Two changes would make this reaction reactant-favored
C. Increasing the temperature
D. Reducing the pressure
<h3>Further explanation</h3>
Given
Reaction
2H₂ + O₂ ⇒ 2H₂0 + energy
Required
Two changes would make this reaction reactant-favored
Solution
The formation of H₂O is an exothermic reaction (releases heat)
If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)
While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient
in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2
As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)
And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater
Answer:
60%
Explanation:
M(NH4NO3) = 2*14 +4*1 +3*16 = 80 g/mol
M(3O) = 3*16 = 48 g/mol
(48/80) *100 % =60% oxygen by mass.
I don’t know about 1 But number 2 is AB + xH2O = AB.xH2O
Answer:

Explanation:
Volume of a cone:
We have
and we want to find
when the height is 2 cm.
We can see in our equation for the volume of a cone that we have three variables: V, r, and h.
Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.
We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.
Plug this value for r into the volume formula:
Differentiate this equation with respect to time t.
Plug known values into the equation and solve for dh/dt.
Divide both sides by 100π to solve for dh/dt.
The height of the cone is increasing at a rate of 1/10π cm per second.