To find the Percent Composition of an atom, you use this formula:
Mass of element in the compound you're studying on ( in this case it's 5 since there is 5 Hydrogens) over the mass of the compound (which is here 79), Multiplied by 100 since you want a percent.
So we get:
So you get about:
So, the percent composition of Hydrogen in NH4HCO3 is 6.3%
Hope this Helps! :D
Answer:
p = 260 kilogram/cubic meter
Explanation:
ρ = 
= 
= 0.26 gram/milliliter
= 260 kilogram/cubic meter
Answer:
Explanation:
First, we find in the tables the ΔH of formation of each compound. As you can see in the (image 1)
Then we solve the ecuation for ΔH°reaction
ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(Reactants)
ΔH°reaction= (-2* 393.5 - 2*285.8) - (52.4 + 0) kJ/mol
ΔH°reaction = -1.41 *10^3 kJ/mol
The Elements (The Periodic Table.)
Answer:
In this case, the system doesn't be affected by the pressure change. This means that nothing will happen
Explanation:
We can answer this question applying the Le Chatelier's Principle. It says that changes on pressure, volume or temperature of an equilibrium reaction will change the reaction direction until it returns to the equilibrium condition again.
The results of these changes can define as:
Changes on pressure: the reaction will move depending the quantity of moles on each side of the reaction
Changes on temperature: The reaction will move depending on if it's endothermic or exothermic
Changes on volume: The reaction will move depending the limit reagent and the quantity of moles on each side of the reaction
In the exercise, they mention a change on pressure of the system at constant temperature (that means the temperature doesn't change). As Le Chatelier Principle's says, we must analyze what happens if the pressure increase or decrease. If pressure increase the reaction will move on the side that have less quantity of moles, otherwise, if the pressure decreases the reaction will move to the side that have more quantity of moles. In this case, we can see that both sides of the equation have the same number of moles (2 for the reactants and 2 for the products). So, in this case, we can conclude that, despite the change on pressure (increase or decrease), nothing will happen.
