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4 years ago

Question 16 (6 points)

Chemistry

1 answer:

MakcuM [25]4 years ago

6 0

Answer:

1.) Fluorine

2.) Francium

3.) Nonmetals

4.) Metals

5.) Aluminum

6.) Helium

Explanation:

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Convert a volume of 2250cm3 to cubic meters

Taya2010 [7]

2250cm3= 2250/1000000m3=.00225m3

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3 years ago

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The heat of combustion of propane, C3H8 (g) is -2057 kJ/mol. What would be the enthalpy change if enough propane was burned to g

Digiron [165]

Considering the reaction stoichiometry, the enthalpy change if enough propane was burned to give off 12 moles of carbon dioxide gas is 8228 kJ.

The balanced reaction is:

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)

The heat of combustion of propane, C₃H₈, is -2057 kJ/mol. This is, 2057 kJ is released for every 1 mol C₃H₈.

So to determine the enthalpy change if enough propane was burned to emit 12 moles of carbon dioxide, you must take into account the stoichiometry of the reaction.

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

Then you can apply the following rule of three: if by stoichiometry 3 moles of CO₂ are produced by 1 mole of C₃H₈, 12 moles of CO₂ are produced by how many moles of C₃H₈?

$amount of moles of C_{3} H_{8} =\frac{12 moles of CO_{2}x1 mole of C_{3} H_{8} }{3 moles of CO_{2}}$

amount of moles of C₃H₈= 4 moles

So to determine the enthalpy change, you can apply the following rule of three: If for each mole of C₃H₈ 2057 kJ are released, for 4 moles of C₃H₈ how much heat is released?

$Heat released=\frac{4 molesx2057 kJ}{1 mole}$

Heat released= 8228 kJ

The enthalpy change if enough propane was burned to give off 12 moles of carbon dioxide gas is 8228 kJ.

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jek_recluse [69]

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3 years ago

The average molecular weight for element X is 59.97 g/mol. There are two known isotopes of element X, one weighing 59 g/mol, and

Temka [501]

Answer:

The correct answer is:48.5% X-59, 51.5% X-61

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of X-59 isotope be 'x'. So, fractional abundance of X-61 isotope will be '1 - x'

For X-59 isotope:

Mass of X-59 isotope =59 g/mol

Fractional abundance of X-59 isotope = x

For X-61 isotope:

Mass of X-61 isotope = 61 g/mol

Fractional abundance of X-61 isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$59.97 g/mol=[(59 g/mol\times x)+(61 g/mol\times (1-x))]\\\\x=0.515$

Percentage abundance of X-59 isotope = $0.515\times 100=51.5\%$

Percentage abundance of X-61 isotope = $(1-0.515)=0.485\times 100=48.5\%$

Hence, the percentage abundance of both the isotopes X-59and X-61 are 51.5% and 48.5% respectively.

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Moseley made revisions to the periodic table that resolved some of the problems with Mendeleev’s version. Which of these was a r

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Since you provide no options, Henry Moseley measured a property linked to Periodic Table position. After his revisions to the periodic table, Atomic number became more meaningful and the three pair of elements that seemed to be in the wrong order could be explained

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