Answer:
8.44 atm
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 2.25 L
Initial temperature (T₁) = 350 K
Initial pressure (P₁) = 1.75 atm
Final volume (V₂) = 1 L
Final temperature (T₂) = 750 K
Final pressure (P₂) =?
The final pressure of the gas can be obtained as illustrated below:
P₁V₁/T₁ = P₂V₂/T₂
1.75 × 2.25 / 350 = P₂ × 1 / 750
3.9375 / 350 = P₂ / 750
Cross multiply
350 × P₂ = 3.9375 × 750
350 × P₂ = 2953.125
Divide both side by 350
P₂ = 2953.125 / 350
P₂ = 8.44 atm
Thus, the final pressure of the gas is 8.44 atm.
Answer:
Higher concentration to an area of lower concentration
Explanation:
When you open a perfume bottle at a corner of a room, after a while, its fragrance can be perceived across a distance at the other end of the room. This is because, molecules of the compound in the fragrance have moved from the area of higher concentration in the perfume bottle, across a concentration gradient to a region of lower concentration at the other end of the room. This is diffusion.
A 25.00 ml sample of hydrochloric acid solution, HCl, is titrated with 0.0512 m NaOH solution. the volume of NaOH solution required is 21.68 ml then the molarity of the HCl solution is 0.044 M .
Calculation ,
Formula used : 
...( i )
Where M is the molarity or concentration and V is the volume in ml .
concentration of hydrochloric acid solution ( 
) = ?
concentration of NaOH ( 
) = 0.0512 M
volume of hydrochloric acid solution ( 
) = 25.00 ml
volume of NaOH ( 
) = 21.68 ml
Putting the value of concentration , volume of both in equation ( i ) we get .
× 25.00 ml = 0.0512 × 21.68 ml
= 0.0512 × 21.68 ml / 25.00 ml= 0.044 M
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Answer:
The volume will decrease.
Explanation:
According to Charles' Law, the volume of a gas is directly proportional to its temperature.
V = kT
Thus, if T decreases, V decreases.
The amount of energy in kilocalories released from 49 g of glucose given the data is -4.4 Kcal
How to determine the mole of glucose
Mass of glucose = 49 g
Molar mass of glucose = 180.2 g/mol
Mole of glucose = ?
Mole = mass / molar mass
Mole of glucose = 49 / 180.2
Mole of glucose = 0.272 mole
How to determine the energy released
C₆H₁₂O₆ →2C₂H₆O + 2CO₂ ΔH = -16 kcal/mol
From the balanced equation above,
1 mole of glucose released -16 kcal of energy
Therefore,
0.272 mole of glucose will release = 0.272 × -16 = -4.4 Kcal
Thus, -4.4 Kcal were released from the reaction
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