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Kisachek [45]
3 years ago
14

An element has three naturally occurring isotopes. Use the information below to calculate the weighted average atomic mass of th

e element, showing both the setup and the final answer for the calculation.
Isotope

Atomic Mass

Percent Abundance

X 1.01 u 99.984%

Y 2.01 u 0.014%

Z 3.02 u 0.002%
Chemistry
1 answer:
puteri [66]3 years ago
5 0
Weight average = 1.01 * 0.99984 + 2.01* 0.00014 + 3.02 * 0.00002

answer is 1 .0101802
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In carbon dioxide (CO2), there are two oxygen atoms for each carbon atom. Each oxygen atom forms a double bond with carbon, so t
SashulF [63]

Answer:

The Answer is eight.

Explanation:

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4 0
3 years ago
Determine the molarity and mole fraction of a 1.09 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of a
Juli2301 [7.4K]

Answer:

Molarity = 0.809 M

mole fraction = 0.047

Explanation:

The complete question is

Calculate the molarity and mole fraction of acetone in a 1.09-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

Solution -

Solution for molarity:

1.09-molal means 1.09 moles of acetone in 1.00 kilogram of ethanol.

1)  

Mass of 1.09 mole of acetone

= 1.09  mol x 58.0794 g/mol = 63.306 g

Density of acetone = 0.788 g/cm3  

Thus, volume of 1.09 moles of acetone = 63.306 g/0.788 g/cm3 = 80.34 cm3

For ethanol

1000 g divided by 0.789 g/cm3 = 1267.427 cm3

Total volume of the solution = Volume of acetone + Volume of ethanol = 80.34 cm3 + 1267.427 cm3 = 1347.765 cm3  = 1.347 L

a) Molarity:

1.09 mol / 1.347 L = 0.809 M

Mole Fraction  

a) moles of ethanol:

1000 g / 46.0684 g/mol = 21.71 mol

b) moles of acetone:

1.09 / (1.09 + 21.71) = 0.047

3 0
3 years ago
"Reduce, reuse, recycle." Give examples of how one or more of
Anarel [89]

Answer:

Examples of how the ideas can be applied to the issues and practices of hydraulic fracturing for the acquisition of shale gas are;

Reuse; The produced water in obtained from oil and gas well production are reused for fracking, drilling, and if the water is good enough, it can be used for farming

Reduce; The use of recycled brine and water in drilling and fracking process reduces the application of freshwater in the those processes and reduces pollution of natural water sources

Recycle; Recycling involves creating products from waste. In the hydraulic fracturing process approximately 13 percent of the water produced and the flowback water are recycled to be used more than once thereby reducing the net consumption of freshwater

Explanation:

In hydraulic fracturing, also known informally as fracking, is the drilling method used in oil and gas well development process that makes use of water sand and chemical injection through the well bore to open and widen cracks in the bedrock formations in the areas around the wellbore.

6 0
3 years ago
If your creating a machine that will clean up trash from the ocean is that adding or removing thermal energy?
bogdanovich [222]

Answer: adding

Explanation:

Because it takes thermal energy to power it

6 0
4 years ago
How many zinc atoms are contained in 3.75 moles of zinc?
bija089 [108]
<h3>Answer:</h3>

2.26 × 10²⁴ atoms Zn

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.75 mol Zn (Zinc)

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 3.75 \ mol \ Zn(\frac{6.022 \cdot 10^{23} \ atoms \ Zn}{1 \ mol \ Zn})
  2. [DA] Multiply/Divide [Cancel out units]:                                                        \displaystyle 2.25825 \cdot 10^{24} \ atoms \ Zn

<u>Step 4: Check</u>

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

2.25825 × 10²⁴ atoms Zn ≈ 2.26 × 10²⁴ atoms Zn

3 0
3 years ago
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