of \  \ Ca(NO_3)_2}{molar \  mass of \ Ca(NO_3)_2} }{kg \ of \  water}

molality = \dfrac{\dfrac{11.3 \ g }{164 \ g/mol} }{0.115 \ kg }

molality = 0.599 m

∴

The freezing point (°C) of a solution can be prepared by using the formula:

\Delta T = iK_fm

\Delta T =3 \times (1.86 \ ^0C/m) \times (0.599 \ m)

\Delta T =3.34^0 \ C

\Delta T = the freezing point of water - freezing point of the solution

3.34° C = 0° C -  freezing point of the solution

freezing point  (°C) of the solution =  0° C - 3.34° C

freezing point   (°C) of the solution =  - 3.34° C

3 0
3 years ago
What is the molality of a solution made by dissolving 0.0816 moles of
Paul [167]

Answer:

m = 0.544 moles/kg

Explanation:

To do this, we need to use the expression for molality of a solution which is:

m = n / kg water

Where:

n: moles of solute

m: molality of solution

In this case, we have the moles of the solute, which is the C6H12O6 and the grams of water, and these grams would have to be turn into kg. Let's do this first:

kg of water = 150 g * 1 kg / 1000 g = 0.150 kg of water.

Now, replacing in the above expression of molality, we can calculate the molality of the solution:

m = 0.0816 / 0.150

<h2>m = 0.544 moles/kg</h2>
3 0
3 years ago
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