The oxidation number of elements in equation below are,
4NH₃ + 3Ca(ClO)₂ → 2N₂ + 6H₂O + 3CaCl₂
O.N of N in NH₃ = -3
O.N of Ca in Ca(ClO)₂ and CaCl₂ = +2
O.N of N in N₂ = 0
O.N of Cl in Ca(ClO)₂ = +1
O.N of Cl in CaCl₂ = -1
Oxidation:
Oxidation number of Nitrogen is increasing from -3 (NH₃) to 0 (N₂).
Reduction:
Oxidation number of Cl is decreasing from +1 [Ca(ClO)₂] to -1 (CaCl₂).
Result:
<span>N is oxidized and Cl is reduced.</span>
Answer:
Yes.
The nuclear equation {226/88 Ra → 222/26 Rn + 4/2 He} is balanced. As we know that an alpha particle is identical to a helium atom. This implies that if an alpha particle is eliminated from an atom's nucleus, an atomic number of 2 and a mass number of 4 is lost.
Therefore, the equation will be reduced to:
226 - 4 = 222
88 - 2 = 86
Hence, the equation is balanced.
Explanation:
Answer: 19.4 g/cm3
Explanation: density is the relationship between mass over volume.
So density of gold is 15.7g/0.81cm3 = 19.4 g/cm3
Answer:
The ideal gas law is expressed mathematically by the ideal gas equation as follows;
P·V = n·R·T
Where;
P = The gas pressure
V = The volume of the gas
n = The number of moles of the gas present
R = The universal gas constant
T = The temperature of the gas
A situation where the ideal gas law is exhibited is in the atmosphere just before rainfall
The atmospheric temperature of the area expecting rainfall drops, (when there is appreciable blockage of the Sun's rays by cloud covering) followed by increased wind towards the area, which indicates that the area was in a state of a low pressure, 'P', and or volume, 'V', or a combination of both low pressure and volume P·V
When the entry flow of air into the area is observed to have reduced, the temperature of the air in the area is simultaneously sensed to have risen slightly, therefore, the combination of P·V is seen to be proportional to the temperature, 'T', and the number of moles of air particles, 'n' in the area
Explanation:
Answer:
88,88 % de O y 11,11 % de H
Explanation:
La composición porcentual se define como la masa que hay de cada mol de átomo en 100g. Las moles de agua en 100g son:
<em>Masa molar agua:</em>
2H = 2*1g/mol = 2g/mol
1O = 1*16g/mol = 16g/mol
Masa molar = 2 + 16 = 18g/mol
100g H2O * (1mol / 18g) = 5.556 moles H2O.
Moles de hidrógeno:
5.556 moles H2O * (2mol H / 1mol H2O) = 11.11 moles H
Moles Oxígeno = Moles H2O = 5.556 moles
La masa de hidrógeno es:
11.11mol * (1g/mol) 11.11g H
La masa de oxígeno es:
5.556 mol * (16g / 1mol) = 88.89g O
Así, el porcentaje de O es 88.89% y el de H es 11.11%. La opción correcta es:
<h3>88,88 % de O y 11,11 % de H</h3>
