The answer to this question is 5
Answer:
24g of NaOH are required
Explanation:
Molarity, M, is an unit of concentration widely used in chemistry defined as the ratio between moles of solute (In this case, NaOH), and volume of solution in liters.
We can find the moles of NaOH and its mass with the volume and desired concentration as follows:
<em>Moles NaOH:</em>
400.0mL = 0.400L * (1.50mol / L) = 0.600 moles NaOH
<em>Mass NaOH -Molar mass: 40.0g/mol-:</em>
0.600 moles * (40.0g / mol) =
<h3>24g of NaOH are required</h3>
When we describe the energy of a particle as quantized, we mean that only certain values of energy are allowed. ... In this case, whenever we measure the particle's energy, we will find one of those values. If the particle is measured to have 4 Joules of energy, we also know how much energy the particle can gain or lose. Quantized energy means that the electrons can possess only certain discrete energy values; values between those quantized values are not permitted
Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,
CO + 2 H₂ → CH₃OH
Calculating Moles of CO:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 1 Mole of CO
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of CO
Solving for X,
X = (3.60 × 10² g × 1 Mole) ÷ 32 g
X = 11.25 Moles of CO
Calculating Moles of H₂:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 2 Mole of H₂
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of H₂
Solving for X,
X = (3.60 × 10² g × 2 Mole) ÷ 32 g
X = 22.5 Moles of H₂
Result:
3.60 × 10² g of CH₃OH is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.
The balanced equation for the neutralisation reaction is as follows
2H₃PO₄ + 3Mg(OH)₂ --> Mg₃(PO₄)₂ + 6H₂O
stoichiometry of H₃PO₄ to H₂O is 2:6
number of H₃PO₄ moles reacted - 0.24 mol
if 2 mol of H₃PO₄ form 6 mol of H₂O
then 0.24 mol of H₃PO₄ forms - 6/2 x 0.24 = 0.72 mol of H₂O
therefore 0.72 mol of H₂O are formed
