If the mass of both the reactants is 10kg then the mass of the products also equals 10kg.
It is due to the law of conservation of mass.
Mass can neither be created nor be destroyed.
Answer:
The answer to your question is pH = 1.45
Explanation:
Data
pH = ?
Volume 1 = 200 ml
[HCl] 1 = 0.025 M
Volume 2 = 150 ml
[HCl] 2 = 0.050 M
Process
1.- Calculate the number of moles of each solution
Solution 1
Molarity = moles / volume
-Solve for moles
moles = 0.025 x 0.2
result
moles = 0.005
Solution 2
moles = 0.050 x 0.15
-result
moles = 0.0075
2.- Sum up the number of moles
Total moles = 0.005 + 0.0075
= 0.0125
3.- Sum up the volume
total volume = 200 + 150
350 ml or 0.35 l
4.- Calculate the final concentration
Molarity = 0.0125 / 0.35
= 0.0357
5.- Calculate the pH
pH = -log [H⁺]
-Substitution
pH = -log[0.0357]
-Result
pH = 1.45
Answer: Adenine and guanine are the two purines and cytosine, thymine and uracil are the three pyrimidines. The main difference between purines and pyrimidines is that purines contain a sixmembered nitrogencontaining ring fused to an imidazole ring whereas pyrimidines contain only a sixmembered nitrogencontaining ring. They both are types or categories of nitrogen containing bases present in nuclei acids of DNA and RNA.
Purines are 2 Ring or Carbon Ring, Nitrogen containing bases. That consist of these 2 rings next placed next to each other. These examples include - Adenine and Guanine.
Pyrimidines are 1 or single Ring Nitrogen containing structures. There are 3 nitrogenous bases that are categorized as pyrimidines. Cytosine, Thymine, and Uracil.
Answer:
Double bond
Explanation:
an alkene is a unsaturated hydrocarbon which means that it contain at least one double bond
Answer:
THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O
Explanation:
The empirical formula for the unknown compound can be obtained by following the processes below:
1 . Write out the percentage composition of the individual elements in the compound
C = 75.68 %
H = 8.80 %
O = 15.52 %
2. Divide the percentage composition by the atomic masses of the elements
C = 75 .68 / 12 = 6.3066
H = 8.80 / 1 = 8.8000
O = 15.52 / 16 = 0.9700
3. Divide the individual results by the lowest values
C = 6.3066 / 0.9700 = 6.5016
H = 8.8000 / 0.9700 = 9.0722
O = 0.9700 / 0.9700 = 1
4. Round up the values to the whole number
C = 7
H = 9
O = 1
5 Write out the empirical formula for the compound
C7H90
In conclusion, the empirical formula for the unknown compound is therefore C7H9O