speed equals wavelength times frequency so
.7 x 500 = 350
The correct option is C.
It is very important as a nurse to ensure that one always create a complete and comprehensive documentation of the care given to a client. This has many advantages and one of it is that it will the give health providers that will be taken care of the patient in the future the necessary background information about the client. This will enable them to attend to the client properly based on his previous medial history.
Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
Answer:
Explanation:
A chemical formula can be defined as a notation that is used to show which element and how many is contained in a chemical compound.
Also, in chemistry, the sum of charges of the anion and the cation of any ionic compound is always equal to zero.
A chemical equation is considered to be balanced when the amount of reactants on the left is equal to the amount of products on the right.
Therefore;
[2]FeBr3 + [3]Na2S → [1]Fe2S3 + [6]NaBr
In the above chemical equation, we will balance the reactants in the chemical equation with the smallest coefficients possible;
Two (2) moles of Iron (III) Bromide reacts with two (2) moles of Sodium Sulfide to form Iron (III) Sulfide and Sodium Bromide.
