I believe you mean KO2 reacting with H2O. The reaction is 4KO2+2H2O->4KOH +3O2. The mole ratio O2:KO2 is 3:4. Thus moles of O2 produced = 0.500/4*3 = 0.375 mol.
Given:
Mass of the ionic compound = 10.00 g
Mass of water = 75.0 g
Initial temperature of water T1= 23.2 C
Final temperature of water T2 = 31.8 C
Specific heat of water c = 4.18 J/gC
To determine:
Enthalpy of dissolution of the ionic compound
Heat gained by water equation:
Q = mcΔT
m = mass of water
c = specific heat
ΔT = change in temperature (T2-T1)
Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J
Thus, the heat gained by water equals heat lost by the ionic compound (enthalpy of dissolution)
Therefore, q(ionic) = 2696 J
ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g
Answer: A) enthalpy change = 2.7*10² J/g
Explanation:
Elements provided:
F, Sr, P, Ca, O, Br, Rb, Sb, Li, S
Elements sharing similar reactivity belong to the same group in the periodic table, indicating that those in the same column exhibit comparable reactivity. Here are the identified groupings:
Li and Rb are alkali metals in group 1
Ca and Sr are alkaline earth metals in group 2
F and Br are halogens in group 7
O and S belong to group 6
P and Sb are classified in group 5 of the periodic table
Thus, these classifications illustrate elements with the same chemical characteristics.
M = 81.50g, mm = m/n
n =???
PV = nRT --> n = PV/RT
n = (1.75)(4.92)/(.0821)(307)
n = 8.61/25.20 =.342
--> mm = m/n = 81.5/.342 = 238.58
Answer:
Explanation:
The oxidation state corresponds to the charge of each atomic ion. An increase indicates oxidation of the element while a decrease reflects reduction of the element.
2AgCl+Zn⟶2Ag+ZnCl2
Zinc undergoes oxidation, while Ag experiences reduction.
Ag⁺ changes to Ag (oxidation state decreases), thus Ag is reduced.
Zn alters to Zn⁺² (oxidation state increases), hence Zn is oxidized.
4NH₃+3O₂⟶2N₂+6H₂O
The oxidation state of nitrogen in ammonia is -3
whereas it is zero in elemental nitrogen.
An increase in the oxidation state indicates nitrogen is oxidized.
The oxidation state of oxygen is zero when in molecular oxygen and -2 when in water. Therefore, the oxidation state decreases, indicating oxidation is reduced.
Fe₂O₃+2Al⟶Al₂O₃+2Fe
The oxidation state of Fe in Fe₂O₃ is +3, switching to zero in Fe, so iron is reduced.
Aluminum's oxidation state is zero in Al, rising to +3 in Al₂O₃, indicating it is oxidized.
