Answer:

Explanation:
In this question, we wish to find the missing nuclei for the equation:

In order to find the missing species, we need to use the charge and mass balance law. That is, the mass should be conserved: the total mass on the left-hand side with respect to the arrow should be equal to the total mass on the right-hand side with respect to the arrow:

Notice from here that:

So far we know that the mass of X is 4. Similarly, we apply the law of charge conservation. The total charge should be conserved:

From here:

We have a particle:

Looking at the periodic table, an atom with Z = 2 corresponds to helium. This can also be written as an alpha particle:

Limiting reactant : O₂
Mass of N₂O₄ produced = 95.83 g
<h3>Further explanation</h3>
Given
50g nitrous oxide
50g oxygen
Reaction
2N20 + 302 - 2N204
Required
Limiting reactant
mass of N204 produced
Solution
mol N₂O :

mol O₂ :

2N₂O+3O₂⇒ 2N₂O₄
ICE method
1.136 1.5625
1.0416 1.5625 1.0416
0.0944 0 1.0416
Limiting reactant : Oxygen-O₂
Mass N₂O₄(MW=92 g/mol) :

Answer:
Heterogenous
Explanation:
Mud is a mixture of water and soil
In one mole of talc, we observe that there are:
3 moles of Mg
4 moles of Si
2 moles of H
12 moles of O
The molar ratio of O to Mg is then:
12 moles of O : 3 moles of Mg = 4 : 1
Therefore, if 6.1 moles of Mg are present, the moles of O are:
4 * 6.1 = 24.4 moles of O