Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
<em />
To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
Answer:
The empirical formula is, C4H4S
Explanation:
Number of moles of carbon = 1.119 g/ 44g/mol = 0.025 moles
Mass of Carbon= 0.025 moles × 12 g/ mole = 0.3 g
Number of moles of hydrogen = 0.229/18g/mol × 2 = 0.025 moles
Mass of hydrogen = 0.025 moles × 1 = 0.025 g
Number of moles of sulphur = 0.407g/ 64 g/mol = 0.0064 moles
Mass of sulphur= 0.0064 moles ×32 = 0.2 g
Now we obtain the mole ratios by dividing through by the lowest ratio.
C- 0.025 moles/ 0.0064 moles, H- 0.025 moles/ 0.0064 moles, S- 0.0064 moles/0.0064 moles
C4H4S
The ground could sink and sink holes could occur, otherwise the parking lot could simply break apart and wear faster
Answer:
Head loss in turbulent flow is varying as square of velocity.
Explanation:
As we know that head loss in turbulent flow given as
Where
F is the friction factor.
L is the length of pipe
V is the flow velocity
D is the diameter of pipe.
So from above equation we can say that
It means that head loss in turbulent flow is varying as square of velocity.
We know that loss in flow are of two types
1.Major loss :Due to surface property of pipe
2.Minor loss :Due to change in momentum of fluid.
Initial buret reading means the volume of acid taken in the buret and final reading means the remaining volume of acid after experiment
