Answer:
The Bohr model and all of its successors describe the properties of atomic electrons in terms of a set of allowed (possible) values. Atoms absorb or emit radiation only when the electrons abruptly jump between allowed, or stationary, states. Direct experimental evidence for the existence of such discrete states was obtained (1914) by the German-born physicists James Franck and Gustav Hertz.
Explanation:
Answer:
Explanation:
The substances that are present at the beginning are called reactants and the substances present at the end are called products. Sometimes when reactants are put into a reaction vessel, a reaction will take place to produce products. Reactants are the starting materials, that is, whatever we have as our initial ingredients.Oxygen, a colorless, odorless, tasteless gas essential to living organisms, being taken up by animals, which convert it to carbon dioxide; plants, in turn, utilize carbon dioxide as a source of carbon and return the oxygen to the atmosphere. Oxygen forms compounds by reaction with practically any other element.
Answer:
1.73 M
Explanation:
Molarity is moles per liter, so we need to divide 7.47 moles by 4.31 liters to get the molarity of the solution.
7.47/4.31 ≈ 1.73 M
Answer:
Empirical formula: CH₃O
Empirical formula mass = 31 g/mol
Explanation:
Data Given:
Molecular Formula = C₁₀H₃₀O₁₀
Empirical Formula = ?
Empirical Formula mass =
Solution
Empirical Formula:
Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.
So,
The ratio of the molecular formula should be divided by whole number to get the simplest ratio of molecule
As
C₁₀H₃₀O₁₀ Consist of 10 Carbon (C) atoms, 30 Hydrogen (H) atoms, and 10 Oxygen (O) atoms.
Now
Look at the ratio of these three atoms in the compound
C : H : O
10 : 30 : 10
Divide the ratio by two to get simplest ratio
C : H : O
10/10 : 30/10 : 10/10
1 : 3 : 1
So for the empirical formula the simplest ratio of carbon to hydrogen to oxygen is 1:3:1
So the empirical formula will be
Empirical formula of C₁₀H₃₀O₁₀ = CH₃O
Now
To find the empirical formula mass in g/mol
Formula mass:
Formula mass is the total sum of the atomic masses of all the atoms present in a formula unit.
**Note:
if we represent the molar mass of the empirical formula for one mol in grams then it is written as g/mol
So,
As the empirical formula of C₁₀H₃₀O₁₀ is CH₃O
Then Its empirical formula mass will be
CH₃O
Atomic Mass of C = 12
Atomic Mass of H = 3
Atomic Mass of O = 16
Total Molar mass of CH₃O
CH₃O = 12 + 3(1) + 16
CH₃O = 12 + 3 + 16
CH₃O = 31 g/mol
