Answer:
saturated fats
Explanation:
bcoz it is solid at room temperature and does not have double bonds between carbons
In the given situation, the reaction is-
NO + H2 ↔ Products
The rate of the reaction can be expressed (in terms of the decrease in the concentration of the reactants) as-
Rate = -dΔ[NO]/dt = -dΔ[H2]/dt
Now, if the concentration of NO is decreased there will be fewer molecules of the reactant NO which would decrease the its collision with H2. As a result the rate of the forward reaction would also decrease.
Ans) A decrease in the concentration of nitrogen monoxide decreases the collisions between NO and H2 molecules. the rate of the forward reaction then decreases.
The answer is c it is Carbon Dioxide
Answer:
The mass stays the same only volume changes, the volume decreases
Explanation:
The ice shrinks (decreases volume) and becomes more dense. The weight will not (and cannot) change.
Explanation:
In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.
Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.
STP conditions are defined as a pressure of
100 kPa
and a temperature of
0
∘
C
. Under these conditions for pressure and temperature, one mole of any ideal gas occupies
22.7 L
- this is known as the molar volume of a gas at STP.
So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.
The balanced chemical equation for this decomposition reaction looks like this
2
KClO
3(s]
heat
×
−−−→
2
KCl
(s]
+
3
O
2(g]
↑
⏐
⏐
Notice that you have a
2
:
3
mole ratio between potassium chlorate and oxygen gas.
This tells you that the reaction will always produce
3
2
times more moles of oxygen gas than the number of moles of potassium chlorate that underwent decomposition.
Use potassium chlorate's molar mass to determine how many moles you have in that
231-g
sample
231
g
⋅
1 mole KClO
3
122.55
g
=
1.885 moles KClO
3
Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate
1.885
moles KClO
3
⋅
3
moles O
2
2
moles KClO
3
=
2.8275 moles O
2
So, what volume would this many moles occupy at STP?
2.8275
moles
⋅
22.7 L
1
mol
=
64.2 L
