Answer:
Isostasy controls the regional elevations of continents and ocean floors in accordance with the densities of their underlying rocks. ... The concept of isostasy played an important role in the development of the theory of plate tectonics.
<span>They should all be eukaryotic.</span>
The correct option is B. To get the number of atom for each compound, each element in the compound will be counted as an atom. For instance, for Fe[ClO4]2, there are 1 atom of Fe, 2 atoms of Cl, and 8 atoms of O, making a total of 11 atoms [1 + 2 + 8= 11]. The other options have less than 11 atoms.
Answer:
a) The lewis dot structure is shown in the image attached to this answer
b) The formal charge on each of the atoms is zero
c) bromine has an oxidation state of +5 while fluorine has an oxidation state of -1
d) 90 degrees
e) Square Pyramidal
f) polar bonds
g) polar molecule
Explanation:
The molecule BrF5 has a formal charge of zero. It exhibits an sp3d2 hybridization state with a square pyramidal geometry. The bond angle in the molecule is 90 degrees. It is a molecule of the type AX5E. The oxidation state of bromine is +5 while that of fluorine is -1.
The Br-F bonds are polar. The overall molecule is polar due to asymmetric charge distribution concentrating on the central atom since the molecule is square pyramidal.
Exothermic reactions:
1) release heat to the surroundings
2) the change of enthalpy, ΔH, is negative: ΔH < 0
3) the temperature of the system increases
With that you can conclude about every equation given:
<span>NH3(g) + 12.0 kcal → ½N2(g) + 3/2 H2(g) : is not exothermic because heat in the side of the reactants means that heat is being used, not released.
C(graphite) → C(diamond), ΔH = - 0.45 kcal : is exothermic because ΔH is negative
C + 2S → CS2, ΔH = 27,550 cal: is not exothermic because ΔH is positive
CH4 + 2O2 → CO2 + 2H2O + 212,800 cal : is exothermic because heat appears as a product of the reaction, which means that it is released.
2H2O → 2H2 + O2, ΔH = +58 kcal : is not exothermic because ΔH is positive.</span>
