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2 years ago

Which model is the best for identifying geometric isomers?

Chemistry

2 answers:

melisa1 [442]2 years ago

6 0

The correct answer is <span>ball-and-stick model I just take it</span>

Komok [63]2 years ago

6 0

Ball-and-stick model. Just got it right

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What happens to metal when rapidly heated then cooled

Anna [14]

Answer:

It is called tempering. Its tensile strength may reduce but it will become more rigid and hard to break compared to the original metal.

Explanation:

( Source : Quora )

4 0

2 years ago

Use the iupac nomenclature rules to give the name for this compound - n2o3.

wel

By use IUPAC nomenclature rule compound N2O3 is named as

Dinitrogen trioxide ( answer d)

Nitrogen (N2) is named before oxygen(O3) since they are arranged alphabetically. In addition the prefix Di is used infront of nitrogen since they are two nitrogen atoms while prefix tri is used infront of oxide since they are 3 oxygen atoms

8 0

2 years ago

Read 2 more answers

How can we tell when something is polar or non-polar in an Inter-molecular Force?

Montano1993 [528]

If the substance mixes with water it's polar. If it doesn't it ms non polar.

7 0

2 years ago

Given the chemical reaction: Hg2+(aq) + Cu(s) + Hg(s) + Cu2+(aq)

Pavel [41]

Answer: When the reaction reaches equilibrium, the cell potential will be 0.00 V

Explanation:

Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.

The equilibrium is dynamic in nature and the reactions are continuous in nature. Rate of forward reaction is equal to the rate of backward reaction.

The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G=-nFE^0$

The Gibbs free energy is related to equilibrium constant by following relation:

$\Delta G=-2.303RTlog K$

For equilibrium $\Delta G=0$

Thus $0=-nFE^0$

$E^0=0$

Thus When the reaction reaches equilibrium, the cell potential will be 0.00 V

5 0

2 years ago

15.89 percent carbon, 21.18 percent oxygen, and 62.93 percent osmium. what is the empirical formula

otez555 [7]

Answer:

OsCO or COOs

Explanation:

Data given

Carbon = 15.89 %

Oxygen = 21.18 %

Osmium = 62.93%

Empirical formula = ?

Solution:

First find the masses of each component

Consider total compound is 100g

As we now

mass of element = % of component

So,

15.89 g of C = 15.89 % Carbon

21.18 g of O = 21.18 % Oxygen

62.93 g of Os = 62.93% Osmium

Now convert the masses to moles

For Carbon

Molar mass of C = 12 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 15.89 g/ 12 g/mol

no. of mole = 1.3242

mole of C = 1.3242

For Oxygen

Molar mass of O = 16 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 21.18 g/ 16 g/mol

no. of mole =

mole of O = 1.3238

For Os

Molar mass of Os = 190 g/mol

no. of mole = mass in g / molar mass

Put value in above formula

no. of mole = 62.93 g/ 190 g/mol

no. of mole =

mole of Os = 1.3312

Now we have values in moles as below

C = 1.3242

O = 1.3238

Os = 1.3312

Divide the all values on the smalest values to get whole number ratio

C = 1.3242 /1.3238 = 1.0003

O = 1.3238 /1.3238 = 1

Os = 1.3312 /1.3238 = 1.0056

So all have round value 1 mole

C = 1

O = 1

Os = 1

So the empirical formula will be (OsCO) i.e. all 3 atoms in simplest small ratio

7 0

3 years ago