Answer:
A small amount of solute dissolved in a larger amount of solvent.
Take this analogy to help you understand, if you were to put a teaspoon of sugar in a liter of water it would dissolve, but if you put a sack of sugar in it it would not dissolve! The solute is what is being dissolved and the solvent is what dissolves the solute, so that eliminates some of the options.
Answer:
V₂ =31.8 mL
Explanation:
Given data:
Initial volume of gas = 45 mL
Initial temperature = 135°C (135+273 =408 K)
Final temperature = 15°C (15+273 =288 K)
Final volume of gas = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 45 mL × 288 K / 408 k
V₂ = 12960 mL.K / 408 K
V₂ =31.8 mL
Answer:
pH = 5.76
Explanation:
We can solve this problem by using<em> Henderson-Hasselbach's equation</em>:
pH = pKa + log
We are already know all the required information, thus we<u> input the data given by the problem</u>:
pH = 4.76 + log(20/2)
And finally <u>calculate the pH</u>:
pH = 5.76
The pH of that acetic acid solution is 5.76.
Answer:
Answer A
Explanation:
2 H2 + O2 = 2 H2O
knowing the mole weights of these elements shows O2 is the limiting reactant in this situation.
4 g oxygen is 4 /31.998 = .125 mole O2
We should get two times these moles of water = .250 moles
weight = .250 (1.008 * 2 + 15.999) = 4.50 gm H20 (theoretical)
....but we only got 4 gm
4/4.50 = 88.8%