Answer: 1.27 bar
Explanation:
1 atm = 1.01325 bar
1.25 atm = Z (let Z be the unknown value)
To get the value of Z, cross multiply
Z x 1 atm = 1.25 atm x 1.01325 bar
1 atm•Z = 1.2665625 atm•bar
To get the value of Z, divide both sides by 1 atm
1 atm•Z/1 atm = 1.2665625 atm•bar/1atm
Z = 1.2665625 bar
(Round up Z to the nearest hundredth as 1.27 bar)
Thus, 1.25 atm when coverted gives 1.27 bar
Answer:
a) increases
b) decreases
c) does not change
d) increases
Explanation:
The vapour pressure of a liquid is dependent on;
I) the magnitude of intermolecular forces
II) the temperature of the liquid
Hence, when any of these increases, the vapour pressure increases likewise.
Similarly, the boiling point of a liquid depends on the magnitude of intermolecular forces present because as intermolecular forces increases, more energy is required to break intermolecular bonds.
Lastly, increase in surface area of a liquid does not really affect it's vapour pressure.
Answer:
1.16 moles CO₂
Explanation:
To find the moles of CO₂, you need to (1) convert grams C₈H₁₈ to moles (via the molar mass) and then (2) convert moles C₈H₁₈ to moles CO₂ (via the mole-to-mole ratio from equation coefficients). It is important that the conversions/ratios are arranged in a way that allows for the cancellation of units. The final answer should have 3 significant figures like the given value.
Molar Mass (C₈H₁₈): 8(12.011 g/mol) + 18(1.008 g/mol)
Molar Mass (C₈H₁₈): 114.232 g/mol
2 C₈H₁₈ + 25 O₂ -----> 16 CO₂ + 18 H₂O
^ ^
16.6 g C₈H₁₈ 1 mole 16 moles CO₂
-------------------- x ----------------- x ------------------------- = 1.16 moles CO₂
114.232 g 2 moles C₈H₁₈
Answer:
Plants use carbon dioxide for photosynthesis
Explanation:
They take it in, and chemically combined it with water to form glucose and oxygen
Answer: A solute dissolves because its particles interact with the particles of a solvent. Anything that allows more solvent to touch more solute will cause a solute to dissolve more quickly. Small pieces of a substance dissolve faster than large pieces.
