Answer:
2NaCN + CaCO3 --> Na2CO3 + Ca(CN)2
Explanation:
Knowing the names gets us: NaCN + CaCO3 --> Na2CO3 + Ca(CN)2
Balance: there are two sodiums and cyanides on the product side so add a 2 to the reactant side.
2 mole Na needs 1 mole Cl2
4 mole Na = 2 mole Cl2
Answer:
Two electrons each!
Explanation:
The question pretty much requires us to find the oxidation number of Zinc in the compound.
Zn3P2
Following oxidation number rules;
O.N of Zn3P2 = 0
.Since phosphorus has valence of 5, it needs three more electrons to achieve its octet state. Hence;
Oxidation number of P = -3
Let oxidation number of Zn = x
We have;
3x + 2 (-3) = 0
3x + (-6) = 0
3x = 6
x = 6/3 = 2
This means each zn electrons lost 2 electrons.
Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
I think they oxidize the stain
