Answer:
Alkali and alkaline-earth cations are energetically stable with an empty valence shell. Atoms of alkali and alkaline-earth elements achieve octets after losing one (alkali) or two (alkaline-earth) electrons such that they minimize the potential energy of the system.
Answer:
First, let's determine how many moles of oxygen we have.
Atomic weight oxygen = 15.999
Molar mass O2 = 2*15.999 = 31.998 g/mol
We have 3 drops at 0.050 ml each for a total volume of 3*0.050ml = 0.150 ml
Since the density is 1.149 g/mol,
we have 1.149 g/ml * 0.150 ml = 0.17235 g of O2
Divide the number of grams by the molar mass to get the number of moles 0.17235 g / 31.998 g/mol = 0.005386274 mol
Now we can use the ideal gas law. The equation PV = nRT where P = pressure (1.0 atm) V = volume n = number of moles (0.005386274 mol) R = ideal gas constant (0.082057338 L*atm/(K*mol) ) T = Absolute temperature ( 30 + 273.15 = 303.15 K)
Now take the formula and solve for V, then substitute the known values and solve.
PV = nRT V = nRT/P V = 0.005386274 mol * 0.082057338 L*atm/(K*mol) * 303.15 K / 1.0 atm V = 0.000441983 L*atm/(K*) * 303.15 K / 1.0 atm V = 0.133987239 L*atm / 1.0 atm V = 0.133987239 L
So the volume (rounded to 3 significant figures) will be 134 ml.
Answer:
Moles of NO₂ = 0.158
Explanation:
SO 2 ( g ) + NO 2 ( g ) ⇄ SO 3 ( g ) + NO ( g )
According to the law of mass equation
= ![\frac{[SO_{3} ][NO]}{[SO_{2}][NO_{2} ]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSO_%7B3%7D%20%5D%5BNO%5D%7D%7B%5BSO_%7B2%7D%5D%5BNO_%7B2%7D%20%20%5D%7D)
⇒ 3.10 = ![\frac{(1.00)(1.00)}{(2.30) [NO_{2} ]}](https://tex.z-dn.net/?f=%5Cfrac%7B%281.00%29%281.00%29%7D%7B%282.30%29%20%5BNO_%7B2%7D%20%5D%7D)
At equilibrium [SO₃] = [NO]
⇒ [NO₂] = 
⇒ [NO₂] = 0.158
So. number of moles of NO₂ at equilibrium added = 0.158
The weathering process caused by cycles of freezing and thawing of water in surface pores, cracks, and other openings
1 mole ------------- 22.4 L ( at STP )
?? mole ---------- 12 L
12 x 1 / 22.4 => 0.5357 moles
hope this helps!
