Answer:
B - What we change
Explanation:
Dependent Variable - What we measure
Control Variable - what stays the same
Conclusion - what we conclude
<em>Hope</em><em> </em><em>this</em><em> </em><em>can</em><em> </em><em>Help</em><em>!</em>
<em>:</em><em>D</em>
Hello. This question is incomplete. The full question is:
"Consider the following reaction. 2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
A proposed reaction mechanism is: NO(g) + NO(g) N2O2(g) fast N2O2(g) + H2(g) → N2O(g) + H2O(g) slow N2O(g) + H2(g) → N2(g) + H2O(g) fast
What is the rate expression? A. rate = k[H2] [NO]2 B. rate = k[N2O2] [H2] C. rate = k[NO]2 [H2]2 D. rate = k[NO]2 [N2O2]2 [H2]"
Answer:
A. rate = k[H2] [NO]2
Explanation:
A reaction mechanism is a term used to describe a set of phases that make up a chemical reaction. In these phases a detailed sequence of each step is shown, composed of several complementary reactions, which occur during a chemical reaction.
These mechanisms are directly related to chemical kinetics and allow changes in reaction rates to be observed in advance.
Reaction rate, on the other hand, refers to the speed at which chemical reactions occur.
Based on this, we can observe through the reaction mechanism shown in the question above, that the action "k [H2] [NO] 2" would have no changes in the reaction rate.
Answer:
C) 4.24 x 
Explanation:
E = Δm
Δm = 0.0284 x 1.66 x
kg = 4.714x
kg
putting value in above equation
E = 4.714x
kg x (3x
= 4.24 x 
Answer:
Explanation:
Principal quantum no "n" = 3
Azimuthal quantum no "l"= 1
Magnetic quantum no "m"= +1/2
Over all is 3pz
Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol