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gayaneshka [121]
3 years ago
13

PLS HELPPP!!!!

Chemistry
1 answer:
prisoha [69]3 years ago
5 0

Answer:

D.

Explanation:

Bohr's model represented electrons in their respective e⁻ shells. Only D fits the description.

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A molecule of butane and a molecule of 2-butene both have the same total number of
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The correct answer is option 1. Butane and 2-butene have the same total number of carbon atoms. They both have four carbon atoms. They differ in there structure since the latter has double bonds on it. As a result of the different structure, they also have different properties.
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Can anybody define hypothesis in their own words? (its for school)
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Answer:

An educated guess based on what you already know.

Explanation:

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A. True <br> b. False a chemical species is either an acid or a base, it cannot function as both
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False They can function as both. An example is Aluminium Oxide. These kind of substances are called "Amphoteric", they can behave as both acids and bases.
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Iron has a density of 7.86 g/cm3. Calculate the volume (in dL) of a piece of iron having a mass of 4.07 kg . Note that the densi
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Answer:

The volume of the piece of iron is 5.18dL.

Explanation:

The density (ρ) is equal to the mass (m) divided the volume (V).

\rho = \frac{m}{V}

If we rearrange it, we have:

V=\frac{m}{\rho }

To express the volume in dL we will need the following relations:

  • 1 dL = 0.1 L
  • 1 kg = 10³ g
  • 1 cm³ = 1 mL
  • 1mL = 10⁻³L

Then,

\rho = \frac{7.86g}{cm^{3} } .\frac{1cm^{3} }{1mL} .\frac{1mL}{10^{-3}L } .\frac{1kg}{10^{3}g } =7.86kg/L

Finally,

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If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, h
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Answer:

There were 0.00735 moles Pb^2+ in the solution

Explanation:

Step 1: Data given

Volume of the KI solution = 73.5 mL = 0.0735 L

Molarity of the KI solution = 0.200 M

Step 2: The balanced equation

2KI + Pb2+ → PbI2 + 2K+

Step 3: Calculate moles KI

moles = Molarity * volume

moles KI = 0.200M * 0.0735L = 0.0147 moles KI

Ste p 4: Calculate moles Pb^2+

For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+

For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+

There were 0.00735 moles Pb^2+ in the solution

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3 years ago
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