The answer is C,growth spurts,puberty,& sexual maturity
The amount of grams that are in 2.3 moles of N = 32.223 or 32/100
Because there are 14.01 grams per mile of nitrogen atoms.
So…
14.01 x 2.3= 32.223
Hope this helps :)
Answer:
A. percentage mass of iron = 5.17%
percentage mass of sand = 8.62%
percentage mass of water = 86.205%
B. (Iron + sand + water) -------> ( iron + sand) ------> sand
C. The step of separation of iron and sand
Explanation:
A. Percentage mass of the mixtures:
Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g
percentage mass of iron = 15/290 * 100% = 5.17%
percentage mass of sand = 25/290 * 100% = 8.62%
percentage mass of water = 250/290 * 100% = 86.205%
B. Flow chart of separation procedure
(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand
C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.
Answer:
Just here for the points sorry
Explanation:
Minecraft I will not be able to make the weekend of this trip until Sunday evening and I will be away for the rest of the of the week weekend and I will will be back from London tomorrow for lunchtime a week or so if to for for if to go for the it a the the it a couple of rest in the morning the other week night if and time as I we are have the first one in the evening morning so I'll we have an appointment early for in a the class morning and for and then
Answer:
π = 14.824 atm
Explanation:
wt % = ( w NaCL / w sea water ) * 100 = 3.5 %
assuming w sea water = 100 g = 0.1 Kg
⇒ w NaCl = 3.5 g
osmotic pressure ( π ):
∴ T = 20 °C + 273 = 293 K
∴ C ≡ mol/L
∴ density sea water = 1.03 Kg/L....from literature
⇒ volume sea water = 0.1 Kg * ( L / 1.03 Kg ) = 0.097 L sln
⇒ mol NaCl = 3.5 g NaCL * ( mol NaCL / 58.44 g ) = 0.06 mol
⇒ C NaCl = 0.06 mol / 0.097 L = 0.617 M
⇒ π = 0.617 mol/L * 0.082 atm L / K mol * 293 K
⇒ π = 14.824 atm