I don't know but I'm wasting 5 seconds of your time you can't take back sorry
Answer:
We have to take 37.5 mL of a 0.400 M solution
Explanation:
Step 1: Data given
Stock volume = 100 mL = 0.100L
Stock concentration 0.400 M
Volume of solution he wants to make = 100 mL = 0.100L
Concentration of solution he wants to make = 0.150 M
Step 2: Calculate the volume of 0.400 M CuSO4 needed
C1*V1 = C2*V2
⇒with C1 = the stock concentration = 0.400M
⇒with V1 = the volume of the stock = TO BE DETERMINED
⇒with C2 = the concentration of the solution he wants to make = 0.150 M
⇒with V2 = the volume of the solution made = 0.100 L
0.400 M * V1 = 0.150M * 0.100L
V1 = (0.150M*0.100L) / 0.400 M
V1 = 0.0375 L = 37.5 mL
We have to take 37.5 mL of a 0.400 M solution
Answer:
The disadvantages of each of the given model of electron configuration have been mentioned below:
1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.
2). Arrow and line diagrams make the counting of electrons and take up too much space.
3). Written Configurations do not display the electron distribution in orbitals and help in lose counting of electrons easily.
From the periodic table:
molar mass of carbon = 12 gm
molar mass of hydrogen = 1 gm
molar mass of oxygen = 16 gm
molar mass of <span>acetylsalicylic acid = 9(12) + 8(1) + 4(16) = 180 gm
Number of moles = mass / molar mass = 0.5 / 180 = 2.778 x 10^-3 moles
To get the number of molecules, we multiply the number of moles by Avogadro's number as follows:
number of molecules = </span>2.778x10^-3x6.02x10^23 =1.672 x 10^21 molecules
<span>Ca(NO3)2 + Na2CO3 = CaCO3 + 2NaNO3
Yes a precipitate of Calcium Carbonate is formed since it is insoluble in water.
Mol Wt of Calcium Nitrate is 164. And that of Calcium Carbonate is 100.
One mole of Calcium Nitrate produces one mole of Calcium Carbonate.
i.e. 164 gms will produce 100gms of precipitate
So, 1.74gms of Calcium Carbonate will be obtained from 2.85gms Calcium Nitrate present in the original solution.</span>
