If you don't practice enough it's obviously going to be hard but if you practice enough it's going to be a piece of cake so don't think if it's going to be hard or not just think it's going to be worth the try at the very end
Answer:
Beta emission
Explanation:
In beta emission, a neutron is converted into a proton thereby emitting an electron and a neutrino. A neutrino is a particle that serves to balance the spins.
When a nucleus undergoes beta emission, the mass number of the parent and daughter nuclei remain the same while the atomic number of the daughter nucleus is greater than that of its parent by one unit.
Hence, in beta emission, the daughter nucleus is found one pace to the right of the parent in the periodic table.
Answer is: sodium (Na) and iodine (I₂).
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First ionic bonds in this salt are separeted
because of heat:
</span>NaI(l) → Na⁺(l) + I⁻(l).
Reaction of reduction
at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.
2Na⁺(l) + 2e⁻ → 2Na(l).
Reaction of oxidation
at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.
The anode is positive
and the cathode is negative.
Answer:
0.453 moles
Explanation:
The balanced equation for the reaction is:
2Fe(s) + 3O2(g) ==> 2Fe2O3
From the equation, mass of O2 involved = 16 x 2 x 3 = 96g
mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2
= 100g
Therefore 96g of O2 produced 100g of Fe2O3
32.2g of O2 Will produce 100x32.2/96
= 33.54g of Fe2O3
Converting it to mole using number of mole = mass/molar mass
but molar mass of Fe2O3 = 26 + (16 X 3)
= 74g/mole
Therefore number of mole of 33.54g of Fe2O3 = 33.54/74
= 0.453 moles
Answer:
28.75211 kj
Explanation:
Given data:
Mass of iron bar = 841 g
Initial temperature = 84°C
Final temperature = 7°C
Heat released = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
specific heat capacity of iron is 0.444 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 7°C - 84°C
ΔT = -77°C
By putting values,
Q = 841 g × 0.444 j/g.°C × -77°C
Q = 28752.11 j
In Kj:
28752.11 j × 1 kJ / 1000 J
28.75211 kj
