Yes, the correct answer is b
Answer:
KO is the limiting reactant.
0.11 mol O₂ will be produced.
Explanation:
4 KO₂ + 2 H₂O ⇒ 4 KOH + 3 O₂
Find the limiting reagent by dividing the moles of the reactant by the coefficient in the equation.
(0.15 mol KO₂)/4 = 0.0375
(0.10 mol H₂O)/2 = 0.05
KO₂ is the limiting reagent.
The amount of product produced depends on the limiting reagent. To find how much is produced, take moles of limiting reagent and multiply it by the ratio of reagent to product. You can find the ratio by looking at the equation. For every 4 moles of KO₂, 3 moles of O₂ are produced.
0.15 mol KO₂ (3 mol O₂)/(4 mol KO₂) = 0.1125 mol O₂
0.11 mol O are produced.
Answer: Molarity of a physiological saline solution is 0.15 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
n = moles of solute
=volume of solution in ml
Given : 0.89 g of sodium chloride in 100 ml of solution.
To calculate the moles, we use the equation:
Volume of solution = 100 ml
Molarity of physiological saline solution is 0.15 M
Answer:
B.Carbohydrates and D. water
Answer:
The student should weigh out 61.2g of ethanolamine [6.12 * 10]
Explanation:
In this question, we are expected to calculate the mass of ethanolamine needed to make 60.0ml of it given that the density of the ethanolamine in question is 1.02g/cm^3
Mathematically, it has been shown that mass = density * volume
Hence, by multiplying the density by the volume, we get the mass.
Now, from the question we can see that we have the values for the density and the volume. We now need to get the mass.
Since cm^3 is same as ml, we need not perform any conversion.
Hence, the needed mass is:
60 * 1.02 = 61.2g
