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3 years ago

Which of the following examples are the result of unbalanced forces? Select all the apply

Chemistry

1 answer:

tankabanditka [31]3 years ago

3 0

The examples of unbalanced forces are when an object is likely to weigh more than the other.

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At 500°C the equilibrium constant, Kp, is 4.00 × 10–4 for the equilibrium: 2HCN(g) H2(g) + C2N2(g) What is Kp for the following

Andrews [41]

<u>Answer:</u> The value of $K_p$ for the equation is $2.5\times 10^3$

<u>Explanation:</u>

The given chemical reaction is:

$2HCN(g)\rightleftharpoons H_2(g)+C_2N_2(g);K_1$

The chemical equation for which the equilibrium constant is to be calculated follows:

$H_2(g)+C_2N_2(g)\rightleftharpoons 2HCN(g);K_p$

As, the equation is the result of the reverse of given reaction. So, the equilibrium constant for the equation will be the inverse of equilibrium constant for the given reaction.

The value of equilibrium constant for the equation is:

$K_p=\frac{1}{K_1}$

We are given:

$K_1=4.00\times 10^{-4}$

Putting values in above equation, we get:

$K_p=\frac{1}{4.00\times 10^{-4}}=2.5\times 10^3$

Hence, the value of $K_p$ for the equation is $2.5\times 10^3$

3 0

3 years ago

I don't understand this problem<br>?Mg+?02=?Mg0

Ghella [55]

Explanation:

2Mg + O2 ➖ 2MgO

hope it helps

3 0

3 years ago

A gas with a volume of 4.0 liters at a pressure of 205 kilopascals is allowed to expand to a volume of 12.0 liters. What will th

Ksenya-84 [330]

Boyle's law p1V1=p2V2

p2=(p1V1)/V2

p2=(205*10^3 Pa * 4*10^-3 m^3 ) / (12*10^-3 m^3)

p2= 68333 Pa

8 0

3 years ago

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

<u>Calculate the moles of potassium hydroxide.</u>

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0

3 years ago

In the context of small molecules with similar molar masses, arrange these intermolecular forces by strength (hydrogen bonding -

Katarina [22]

<h2>Answer:</h2>

Arrangement of inter molecular forces from strongest to weakest.

Hydrogen bonding
Dipole-dipole interactions
London dispersion forces.

<h3>Explanation:</h3>

Intermolecular forces are defined as the attractive forces between two molecules due to some polar sides of molecules. They can be between nonpolar molecules.

Hydrogen bonding is a type of dipole dipole interaction between the positive charge hydrogen ion and the slightly negative pole of a molecule. For example H---O bonding between water molecules.

Dipole dipole interactions are also attractive interactions between the slightly positive head of one molecule and the negative pole of other molecules.

But they are weaker than hydrogen bonding.

London dispersion forces are temporary interactions caused due to electronic dispersion in atoms of two molecules placed together. They are usually in nonpolar molecules like F2, I2. they are weakest interactions.

5 0

3 years ago