Answer:
Explanation:
Isotopes are atoms of elements that have the same atomic number but different mass number hence ISOTOPY. Radioactive Isotopes on the other hand are unstable as they either undergo Alpha decay, beta decay or gamma decay compared to stable isotopes.
Radioactive elements decay at varyinf rates as such the rate of radioactive decay is used in the characterization of radioactive element and mostly expressed in terms of the half life of the radioactive elements.
Half life is the time taken for half of the total atoms of an elements to decay into half of its initial sizes. for example, the half life of radium-226 is 1622 years, it implies that if we have 1000000 radium atoms at the beginning, then at the end of 1622years, 500000 would have disintegrated. These phenomenon can never be experienced by stable isotopes as such they can not be used in carbon dating techniques. reason why uranium-238 is mostly and commonly used in the earth crust to estimate the ages of rocks because it has a half life of 4.5 x 10^9 years.
And also, the radioactive isotopes of most common light element are short, they have very short half life (few days or weeks) and they decay rapidly to vanshing point, as such, they are not found in nature to any reasonable extent.
Answer:
Three pairs of walking legs, wings, body divided into three segments, pair of sensory antennae.
Explanation:
Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M
