Answer:
The nucleus of an atom consists of Protons and Neutrons.-A.
First, we use avogadro's number to convert atoms into moles. Then, relate the number of moles from elemental to the compound. Lastly, we use conditions at STP to calculate the volume. We do as follows:
<span>9.86 x 10²⁸ O-atoms ( 1 mol / 6.022x10^23 atoms O) ( 1 mol N2O2 / 2 mol O ) ( 22.4 L / 1 mol ) = 1833809.37 L needed</span>
This answer is 24 because 2.17 x 10 -8 is 24 so that would be your answer
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:
60 grams of ice will require 30.26 calories to raise the temperature 1°C.
Explanation:
The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released or absorbed by the system.
m is the mass of the ice (m = 60.0 g).
c is the specific heat capacity of ice (c = 2.108 J/g.°C).
ΔT is the temperature difference (ΔT = 1.0 °C).
∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.
<em>It is known that 1.0 cal = 4.18 J.</em>
<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>
