Answer:
305 litres of NO gas will be produced from 916 L of NO₂
Explanation:
Given the balanced equation of the chemical reaction as follows:
3 NO₂ (g) + H₂O( l) —— 2 HNO₃ (l) + NO (g)
Under standard conditions, 3 moles of No₂ will react with 1 mole of water to produce 1 mole of NO gas.
Molar volume of a gas at STP is 22.4 L
Number of moles of NO₂ gas present in 916 L = 916/22.4 = 40.893 moles of NO₂ gas
From the mole ratio of NO₂ to NO in the equation of reaction,
Number of moles of NO that will be produced = 1/3 × 40.893 moles = 13.631 moles of NO gas
Volume of 13.631 moles of NO gas = 13.631 × 22.4
Volume of NO gas produced = 305.334L
Therefore, Volume of NO gas produced from the reaction of 916 L of NO₂ with water = 305 L
Explanation:
Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.
It is known that at standard condition, vapor pressure is 760 mm Hg.
And, it is given that methanol vapor pressure in air is 88.5 mm Hg.
Hence, calculate the volume percentage as follows.
Volume percentage = 
= 
= 11.65%
Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.
16-18= -2 so it has a negative charge. Just subtract the electrons from the protons if you get a positive number it will have a positive charge and vice versa.
Less friction to stop the wheel from turning
Answer:
The total energy, i.e. sum of kinetic and potential energy, is constant.
i.e. E = KE + PE
Initially, PE = 0 and KE = 1/2 mv^2
At maximum height, velocity=0, thus, KE = 0 and PE = mgh
Since, total energy is constant (KE converts to PE when the ball is rising),
therefore, KE = PE
or, 1/2 mv^2 = mgh
or, h = v^2 /2g = 13^2 / (2x9.8) = 8.622 m
Hope this helps.
