Hi! the english version of the given question is "A helium balloon is inflated to the volume of 0.045 m3 at a temperature of 2 ° C. If the balloon is cooled to -12 ° C. What will its new volume be? Consider that the pressure does not vary". The answer is given in english.
Answer:
The new volume of balloon is .
Explanation:
Let's assume the helium gas inside the balloon behaves ideally.
The total number of moles of helium gas inside the balloon remains constant.
Applying combined gas law, we get:
Where:
are the initial and final pressure of the balloon respectively.
are the initial and final volume of the balloon respectively.
are the initial and final temperature in the kelvin scale respectively.
Given:
Substituting the above values, we get:
Answer:
Generally the ice should be more than 4 inches thick to skate on it safely. However, the ice thickness is not always even and there can be thin spots, especially near springs or near river inlets or outlets. Most lakes and ponds don't completely freeze because the ice (and eventually snow) on the surface acts to insulate the water below. Our winters aren't long or cold enough to completely freeze most local water bodies. This process of lakes turning over is critically important to the life in the lake.
Explanation:
Answer:
HI(aq) + NaOH(aq)-------------> NaI(aq) + H2O(l)
Explanation:
The molecular reaction between hydroiodic acid and aqueous sodium hydroxide is shown above. It is a reaction of one mole of hydroiodic acid with one mole of sodium hydroxide to yield salt and water only. It is a neutralization reaction. Hydrogen iodide dissolves in water to produce hydronium ions which agrees with Arrhenius description of acids. Hydroiodic acid is a strong acid with PKa of -9.3
Answer:
0.674M of NaOH
Explanation:
Molarity, M, is an unit of concentration defined as ratio between moles of solute and liters of solution.
In the problem, sodium hydroxide (NaOH) is the solute (Less quantity) that is dissolved in 3.77 liters of water, probably.
Thus, molarity of 2.54 mol NaOH in 3.77L is:
2.54mol NaOH / 3.77L = <em>0.674M of NaOH</em>