Half life is the time taken for a radioactive isotope to decay by half its original mass. In this case the half life of carbon-14 is 5.730 years.
Using the formula;
New mass = original mass × (1/2)^n; where n is the number of half lives (in this case n=1 )
New mass = 2 g × (1/2)^1
= 1 g
Therefore; the mass of carbon-14 that remains will be 1 g
Answer:
C3H6O
Explanation:
The percentage composition of the elements in the compound are given as follows:
62.1 % carbon = 62.1g of C
10.5 % hydrogen = 10.5g of H
27.6 % oxygen = 27.6g of O
Next, we convert each mass to mole by dividing by their molar/atomic mass
C = 62.1/12 = 5.175mol
H = 10.5/1 = 10.5mol
O = 27.6/16 = 1.725mol
Next, we divide each mole value by the smallest mole value (1.725)
C = 5.175mol ÷ 1.725 = 3
H = 10.5mol ÷ 1.725 = 6.086
O = 1.725mol ÷ 1.725 = 1
The empirical ratio approximately of C:H:O is 3:6:1, hence, the empirical formula is C3H6O
Answer: The molarity of 
in this solution is 
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
Given : pH = 10.66
Putting in the values:
![10.66=-\log[H^+]](https://tex.z-dn.net/?f=10.66%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-10.66}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-10.66%7D)
![[H^+]=2.2\times 10^{-11}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D2.2%5Ctimes%2010%5E%7B-11%7D)
Thus the molarity of in this solution is
Answer:
Its either with itself to become fluoride Fl2 or with an alkali metal (elements in group 1) to achieve a stable electronic configuration or octet structure
Explanation:
Answer:
Photic Zone
Explanation:
The photic zone is the uppermost layer of the sea that recieves sunlight allowing phytoplankton to perform photosynthesis.
