How many molecules are in 3.01 g of H2O
1 answer:
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Answer:
A breakdown of the breaking buffer was first listed with its respective component and their corresponding value; then a table was made for the stock concentrations in which the volume that is being added was determined by using the formula . It was the addition of these volumes altogether that make up the 0.25 L (i.e 250 mL) with water
Explanation:
Given data includes:
Tris= 10mM
pH = 8.0
NaCl = 150 mM
Imidazole = 300 mM
In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.
Stock Concentration Volume to be Final Concentration
added
1 M Tris 2.5 mL 10 mM
5 M NaCl 7.5 mL 150 mM
1 M Imidazole 75 mL 300 mM
. is the formula that is used to determine the corresponding volume that is added for each stock concentration
The stock concentration of Tris ( 1 M ) is as follows:
.
The stock concentration of NaCl (5 M ) is as follows:
.
The stock concentration of Imidazole (1 M ) is as follows:
.
Hence, it is the addition of all the volumes altogether that make up 0.25L (i.e 250 mL) with water.
Explanation:
Charges on both magnesium and oxygen is 2. Though opposite in sign, they have equal charges so, both of them will be cancelled by each other.
As a result, formula of magnesium oxide is MgO and not .
The student write the equation as , it is not correct.
Therefore, given equation will be balanced as follows.
Since, number of atoms on both reactant and product side are equal. Hence, this equation is completely balanced.