Answer:
81.97 g of NaAl(OH)₄
Solution:
The reaction for the preparation of Sodium Aluminate from Aluminium Metal and NaOH is as follow,
2 NaOH + 2 Al + 6 H₂O → 2 NaAl(OH)₄ + 3 H₂
According to this equation,
2 Moles of NaOH produces = 163.94 g (2 mole) of NaAl(OH)₄
So,
1 Mole of NaOH will produce = X g of NaAl(OH)₄
Solving for X,
X = (1 mol × 163.94 g) ÷ 2 mol
X = 81.97 g of NaAl(OH)₄
Answer:
Two immiscible liquids, oil and water, can be separated by using Separating Funnel. The mixture of oil and water forms two separate layers because they are completely insoluble in each other.
The number of particles= 1.43 x 10²⁴ particles
<h3>Further explanation</h3>
Given
53.2 L volume of gas at STP
Required
The number of particles
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.
So mol for 53.2 L :
= 53.2/22.4
= 2.375 moles
1 mol = 6.02 x 10²³ particles (molecules, atoms, ions)
The number of paricles for 2.375 moles gas :
= 2.375 x 6.02 x 10²³
= 1.43 x 10²⁴ particles
Explanation:
Yes carbon dioxide found in each case
Answer:
According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants. The law of conservation of mass is useful for a number of calculations and can be used to solve for unknown masses, such the amount of gas consumed or produced during a reaction.
