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
"Physical changes occur when objects or substances undergo a change that does notchange their chemical composition. This contrasts with the concept of chemical change in which the composition of a substance changes or one or more substances combine or break up to form new substances."
Answer:
False
Explanation:
The primary structure of a protein refer to the amino acid sequence. The secondary structure of a protein refer to the alpha helices, beta sheets and turns, while the tertiary structure refer to folding of the sheets due to hydrogen bonding or other bonding interaction between them.
One thing to notice in the question is, we are asked about molecular oxygen that has formula O2 not atomic oxygen O.
As we are asked about molecular oxygen, we will answer the question in terms of number of molecules that are present in 16 grams of molecular oxygen.
To get the number of molecules present in 16 grams of O2, we will use the formula:
No. of molecules = no. of moles x Avogadro's number (NA)----- eq 1)
As we know:
The number of moles = mass/ molar mass of molecule
Here we have been given mass already, 16 grams and the molar mass of O2 is 32 grams.
Putting the values in above formula:
= 16/32
= 0.5 moles
Putting the number of moles and Avogadro's number (6.02 * 10^23) in eq 1
No. of molecules = 0.5 x 6.02 * 10^23
=3.01 x 10^23 molecules
or 301,000,000,000,000,000,000,000 molecules
This means that 16 grams of 3.01 x 10^23 molecules of oxygen.
Hope it helps!
Answer and Explanation:
a) The direction is shown in the cube diagram attached to this solution.
b) the angle between two planes (h₁, k₁, l₁) and (h₂, k₂, l₂) is given by the formula,
Cos Φ = (h₁h₂ + k₁k₂ + l₁)/√((h₁² + k₁² + l₁²)(h₂² + k₂² + l₂²))
For (111) and (112)
Cos Φ = (1.1 + 1.1 + 1.2)/√((1² + 1² + 1²)(1² + 1² + 2²))
Cos Φ = (1 + 1 + 2)/√((1+1+1)(1+1+4))
Cos Φ = 4/√(3×6)
Cos Φ = 4/√18
Φ = cos⁻¹ (4/√18) = 19.56°
c) equation 3.3 is missing from the question, I would be back to provide the answers to that as soon as the equation is provided!
Hope this Helps!!
