Details:
The equation to calculate work done is defined as follows.
W = 
where, k = proportionality constant = 
d = separation distance = 0.45 nm = 
Now we will insert the given values into the formula above to compute the work done as follows.
W =
= ![\frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}](https://tex.z-dn.net/?f=%5Cfrac%7B-%5B8.99%20%5Ctimes%2010%5E%7B9%7D%20Jm%2FC%5E%7B2%7D%20%5Ctimes%203.2%20%5Ctimes%2010%5E%7B-19%7D%20C%20%5Ctimes%20-3.2%20%5Ctimes%2010%5E%7B-19%7D%20C%5D%7D%7B0.25%20%5Ctimes%2010%5E%7B-9%7D%20m%7D)
= 
Thus, we can conclude that the work needed to increase the distance between the two ions to infinity is 
.
Answer:
1984
Explanation:
Utilizing the equation;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/2 = half-life of the radioactive isotope
t= age of the wine
Ao= initial activity of the wine
A= activity at time = t
Substituting values, we have 0.693/12.3 = 2.303/t log (5.5/0.688)
0.693/12.3 = 2.079/t
0.056 = 2.079/t
t= 2.079/0.056
t= 37 years
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Answer:
CH4
Explanation:
The ideal behavior of gases generally depends on the strength of intermolecular forces between gas molecules and whether polar bonds are present.
In the case of CCl4, polar bonds exist along with the more electronegative chlorine atom, leading to stronger intermolecular forces at 400K, as opposed to CH4 which contains only non-polar bonds.
Thus, at 400K, CH4 behaves more like an ideal gas compared to CCl4.
Every unicellular organism prospers by executing metabolic activities.
Metabolic activities encompass the set of chemical reactions essential for sustaining life.
Explanation:
Different metabolic pathways maintain an organism's viability. Various metabolic activities occur in all living organisms.
These include processes like cellular respiration, reproduction, excretion, and digestion. Each living cell engages in these activities to survive.
Organisms acquire the energy necessary for these activities through food consumption.
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The reaction can be described as follows: CO + 2H2 = CH3OH. Given the specified quantities of the reactants, we will identify the limiting reactant and compute the remaining excess amount. Calculating, 1.50 x 10^-6 g CO converts to 5.36 x 10^-8 mol CO, while 6.80 x 10^-6 g H2 equals 3.37 x 10^-6 mol H2. Thus, CO is fully consumed in the reaction, leaving 3.37 x 10^-6 - 5.36 x 10^-8 = 3.32 x 10^-6 moles of gas.
