Answer:
Explanation:
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In this case, it is possible to comprehend these mass-particles problems by means of the concept of mole, molar mass and the Avogadro's number because one mole of any substance has 6.022x10²³ particles and have a mass equal to the molar mass.
In such a way, for C₆H₁₂O₆, whose molar mass is about 180.16 g/mol, the referred mass would be:
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There are several ways of expressing concentration of solution. Few of them are listed below
1) mass percentage
2) volume percentage
3) Molarity
4) Normality
5) Molality
In most of the drugs, concentration is expressed either in terms of mass percentage or volume percentage. For, solid in liquid type systems, mass percentage is convenient way of expressing concentration, while for liquid in liquid type solutions, expressing concentration in terms of volume percentage is preferred. Present system is an example of liquid in liquid type solution
Here, concentration of H2O2 is given antiseptic = 3.0 % v/v
It implies that, 3ml H2O2 is present in 100 ml of solution
Thus, 400 ml of solution would contain 4 X 3 = 12 ml H2O2
There is 0.02538502095915 Moles in 5 grams of gold.
Answer:
Assume that 100 grams of C2H4 is present. This means that there are 85.7 grams of carbon and 14.3 grams of hydrogen.
Convert these weights to moles of each element:
85.7 grams carbon/12 grams per mole = 7 moles of carbon.
14.3 grams hydrogen/1 gram per mole = 14 moles of hydrogen.
Divide by the lowest number of moles to obtain one mole of carbon and two moles of hydrogen.
Since we know that there cannot be a stable CH2 molecule, multiply by two and you have C2H4 which is ethylene - a known molecule.
The secret is to convert the percentages to moles and find the ration of the constituents.
