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klemol [59]
2 years ago
10

The mass of a deuterium nucleus 21H is less than its components masses. Calculate the mass defect.________ amu

Chemistry
2 answers:
Eddi Din [679]2 years ago
8 0

Answer:

0.002389

Explanation:

Use the equation the other person used to answer but with the numbers that are given on this assignment!

Inessa [10]2 years ago
5 0

Answer:

The mass defect of a deuterium nucleus is 0.001848 amu.

Explanation:

The deuterium is:

^{A}_{Z}X \rightarrow ^{2}_{1}H  

The mass defect can be calculated by using the following equation:

\Delta m = [Zm_{p} + (A - Z)m_{n}] - m_{a}

Where:

Z: is the number of protons = 1

A: is the mass number = 2      

m_{p}: is the proton's mass = 1.00728 amu  

m_{n}: is the neutron's mass = 1.00867 amu

m_{a}: is the mass of deuterium = 2.01410178 amu

Then, the mass defect is:

\Delta m = [1.00728 amu + (2- 1)1.00867 amu] - 2.01410178 amu = 0.001848 amu

Therefore, the mass defect of a deuterium nucleus is 0.001848 amu.

I hope it helps you!  

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Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100

Given:

Mass of carbon = 1.94 g

Mass of hydrogen = 0.48 g

Mass of sulfur = 2.58 g

First we have to calculate the mass of sample.

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur

Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g

Now we have to calculate the percentage composition of a compound.

\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%

\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%

\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%

Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

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What is the volume of 40.0 grams of argon gas at STP ?
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Answer:

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General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Aqueous Solutions</u>

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<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 40.0 g Ar

[Solve] L Ar

<u>Step 2: Identify Conversions</u>

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<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})
  2. [DA] Divide/Multiply [Cancel out units]:                                                         \displaystyle 24.9235 \ L \ Ar

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

24.9235 L Ar ≈ 24.9 L Ar

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