BaO, Barium Oxide.
Na2SO4, Sodium Sulfate.
CuO, Copper (II) Oxide.
P2O5, Diphosphorus Pentoxide.
HNO3, Nitric Acid.
CO32-, Molecular Formula.
Hope this helps. :)
Answer:
The natural phenomenon used to describe the length of a meter is the speed of light. The length of a meter is the length a light path travels in 1/(299792458) seconds through a vacuum.
The definition is better due to the uncertainty involved in the use of the length of a standard meter stick because the length of the meter stick could change due to atmospheric conditions from place to place
Explanation:
Explanation:
Tc - 99 is technetium 99.
It is a radioactive element that decays spontaneously. It has a half-life of 211,000 years and decays to stable ruthenium.
- On the periodic table, it has an atomic number of 43;
Mass number = 99
Atomic number = number of protons = number of electrons in atom
Number of protons in Tc is 43
electron is 43
Mass number = number of protons + number of neutrons
Number of neutrons = mass number - number of protons = 99-43 = 56
learn more:
Atomic number brainly.com/question/2057656
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Answer : The time passed in years is 
Explanation :
Half-life = 5730 years
First we have to calculate the rate constant, we use the formula :



Now we have to calculate the time passed.
Expression for rate law for first order kinetics is given by:

where,
k = rate constant = 
t = time passed by the sample = ?
a = let initial amount of the reactant = X g
a - x = amount left after decay process = 
Now put all the given values in above equation, we get


Therefore, the time passed in years is 