Answer:
10 Litre
Explanation:
Given that ::
v1 = 25L ; n1 = 1.5 mole ; v2 =? ; n2 = (1.5-0.9) = 0.6 mole
Using the relation :
(n2 * v1) / n1 = (n2 * v2) / n2
v2 = (n2 * v1) / n1
v2 = (0.6 mole * 25 Litre) / 1.5 mole
v2 = 15 / 1.5 litre
v2 = 10 Litre
Answer: D (The abundance percentage of each isotope)
Explanation: hope this helps!
Answer:
The atomic mass of element is 65.5 amu.
Explanation:
Given data:
Abundance of X-63 = 50.000%
Atomic mass of X-63 = 63.00 amu
Atomic mass of X-68 = 68.00 amu
Atomic mass of element = ?
Solution:
Abundance of X-68 = 100-50 = 50%
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (50×63)+(50×68) /100
Average atomic mass = 3150 + 3400 / 100
Average atomic mass = 6550 / 100
Average atomic mass = 65.5 amu.
The atomic mass of element is 65.5 amu.
Carbonates
OPTION A is the correct answer
