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seropon [69]
3 years ago
14

Hypothetical element X has 3 stable isotopes. The relative peak intensities are 79.52 for 180.0 u, 100.0 for 184.0 u, and 82.98

for 186.0 u. What is the average atomic mass for X
Chemistry
1 answer:
vazorg [7]3 years ago
3 0

Answer:

The relative atomic mass of X is 183.4

Explanation:

The relative atomic mass of an element is the weighted average of the masses of the isotopes using the carbon-12 atom has a mass of exactly 12 units as as a standard.

The relative sizes of the peaks gives a direct measure of the relative abundances of the isotopes. The tallest peak is often given an arbitrary height of 100 and is known as the base peak.

To calculate the relative atomic mass of the element X which has 3 stable isotopes with the relative peak intensities as given below:

79.52 for 180.0 u, 100.0 for 184.0 u, and 82.98 for 186.0 u, the sum of the products of the relative peak intensity of each isotope and their isotopic mass is divided by the sum of the peak intensities.

Relative atomic mass of X = (79.52 × 180) + (100 × 184.0) + (82.98 × 186)/(79.52 + 100 + 82.98)

Relative atomic mass of X = (14313.6 + 18400 + 15434.28)/(262.5)

Relative atomic mass of X = 48147.88/262.5 = 183.4

Therefore, the relative atomic mass of X is 183.4

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Explanation:

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3400=m(9.8)(6)\\3400=58.8m\\m=57.8kg

3 0
2 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
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Write a balanced net ionic equation for the following reaction: SrCl2(aq) + H2CO3(aq) → SrCO3(s) + HCl (aq)
max2010maxim [7]
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SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}
 
  Looking at the equation, we can see that it is now fully balanced. Hence, a balanced equation of the reaction is:

SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}
5 0
2 years ago
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Answer:

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Explanation:

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Answer:

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Explanation:

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