Answer:


Explanation:
<u>Calculation of the mass of chromium as:-
</u>
Moles = 1.002 moles
Molar mass of chromium = 51.9961 g/mol
The formula for the calculation of moles is shown below:
Thus,

<u>Calculation of the mass of neon as:-
</u>
Moles =
moles
Molar mass of neon = 20.1797 g/mol
Thus,

<span>Well... first, let's recognize that the chemical formula for chlorodifluoromethane is CHClF2. Count out how many valence electrons there are. C = 4, H = 1, Cl = 7, F (X2) = 14. Total is 26. Let's put C as the central atom, and put the other elements surrounding it. Draw a pair of electrons beach each element and the central atom. Then fill the halogen elements with 3 pairs of electrons each to fill their octets. Count out how many dots you have. There should be 26, making this the correct lewis structure!
Remember, hydrogen doesn't have a full octet, only a maximum of two electrons.</span>
Answer: C
I hope this helped you
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:
6,216.684 kilograms of sodium carbonate must be added to neutralize
of sulfuric acid solution.
Explanation:
Mass of sulfuric acid solution = 

Percentage mass of sulfuric acid = 95.0%
Mass of sulfuric acid = 

Moles of sulfuric acid = 

According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.
Then 58,647.96 moles of sulfuric acisd will be neutralized by :
of sodium carbonate
Mass of 58,647.96 moles of sodium carbonate :

6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg
6,216.684 kilograms of sodium carbonate must be added to neutralize
of sulfuric acid solution.