All categories

4 years ago

2. Using the following data, calculate the average atomic mass of magnesium (give your answer to the nearest

Chemistry

1 answer:

arlik [135]4 years ago

8 0

Answer:

24.32

Explanation:

From the question given above, the following data were obtained:

Isotope A:

Mass of A = 24

Abundance (A%) = 78.70%

Isotope B

Mass of B = 25

Abundance (B%) = 10.13%

Isotope C:

Mass of C = 26

Abundance (C%) = 11.17%

Average atomic mass of Mg =..?

The average atomic mass of Mg can be obtained as illustrated below:

Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]

Average atomic mass = [(24 × 78.70)/100] + [(25 × 10.13)/100] + [(26 × 11.17)/100]

= 18.888 + 2.5325 + 2.9042

= 24.3247 ≈ 24.32

Therefore, the average atomic mass of magnesium (Mg) is 24.32

You might be interested in

Electrons orbit the nucleus in specified energy

pashok25 [27]

Answer:

because the electron is a negatively charge and have high energy

5 0

3 years ago

Read the reaction and the statement. H2(g) + I2(g) ⇌ 2HI(g) The concentration of H2, I2 and HI are measured as 0.04 M, 0.08 M an

hjlf

Answer:

45467777777777yt3457942

4 0

3 years ago

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho

ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= $\frac{59}{760}atm$

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

$n = \frac{PV}{RT}$

$n = \frac{0.078 * 15000}{0.082*290}$

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since numbers of moles = $\frac{mass}{molar mass}$

and mass = density × vollume

Then; we can say ;

number of moles = $\frac{density*volume }{molar mass of ethanol}$

number of moles = $\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}$

number of moles = $\frac{&85}{46.07}$

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0

3 years ago

Consider the chemical equation. 2NBr3 + 3NaOH mc008-1.jpg N2 + 3NaBr + 3HOBr If there are 40 mol of NBr3 and 48 mol of NaOH, wha

Vladimir79 [104]

The correct answer o your amazing question is NBr3

6 0

3 years ago

Read 2 more answers

1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this

LenKa [72]

<h3>Answer:</h3>

0.6 g NaCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

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

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})$
Multiply/Divide: $\displaystyle 0.551373 \ g \ NaCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

5 0

3 years ago