Answer:
2.99 M
Explanation:
In order to solve this problem we need to keep in mind the definition of molarity:
- Molarity = moles of solute / liters of solution
In order to calculate the moles of solute, we <u>convert 125.6 g of NaF into moles</u> using its <em>molar mass</em>:
- 125.6 g NaF ÷ 42 g/mol = 2.99 mol NaF
As the volume is already given, we can proceed to <em>calculate the molarity</em>:
- Molarity = 2.99 mol / 1.00 L = 2.99 M
The formula for kinetic energy is KE=1/2(mv²). Since both mass and velocity are multiplied by each other, particle with a larger mass needs to be moving slower than a particle with less mass if both have the same kinetic energy. You can think of it as 2KE/m=v² or 2KE/v²=m, If you increase the mass the velocity needs to decrease to keep the same KE value.
I hope this helps. Let me know in the comments if anything is unclear.
Answer: Theoretical Yield = 0.2952 g
Percentage Yield = 75.3%
Explanation:
Calculation of limiting reactant:
n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol
pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol
- n-trans-cinnamic acid is the limiting reactant
The molar ratio according to the equation mentioned is equals to 1:1
The brominated product moles is also = 9.584*10⁻⁴ mol
Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)
= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g
Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952
= 75.3%
Mass of KNO₃ : = 40.643 g
<h3>Further explanation</h3>
Given
28.5 g of K₃PO₄
Required
Mass of KNO₃
Solution
Reaction(Balanced equation) :
2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃
mol K₃PO₄(MW=212,27 g/mol) :
= mass : MW
= 28.5 : 212,27 g/mol
= 0.134
Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :
= 6/2 x mol K₃PO₄
= 6/2 x 0.134
= 0.402
Mass of KNO₃ :
= mol x MW KNO₃
= 0.402 x 101,1032 g/mol
= 40.643 g
6.022 x 1023 atoms are in 14 grams of NO2
