-they are inside the nucleus of an atom
- they have a relative charge of +1
- they have a relative mass of 1
Answer:
The number of copper atoms 12.405 ×10²³ atoms.
The number of silver atoms 13.13 ×10²³ atoms.
Beaker B have large number of atoms.
Explanation:
Given data:
In beaker A
Number of moles of copper = 2.06 mol
Number of atoms of copper = ?
In beaker B
Mass of silver = 222 g
Number of atoms of silver = ?
Solution:
For beaker A.
we will solve this problem by using Avogadro number.
The number 6.022×10²³ is called Avogadro number and it is the number of atoms in one mole of substance.
While we have to find the copper atoms in 2.06 moles.
So,
63.546 g = 1 mole = 6.022×10²³ atoms
For 2.06 moles.
2.06 × 6.022×10²³ atoms
The number of copper atoms 12.405 ×10²³ atoms.
For beaker B:
107.87 g = 1 mole = 6.022×10²³ atoms
For 222 g
222 g / 101.87 g/mol = 2.18 moles
2.18 mol × 6.022×10²³ atoms = 13.13 ×10²³ atoms
Answer:
10
Explanation:
In image
Take atomic mass or molar mass
of Al =27
<h2>Natural Abundance for 10B is 19.60%</h2>
Explanation:
- The natural isotopic abundance of 10B is 19.60%.
- The natural isotopic abundance of 11B is 80.40%.
- The isotopic masses of boron are 10.0129 u and 11.009 u respectively.
For calculation of abundance of both the isotopes -
Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.
Determining it as an equation -
10x + 11y= 10.8
x+y=1 (ratio)
10x + 10y = 10
By taking the denominator away from the numerator
we get;
y = 0.8
x + y = 1
∴ x = 0.2
To get percentages we need to multiply it by 100
So, the calculated abundance is 80% for 11 B and 20% 10 B.
Of course they are small
Explanation:
The only way you can see them is by a microscope or a lens and can be anywhere.
