Answer:
Helium, Neon, Argon, Krypton, Xenon, Radon, and Oganesson
Explanation:
they have full outer shells
<span>134 ml
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.</span>
Answer:
412.1kJ
Explanation:
For the reaction , from the question -
4Fe (s) + 3O₂ (g) → 2Fe₂O₃ (s)
Δ Hrxn = Δ H°f (products) - Δ H°f (reactants)
In case the compound is in its standard state , enthalphy of formation is zero
Hence ,
for the above reaction ,
ΔHrxn =( 2 * Δ H° (Fe₂O₃ )) - [ ( 4 *Δ H° Fe ) + (3 * Δ H° O₂ )]
The value for Δ H°(Fe₂O₃ ) = - 824.2kJ/mol
Δ H° Fe = 0
Δ H° O₂ = 0
Putting in the above equation ,
ΔH rxn = ( 2 * Δ H° (Fe₂O₃ )) - 0
ΔHrxn = 2× - 824.2 kJ / mol = - 1648.4 kJ/mol
- 1648.4 kJ/mol , this much heat is released by the buring of 4 mol of Fe.
Hence ,
for 1 mol of Fe ,
- 1648.4 kJ/mol / 4 = 412.1kJ
Answer:
H2SO4
Explanation:
H2SO4 is actually one of the strongest acids, coming to pH value very close to 1 . Acids transfer their hydrogen ions to other substances , such as water , and usually have hydrogen their formula . In this case , H2SO4 has hydrogen in its formula and is acid in the case .
<em>H</em><em>o</em><em>p</em><em>e</em><em> </em><em>i</em><em>t</em><em> </em><em>h</em><em>e</em><em>l</em><em>p</em><em>s</em><em> </em><em>y</em><em>o</em><em>u</em><em> </em><em>!</em><em>!</em>
D all of the above because it has all of them at the crust
