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
Answer:
Cellular respiration.
Explanation:
Through the process of cellular respiration, the energy in food is converted into energy that can be used by the body's cells. During cellular respiration, glucose and oxygen are converted into carbon dioxide and water, and the energy is transferred to ATP.
Webb has calculated the percent composition of a compound. He can check his result by adding them to see if they equal up to 100. Why? Well, percent composition tells the quantity of elements with 100 as a base of total amount. This means that it will have to add to 100 to check the result. You would add up all of the values of percent composition of elements to see if they equal 100, and if they do, the results are accurate.
Your final answer: Webb can check his result by seeing if they add up to 100, considering that is the base total quantity.
Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
Explanation : Given,
Mass of oxygen in sulfur dioxide = 3.49 g
Mass of sulfur in sulfur dioxide = 3.50 g
Mass of oxygen in sulfur trioxide = 9.00 g
Mass of sulfur in sulfur trioxide = 6.00 g
Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.
Mass of oxygen per gram of sulfur for sulfur dioxide = 
Mass of oxygen per gram of sulfur for sulfur dioxide = 
and,
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
Answer: The products of the saponification of glyceryl tripalmitate (tripalmitin) are one molecule of glycerol and three molecules of sodium salt of palmitic acid.
Explanation:
A chemical reaction in which triglycerides react with sodium hydroxide and leads to the formation of one molecule of glycerol and three molecules of a salt of fatty acid is known as saponification.
For example, when tripalmitin reacts with sodium hydroxide then it leads to the formation of one molecule of glycerol and three molecules of sodium salt of palmitic acid.
The reaction equation is as follows.

Thus, we can conclude that the products of the saponification of glyceryl tripalmitate (tripalmitin) are one molecule of glycerol and three molecules of sodium salt of palmitic acid.