1) Chemical reaction 1: 4Cu + O₂ → 2Cu₂O.
n(Cu) = 88,8 ÷ 63,55.
n(Cu) = 1,4.
n(O) = 11,2 ÷ 16.
n(O) = 0,7.
n(Cu) : n(O) = 1,4 : 0,7.
n(Cu) : n(O) = 2 : 1.
Compound is Cu₂O.
2) Chemical reaction 2: 2Cu + O₂ → 2CuO.
n(Cu) = 79,9 ÷ 63,55.
n(Cu) = 1,257.
n(O) = 20,1 ÷ 16.
n(O) = 1,257.
n(Cu) : n(O) = 1,257 : 1,257.
n(Cu) : n(O) = 1 : 1.
Compound is CuO.
Answer:
If you see in the image above, there is an unbalance force applied while playing tug of war. Since it is 1 vs 2, there is a greater net force in the right side then the left side. If it was 2 vs 2 or 1 vs 1, then they are appling balance force. You can also see in the picture that the arrows are pointing outwards (--->) rather then inwards (<---) because you are pulling the rope not pushing the rope. If you add one person on the left side, then the newtons which is 20N will become to 35N and will be balanced, but since there in only 1 person, there is less force on the left side, the newtons gets subtracted having only 20N. Since you are pulling the rope, the friction is opposite (<---). Since you are pulling the rope, you are using Kinetic force and the rope stays in potential force since it stays constant.
Hope this helps, thank you :) and I am not sure about magnitude I think you can that since there is greater force on the right side, there is more magnitude there.
I believe you are referring zero as the exponent. <span>Any number (except 0) with exponent 0 is defined to mean 1.
</span>
For one thing, there is a rule:
<span> a^m/ a^m = a^m-m = a^0
</span>But (when a is not equal to <span>0),
</span>
a^m/ a^m = 1
Therefore, we must define a^0 as 1.
LaChatelier's Principle
LaChatelier's Principle is a principle stating that if a constraint (such as a change in pressure, temperature, or concentration of a reactant) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.
To make a first step you have to know the balanced form for neutralization formula:
According to this, you can <span>calculate what you are being asked :</span>
Then we have :
Hope everything is clear, here is the exact answer you need :
