1) The longer the quarrying goes on the less and less recources we will have, soon leading to no more recources
Answer:The age can be determined by utilizing carbon-14, a radioactive isotope of carbon.
Explanation:
Answer:
The equilbrium constant is 179.6
Explanation:
To solve this question we can use the equation:
ΔG = -RTlnK
<em>Where ΔG is Gibbs free energy = 12.86kJ/mol</em>
<em>R is gas constant = 8.314x10⁻³kJ/molK</em>
<em>T is absolute temperature = 298K</em>
<em>And K is equilibrium constant.</em>
Replacing:
12.86kJ/mol = -8.314x10⁻³kJ/molK*298K lnK
5.19 = lnK
e^5.19 = K
179.6 = K
<h3>The equilbrium constant is 179.6</h3>
<span>Group 1 can be characterized as atoms that have 1 electron in their valence shell. This is valuable when dealing with these questions, because the loss or gain of valence electrons is what defines ionic relationships. When group 1 elements form ionic bonds with other atoms, they are extremely likely to lose their valence electron, since the nucleus has a weaker pull on it than, say, a chlorine atom has on its 7 valence electrons. The weaker pull between the nucleus and the valence electron of group 1 elements means that the radius is high, since the electron is more free to move with less pull on it. This also means that the first ionization energy is low, since it takes relatively little energy for that electron to be pulled away to another atom.</span>
Answer:
2.01 moles of P → 1.21×10²⁴ atoms
2.01 moles of N → 1.21×10²⁴ atoms
4.02 moles of Br → 2.42×10²⁴ atoms
Explanation:
We begin from this relation:
1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br
Then 2.01 moles of PNBr₂ will have:
2.01 moles of P
2.01 moles of N
4.02 moles of Br
To determine the number of atoms, we use the relation:
1 mol has NA (6.02×10²³) atoms
Then: 2.01 moles of P will have (2.01 . NA) = 1.21×10²⁴ atoms
2.01 moles of N (2.01 . NA) = 1.21×10²⁴ atoms
4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms
