0.008 ÷ 51.3 = 0.0002
Sig Figs
1
0.0002
Decimals
4
0.0002
Scientific Notation
2 × 10-4
E-Notation
2e-4
Words
zero point zero zero zero two
I HOPE I HELP
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
Answer:
conditioning-the process of associating a stimulus with a good or bad outcome.
trial-and-error learning-the process of performing a behavior more and more skillfully.
insight learning-the process of using prior knowledge to solve a problem.
imprinting-a bond to an indivisual or object shortly after birth or hatching.
Explanation:
Did this assignment.
Sodium is considered to be a transitional metal
