Answer:
reduce the temperature of the gas
Explanation:
when you reduce the temperature of the gas the pressure will decrease
Given :
A 250 ml beaker weighs 13.473 g .
The same beaker plus 2.2 ml of water weighs 15.346 g.
To Find :
How much does the 2.2 ml of water, alone, weigh .
Solution :
Now, mass of water is given by :

Therefore , mass of 2.2 ml of water alone is 1.873 g .
Hence , this is the required solution .
Answer:
62.5 moles of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O
From the balanced equation above,
2 moles of C₈H₁₈ reacted with 25 moles of O₂.
Finally, we shall determine the number of mole of O₂ needed to react with 5 moles of C₈H₁₈. This can be obtained as shown below:
From the balanced equation above,
2 moles of C₈H₁₈ reacted with 25 moles of O₂.
Therefore, 5 moles of C₈H₁₈ will react with = (5 × 25) / 2 = 62.5 moles of O₂.
Thus, 62.5 moles of O₂ is needed for the reaction.
1 gallon of antifreeze = 60% of mixture. Total mixture:
Vm = 1/0.6
= 1.67 gallons
Volume of water = total vol - antifreeze vol
= 1.67 - 1
= 0.67 gallon of water
Answer:
%age Yield = 85.36 %
Solution:
The Balance Chemical Reaction is as follow,
C₆H₁₂O + Acid Catalyst → C₆H₁₀ + Acid Catalyst + H₂O
According to Equation ,
100 g (1 mole) C₆H₁₂O produces = 82 g (1 moles) of C₆H₁₀
So,
4.0 g of C₆H₁₂O will produce = X g of C₆H₁₀
Solving for X,
X = (4.0 g × 82 g) ÷ 100 g
X = 3.28 g of C₆H₁₀ (Theoretical Yield)
As we know,
%age Yield = (Actual Yield ÷ Theoretical Yield) × 100
%age Yield = (2.8 g ÷ 3.28 g) × 100
%age Yield = 85.36 %