first of all there is only two types of selective breeding and they are hybridization and inbreeding.
Answer:
0.186M
Explanation:
First, we need to obtain the moles of nitric acid that are given for each solution. Then, we need to divide these moles in total volume (120mL + 20mL = 140mL = 0.140L) to obtain molarity:
<em>Moles Nitric acid:</em>
0.0200L * (0.100mol / L) = 0.00200 moles
0.120L * (0.200mol / L)= 0.02400 moles
Total moles: 0.02400moles + 0.00200moles = 0.026 moles of nitric acid
Molarity: 0.026 moles / 0.140L
<h3>0.186M</h3>
Explanation:
1 SS offspring will be there and 2 Ss and 1 ss.
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Aluminum! hope this helps
Basis: 1 L of the substance.
(1.202 g/mL) x (1000 mL) = 1202 g
mass solute = (1202 g) x 0.2 = 240.2 g
mass solvent = 1202 g x 0.8 = 961.6 g
moles KI = (240.2 g) x (1 mole / 166 g) = 1.45 moles
moles water = (961.6 g) x (1 mole / 18 g) = 53.42 moles
1. Molality = moles solute / kg solvent
= 1.45 moles / 0.9616 kg = 1.5 m
2. Molarity = moles solute / L solution
= 1.45 moles / 1 L solution = 1.45 M
3. molar mass = mole solute / total moles
= 1.45 moles / (1.45 moles + 53.42 moles) = 0.0264
