V = maximum capacity of human lung = 3 liter = 3 x 0.001 m³ = 0.003 m³ (Since 1 liter = 0.001 m³)
P = pressure of oxygen = 21.1 kilo pascal = 21.1 x 1000 = 21100 Pa (since 1 kilo = 1000)
T = temperature of air = 295 K
n = number of moles of oxygen
Using the ideal gas equation
PV = n RT
inserting the above values in the equation
(21100) (0.003) = n (8.314) (295)
n = 0.026 moles
<span>Consider the following reaction:
2CH3OH(g)→2CH4(g)+O2(g)ΔH=+252.8kJ
Calculate the amount of heat transferred when 27.0g of CH3OH(g) is decomposed by this reaction at constant pressure.
27.0 g </span>CH3OH(g) (1 mol/ 32.05g) = 0.84 mol CH3OH(g)
ΔH =+252.8kJ = Q = 0.84 mol (252.8kJ) = 213 kJ
The atom's center, or nucleus, is positively charged and the electrons that whirl around this nucleus are negatively charged, so they attract each other. The reason the force is strong is because the atom is so small. The distance between the nucleus and the electrons is about 1 Angstrom (named after a famous scientist); this is 0.00000001 cm (10-8 cm) or about 4 billionths of an
Answer:
+1
Explanation:
Na₂O₂
NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.
Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:
Na₂O₂ = 0 (oxidation number of ground state compound is zero)
2Na + 2O = 0
O = –1
2Na + 2(–1) = 0
2Na – 2 = 0
Collect like terms
2Na = 0 + 2
2Na = 2
Divide both side by 2
Na = 2/2
Na = +1
Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1
