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3 years ago

What is the difference between an empirical and a molecular formula? Give the empirical formulas of (a) C8H18, (b) P4O10.

1 answer:

4 0

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How to find how many electrons are in an element

Murljashka [212]

Explanation:

1) Locate the atomic number in the upper left corner of the square. The atomic number will tell you the number of protons

2) In a neutral state, the element will contain the same number of protons as it will electrons.

--> Boron (B) has an atomic number of 5, so it has 5 protons.

--> Having 5 protons (+1 charge each) means to be balanced, it must have 5 electrons (-1 charge each)

However, what if it's not in a neutral state?

If it's not in a neutral state, it will look something like this:

Ca ^ 2+

<em>Btw, the "^" means that it will be written like it's an exponent. That confuses people sometimes.</em>

Anyways, the "2+" means it's an ion with a positive charge of 2. For us to have a positive charge, it means it's lost two electrons (cause electrons are negative, so taking them away means we are left with a more positive charge).

To get the number of remaining electrons, subtract the 2 from the atomic mass. In this case, that would be 20.

20 - 2 = 18 electrons

Another example:

N ^3-

Atomic Number: 7

7 + 3 = 10 electrons.

Why did I add here instead of subtracting? It's cause the 3- means we have added electrons, resulting in the ion having a negative charge. It's got more electrons than usual.

Hopefully this helps a bit.

5 0

3 years ago

What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm

Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{freezing}}{T_f}$

where,

$\Delta S$ = change in entropy

$\Delta H_{fus}$ = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

$\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol$

$T_f$ = freezing point temperature = $-97.6^oC=273+(-97.6)=175.4K$

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

$\Delta S=\frac{\Delta H_{freezing}}{T_m}$

$\Delta S=\frac{-3170J/mol}{175.4K}$

$\Delta S=-18.07J/mol.K$

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0

3 years ago

Why doesn’t neon readily form an ionic bond?

Goryan [66]

By definition of noble gases, neon does not easily form an ionic bond because it belongs to the group of noble or inert gases, so its reactivity is practically nil.

<h3>Noble gases</h3>

Noble gases are not very reactive, that is, they practically do not form chemical compounds. This means that they do not react with other substances, nor do they even react between atoms of the same gas, as is the case with diatomic gases such as oxygen (O₂).

The chemical stability of the noble gases and therefore the absence of spontaneous evolution towards any other chemical form, implies that they are already in a state of maximum stability.

All chemical transformations involve valence electrons, they are involved in the process of covalent bond formation and the formation of ions. Therefore, the practically null reactivity of the noble gases is due to the fact that they have a complete valence shell, which gives them a low tendency to capture or release electrons.

Since the noble gases do not react with the other elements, they are also called inert gases.

Neon does not easily form an ionic bond because it belongs to the group of noble or inert gases, so its reactivity is practically nil.

Learn more about noble gases:

brainly.com/question/8361108

brainly.com/question/11960526

brainly.com/question/19024000

6 0

2 years ago

Read 2 more answers

6.0L of a gas exert a pressure of 2.5 atm. When the pressure is increased to 10.0atm what is the new volume of the gas

jeka57 [31]

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{6 \times 2.5}{10} = \frac{15}{10} = \frac{3}{2} \\$

We have the final answer as

Hope this helps you

7 0

3 years ago

Ni2+(aq) + 6 NH3(aq) ⇌ Ni(NH3)6 2+(aq)

Jet001 [13]

Answer:

a) ΔGº= -49,9 KJ/mol = - 50 KJ/mol

b) The reaction goes to the right to formation of products

c) ΔG= 84,42 KJ/mol. The direction is to reactive, to the left

Explanation:

a) ΔGº= - RTLnKf

You need to convert Cº to K. 25ºC=298K

Then, ΔGº= - 3,814 J/molK * 298K* Ln(5.6 *10^8)= - 49906 J/mol = -49,9 KJ/mol = - 50 KJ/mol

b) The ΔGº < 0, that means the direct reaction is spontaneous when te reactive and products are in standard state. In other words the reaction goes to the right, to formation of products

c) The general ecuation for chemical reaction is aA + bB → cD + dD. Thus

ΔG=ΔGº + RTLn (([C]^c*[D]^d)/[A]^a*[B]^b)

In this case,

ΔG=ΔGº + RTLn ([Ni(NH3)62+] / [Ni2+]*[NH3]^6 )= 84417 J/mol =84,42 KJ/mol

ΔG >0 means the reaction isn't spontaneous in the direction of the products. Therefore the direction is to reactive, to the left

8 0

3 years ago