Answer:
The concentration of the solution is 1.364 molar.
Explanation:
Volume of perchloric acid = 29.1 mL
Mass of the solution = m
Density of the solution = 1.67 g/mL

Percentage of perchloric acid in 48.597 solution :70.5 %
Mass of perchloric acid in 48.597 solution :
= 
Moles of perchloric acid = 
In 29.1 mL of solution water is added and volume was changed to 250 mL.
So, volume of the final solution = 250 mL = 0.250 L (1 mL = 0.001 L)


The concentration of the solution is 1.364 molar.
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%
Answer:
Specific heat is defined as the amount of heat needed to raise the temperature by one degree celsius. Therefore, in 1 kg there are 100 grams so, 10 grams equal 0.01 Kg. Thus, calculate the specific heat value as follows. Thus, we can conclude that specific heat of the given metal is 0.5 .
Explanation: