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aalyn [17]
3 years ago
5

If you have a sample of a solid with a mass of 48.2 g and a density of 14.3 g/cm3, what would the volume be? (round you answer t

o the nearest hundredth)
Chemistry
1 answer:
Alecsey [184]3 years ago
5 0

Answer:

mass is equal to density into volume

48.2=14.3*v

v=48.2/14.3

v=3.37

so volume is 3.40

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According to Niels Bohr, electrons travel around an atom's nucleus in well-defined orbits.
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Answer: The answer is B

Explanation:

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When an ionic solid is added into a solvent, you can see that the ionic solid dissociates into its respective cations and anions
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The order of the steps, from first to last, is as follows based on the information in the question:

1. Salt is broken down into cations and anions.

2. Anion hydration.

3. Cation hydration.

4. Cations and anions that have been dissolved start to settle as a solid salt.

5. The rate of dissolution and recrystallization are equal.

Without hydration of the ions, the cation and anion cannot get separated. The three processes of ion dissociation, cation hydration, and anion hydration must all take place at once.

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Water is a great solvent that can dissolve a wide variety of compounds due to its polarity and capacity to create hydrogen bonds.

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2 years ago
Arizona was the site of a 400,000-acre wildfire in June 2002. How much carbon dioxide (CO2) was produced into the atmosphere by
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Answer:

2.97 × 10¹³ g

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First, we have to calculate the biomass the is burned. We can establish the following relations:

  • 2.47 acre = 10,000 m²
  • 10 kg of C occupy an area of 1 m²
  • 50% of the biomass is burned

The biomass burned in the site of 400,000 acre is:

400,000acre\times\frac{10,000m^{2} }{2.47acre} \times \frac{10kgC}{m^{2} } \times 50\% = 8.10 \times 10^{9} kgC

Let's consider the combustion of carbon.

C(s) + O₂(g) ⇒ CO₂(g)

We can establish the following relations:

  • The molar mass of C is 12.01 g/mol
  • 1 mole of C produces 1 mole of CO₂
  • The molar mass of CO₂ is 44.01 g/mol

The mass of  produced is CO₂:

8.10 \times 10^{12}gC \times \frac{1molC}{12.01gC} \times \frac{1molCO_{2}}{1molC} \times \frac{44.01gCO_{2}}{1molCO_{2}} =2.97 \times 10^{13} gCO_{2}

4 0
3 years ago
A 2.83 g piece of zinc (density = 7.14 g/ml) is added to a graduated cylinder that contains 12.13 ml H2O. What will be the final
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The final volume reading on the graduated cylinder is 12.52ml

Density = mass / volume

Volume = mass / density

Volume of zinc = 2.83/7.14

Volume of zinc = 0.3963

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Total volume = initial volume+ volume of zinc

Total volume = 12.13 + 0.3963

Total volume = 12.5263 = 12.52 ml

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