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2 years ago

What is height? Give an example.

1 answer:

5 0

Height is defined as the distance from the bottom to the top of something or the highest point or the greatest degree. An example of height is 5'8". An example of height is the top of Mount Everest.

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I'll give brainiest a person who answers first correctly :) I promise

skad [1K]

Answer: Potential energy

Explanation:

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3 years ago

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The electronegativity values of carbon, hydrogen, and nitrogen are compared in the table.

Fofino [41]

Answer:

C. CH₄ is less than NH₃ because the NH bond is more polar than the CH bond

Explanation:

The intermolecular forces between ammonia is far stronger than for methane. Between the molecules of ammonia we have the presence of hydrogen bonds. This bond is absent in methane.

Hydrogen bonds are one of the strongest intermolecular forces. It is as a result of the electrostatic attraction between the hydrogen atom of one molecule and the electronegative atom N, O and F of another molecule.

This strong interaction is absent in methane which has just dipole - dipole attraction.

The strength of the hydrogen bond depends on the electronegativity of the combining atoms.

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3 years ago

What is definition "scientific question"?

Lemur [1.5K]

Answer:

A scientific question is basically a question that can lead to a hypothesis to help us figure out the observation in science. I hope this helps you

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2 years ago

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The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M