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.
Answer:
The answer is Sodium chloride.
Na is sodium and Cl is chlorine.
Answer:
The molarity of this final solution is 0.167 M
Explanation:
Step 1: Data given
Volume of a 0.100 M HNO3 solution = 50.0 mL
Volume of a 0.200 M HNO3 = 100.0 mL
Step 2: Calculate moles
The final molarity must lie between 0.1M and 0.2M
Moles = molarity * volume
Moles HNO3 in 50mL of a 0.100M solution = 0.05 L *0.100 M = 0.005 mol
Moles HNO3 in 100mL of a 0.200M solution = 0.100 L*0.200 = 0.020mol
total moles = 0.005+0.020 = 0.025 moles in 150mL solution = 0.150L
Step 3: Calculate molarity of final solution
Molarity = mol / volume
Molarity 0.025 moles /0.150 L
Molarity = 0.167M
The molarity of this final solution is 0.167 M
