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Answer: The momentum of the child and milk together is 58.125 kg.m/s
Explanation:
Momentum is defined as the product of object's mass and velocity.
Mathematically,

where, p = momentum
m = mass of the object
v = velocity of the object
In the given question, we are given that a child of mass 21.0 kg is carrying a gallon of milk having mass 2.25 kg and running with a velocity of 2.5 m/s. Hence, the momentum by both milk and child will be:
....(1)
Given:

Putting values in equation 1, we get:

Hence, the momentum of the child and milk together is 58.125 kg.m/s
To get the ∆S of the reaction, we simply have to add the ∆S of the reactants and the ∆S of the products. Then, we get the difference between the ∆S of the products and the ∆S of the products. If the <span>∆S is negative, then the reaction spontaneous. If the otherwise, the reaction is not spontaneous.</span>
Answer:
71.372 g or 0.7 moles
Explanation:
We are given;
- Moles of Aluminium is 1.40 mol
- Moles of Oxygen 1.35 mol
We are required to determine the theoretical yield of Aluminium oxide
The equation for the reaction between Aluminium and Oxygen is given by;
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.
Therefore;
1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen
1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium
Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.
4 moles of aluminium reacts to generate 2 moles aluminium oxide.
Therefore;
Mole ratio Al : Al₂O₃ is 4 : 2
Thus;
Moles of Al₂O₃ = Moles of Al × 0.5
= 1.4 moles × 0.5
= 0.7 moles
But; 1 mole of Al₂O₃ = 101.96 g/mol
Thus;
Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol
= 71.372 g