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d1i1m1o1n [39]
3 years ago
7

What type(s) of intermolecular forces exist between two molecules of hexane? (BLB Ch. 11)

Chemistry
1 answer:
kati45 [8]3 years ago
7 0

Answer:

induced dipole-dipole forces or London Dispersion forces / van der Waals forces.

Explanation:

Hexane is non-polar in nature. This is due to :

The bond in the molecule is C-H, which is non-polar in nature because the carbon and the hydrogen having very similar electronegativity values.

Hexane is also symmetric.

The intermolecular force acting in the molecule of the hexane are induced the dipole-dipole forces or London Dispersion forces / van der Waals forces.

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Which change would cause an immediate increase in the rate of the forward reaction
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According to Le Chatelier's Principle, to increase the rate of the forward reaction we add more of the reactants to encourage more collisions and production of more products. If it is endothermic, more heat should be added. If it is endothermic, more heat should be released from the system.
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An atom or ion has 41 neutrons, 36 protons, and 36 electrons. Identify the clement symbol, and determine the mass
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You combine 13 g of magnesium with 5 g of nitrogen to form a compound.
guapka [62]

Answer:

\large \boxed{28 \, \% }

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magnesium  + nitrogen ⟶ Product

      13 g                5 g

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The product contains 5 g of nitrogen .

\text{Percent N} =  \dfrac{\text{5 g}}{\text{18 g}} \times \, 100\% =  \mathbf{28 \, \%}\\\text{The percent by mass of nitrogen is $\large \boxed{\mathbf{28 \, \% }}$}

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3 years ago
The radioactive substance cesium-137 has a half-life of 30 years. The amount At (in grams) of a sample of cesium-137 remaining a
stealth61 [152]

<u>Answer:</u> The amount of sample left after 20 years is 288.522 g and after 50 years is 144.26 g

<u>Explanation:</u>

We are given a function that calculates the amount of sample remaining after 't' years, which is:

A_t(t)=458\times (\frac{1}{2})^{\frac{t}{30}

  • <u>For t = 20 years</u>

Putting values in above equation:

A_t(t)=458\times (\frac{1}{2})^{\frac{20}{30}

A_t(t)=288.522g

Hence, the amount of sample left after 20 years is 288.522 g

  • <u>For t = 50 years</u>

Putting values in above equation:

A_t(t)=458\times (\frac{1}{2})^{\frac{50}{30}

A_t(t)=144.26g

Hence, the amount of sample left after 50 years is 144.26 g

6 0
3 years ago
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