Answer:
= 913.84 mL
Explanation:
Using the combined gas laws
P1V1/T1 = P2V2/T2
At standard temperature and pressure. the pressure is 10 kPa, while the temperature is 273 K.
V1 = 80.0 mL
P1 = 109 kPa
T1 = -12.5 + 273 = 260.5 K
P2 = 10 kPa
V2 = ?
T2 = 273 K
Therefore;
V2 = P1V1T2/P2T1
= (109 kPa × 80 mL × 273 K)/(10 kPa× 260.5 K)
<u>= 913.84 mL</u>
Answer:
b. HCOOH/ NaHCOO.
Explanation:
A buffer system may be formed in one of two forms:
- A weak acid with its conjugate base.
- A weak base with its conjugate acid.
Chose the pairs below that you could use to make a buffered solution.
a. HCI/NaOH. NO. HCl is a strong acid and NaOH is a strong base.
b. HCOOH/ NaHCOO. YES. HCOOH is a weak acid and HCOO⁻ (coming from NaHCOO) is its conjugate base.
c. HNO₂/H₂SO₃. NO. Both are acids and they are unrelated to each other.
d. NaNO₃/ HNO₃. NO. HNO₃ is a strong acid.
Answer: The correct answer is: " endothermic . "
______________________________________
<u>Note</u>: Heat flows <u> into </u> [heat <u> may be </u> absorbed within] an "<u>endothermic</u>" reaction or system
To the contrary, heat flows <u> </u><u>out </u> [heat <u> may </u><em> </em>exit from or <u> may be </u> released from] an "<u>exothermic</u>" reaction or process.
<u>Hint</u>: Think of the "prefixes" of: "<u>endo</u>thermic" and "<u>exo</u>thermic" :
_____________________________________
1) endo- = "within" (as in "endothermic" —heat tends to be absorbed/"within"/"released within"/released within"/into" ;
2) exo- = " outwards"/"exit" (as in "exothermic") —heat tends to '"exit"/leave/escape from/"be released out of/form".
_____________________________________
Hope this is helpful to you!
Best wishes to you in your academic pursuits
—and within the "Brainly" community"!
_____________________________________
In the context of chemistry, yes. Energy input is always equal to the energy output.
Answer: The entropy change of the surroundings will be -17.7 J/K mol.
Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol
Amount of Acetone given = 10.8 g
Number of moles is calculated by using the formula:

Molar mass of acetone = 58 g/mol
Number of moles = 
If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then
0.1862 moles will have = 
To calculate the entropy change for the system, we use the formula:

Temperature = 56.2°C = (273 + 56.2)K = 329.2K
Putting values in above equation, we get
(Conversion Factor: 1 kJ = 1000J)
At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,

