Answer:- 3333 g of solution.
Some of the question part is missing here. It would be like, "Determine the mass in grams of each NaCl solution that contains 1.5 g of NaCl.
(i) 0.045% NaCl by mass
Solution:- 0.045% NaCl by mass means 0.045 g of NaCl are present in 100 g of solution. 1.5 g of NaCl would be present in how many grams of solution?
We could solve this using proportions...
(0.045/100) = (1.5/X)
0.045(X) = 1.5(100)
0.045X = 150
X = 150/0.045 = 3333
So, 1.5 g of NaCl is present in 3333 g of solution.
Answer:
n = 0.573mol
Explanation:
PV = nRT => n = PV/RT
P = 1.5atm
V = 8.56L
R = 0.08206Latm/molK
T = 0°C = 273K
n = (1.5atm)(8.56L)/(0.08206Latm/molK)(273K) = 0.573mol
Answer:
1.08 grams
Explanation:
first, we need to find the number of moles
we divide 12.1/22.4=0.54 moles
we multiply the number of moles with the molecular mass 0.54x2=1.08g
Sulfur has two filled energy levels and six electrons on the third energy level. The corresponding electron configuration is A.
B is incorrect because there are no p orbitals at the first energy level, ie, no 1p orbitals. C is incorrect because the 4s1 electron would spontaneously drop into the 3p orbitals. D is incorrect because the 3d electrons would spontaneously drop into the 3p orbitals.
