The activity of the sample is:
R₀ = 4.5 x 10⁻⁶ Ci (3.70 x 10¹⁰ decays/s / 1 Ci)
= 166500 decays / s
The number of nuclei is:
N₀ = (166500 decays/s) / (5730 yr * 3.154 x 10⁷s /1yr) = 9.2 x 10⁻⁷
The mass of a ₆C¹⁴ source is:
m = N₀m₀ = (9.2 x 10⁻⁷) * (2.34 x 10⁻²⁶) = 2.16 x 10⁻³² kg
Answer:
100J = K.E
Explanation:
K.E =1/2 *m*v^2
= 1/2*0.500*(20.0)^2
= 1/2 *500/1000 *400
So here 2will cancel with 400,and it will be reduced to 200 then the zeros of 200 will cancel with 1000's zeroes and then when the zeros of 200 get cancelled with the zeroes of 1000 u will be left with 500/10 *2 then cancel 2 with 10 then u get 500/5 then reduce that u get 100.
Hope it helps pls mark as brainliest!!!
Answer:
Explanation:
1. First, you want to break up the aqueous solutions into their respective ions (remember to write the charge and number of each element). Note that you can only do this to compounds marked as <em>aqueous</em>, so leave the solid be.
(Technically, both Potassium ions would have a coefficient of 6. I am leaving them as they were originally just for ease)
2. Now, you want to get rid of compounds on each side that are the same. These are called spectator ions, and they are exactly as the name says: they simply "spectate" the reaction, meaning they can be removed from the complete ionic equation.
Both the sulfate and potassium ions would be considered spectator ions and would be canceled out. Once removed the equation would look like:
3. The complete ionic equation would be:
Answer:
The relationship between conductivity and the number of ions in solution means that it can be used to measure solution concentration and give a value for Total Dissolved Solids. This means that measurements are of the total concentration of ions in solution not the concentration of any one ion.
The chemical equation that shows the reaction between nh3 and cuh206 is detailed as: [Cu(H2O)6]2+ (aq) + 2NH3(aq). —> [Cu(OH)2(H2O)4](s) + 2NH4 + (aq). the blue precipitate is Cu(OH)2(H2O)4 in which the blue color is caused by the Cu present in the solid.
