Answer:
Average atomic mass = 63.553 amu.
Explanation:
Given data:
Abundance of Y-63 = 69.17%
Abundance of Y-65 = 100 - 69.17 = 30.83%
Atomic mass of Y-63 = 62.940 amu
Atomic mass of Y-65 = 64.928 amu
Atomic mass of Y = ?
Solution:
Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass= (62.940×69.17)+(64.928×30.83) /100
Average atomic mass = 4353.560 + 2001.730 / 100
Average atomic mass = 6355.29 / 100
Average atomic mass = 63.553 amu.
To calculate the moles of AgNO3 in a solution, we need to know the volume and concentration of the solution.
Moles of AgNO3 = Volume of AgNO3 solution (L) * concentration of AgNO3 solution (M or mole/L) = 1.50 L * 0.050 M = 0.075 mole.
So 0.075 moles of AgNO3 are present in 1.50 L of a 0.050 M solution.
Answer:
6.022 × 10²³ molecules of B₂H₆
Explanation:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
B₂H₆ molecules:
21.63 g = one mole of B₂H₆= 6.022 × 10²³ molecules of B₂H₆
Answer:
134.397 Joules
Explanation:
Using the formula:
E = C × m × Δθ (where E is Energy, C is specific heat capacity and Δθ is change in temperature)
So E = 0.45×5.45×(79.8-25)
So E = 134.397 Joules
Answer: That is true my friend.
Explanation:
Maybe not all substances but it can dissolve more than any other liquid.
