The bigger the atomic radius the easier it is to oxidise the atom. Remember that an atom is oxidized by the loss of an electron.
Explanation:
The bigger the atomic radius the further away the valence electron are from the attractive force of the atomic nucleus. This means that the energy required to remove an electron from the valence shell is easier compared to an atom with a smaller atomic radius. This is because you need to overcome the attractive force of the nucleus on the electron for you to oxidize the atom.
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Answer:
Yes. Weight is the product of mass times gravitational acceleration. So all you have to do is vary the gravitational field and you vary weight.
Explanation:
Answer:
John Dalton
Explanation:
John Dalton in 1808 suggested that all matter consists of tiny particles called atoms and that the atoms of a specific element are identical.
He postulated the Dalton's atomic theory which has the following important parts;
- All matters consists of indivisible particles called atoms
- Atoms of the same element are similar and are different from atoms of other elements.
- Atoms can neither be created nor destroyed.
- Atoms combine in simple whole ratios to form compounds.
Answer:
Explanation:
= Initial volume = 100 mL
= Final volume = 1000 mL
= Initial concentration = 0.5 M
= Final concentration
We have the relation
The new concentration is .
Answer:
ΔG° of reaction = -47.3 x 
J/mol
Explanation:
As we can see, we have been a particular reaction and Energy values as well.
ΔG° of reaction = -30.5 kJ/mol
Temperature = 37°C.
And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:
The first step is to calculate the equilibrium constant for the reaction:
Equilibrium Constant K = ![\frac{[HPO4-2] x [ADP]}{ATP}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO4-2%5D%20x%20%5BADP%5D%7D%7BATP%7D)
And we have values given for these quantities in the biological cell:
[HP04-2] = 2.1 x 
M
[ATP] = 1.2 x 
M
[ADP] = 8.4 x M
Let's plug in these values in the above equation for equilibrium constant:
K = ![\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2.1x10%5E%7B-3%7D%5D%20x%20%5B8.4x10%5E%7B-3%7D%5D%20%7D%7B%5B1.2%20x%2010%5E%7B-2%7D%5D%20%7D)
K = 1.47 x M
Now, we have to calculate the ΔG° of reaction for the biological cell:
But first we have to convert the temperature in Kelvin scale.
Temp = 37°C
Temp = 37 + 273
Temp = 310 K
ΔG° of reaction = (-30.5 ) + (8.314)x (310K)xln(0.00147)
Where 8.314 = value of Gas Constant
ΔG° of reaction = (-30.5 x ) + (-16810.68)
ΔG° of reaction = -47.3 x J/mol
