Answer:
d.- The model does not provide information about the type of rock in which fossils form.
Explanation:
The fossils can be located in different parts like on land, shore and water. But, they were formed at different time, that means, they live in different ages, that is the reason why the model need to include the type of rock. Remember fossil form at different ages and with the pass of time different layers of rock formed.
Hope this info is useful.
Answer:
"The core of the Sun extends from the center to about 20–25% of the solar radius. It has a density of up to 150 g/cm3 (about 150 times the density of water) and a temperature of close to 15.7 million kelvins (K)"
- goggle
3 O2 + CS2 = CO2 + 2 SO2
That is the balanced equation!
I may be late lol!!
Hope it helps
We use the equation for the standard enthalpy change of formation:
ΔHoreaction=∑ΔHof(products)−∑ΔHof(Reactants)
to calculate for the enthalpy for the reaction
2NaOH(s)+CO2(g)→Na2CO3(s)+H2O(l)
We now have
ΔHoreaction = { ΔHfo[Na2CO3(s)] + ΔHfo[H2O(l)] } - { ΔHfo[NaOH(s)] +
ΔHfo[CO2(g)] }
where we use the following Enthalpy of Formation (∆Hfo) values:
Substance ΔHf∘ (kJ/mol)
CO2(g) −393.509
H2O(l) −285.830
Na2CO3(s) −1130.68
NaOH(s) −425.609
and taking note of the coefficients of the products and the reactants,
ΔHoreaction = [1*(−1130.68) + 1*(−285.830)] − [2*(−425.609) + 1*(−393.509)]
= -1416.51 - (-1244.727)
= -171.783 kJ/mol
≈ -171.8 kJ/mol as our enthalpy for the given reaction.
Now considering the reaction
Na2CO3(s)→Na2O(s)+CO2(g) with enthalpy of reaction ΔHoreaction=321.5kJ/mol
we also use the equation for the standard enthalpy change of formation:
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)ΔHoreaction
= { ΔHfo[Na2O(s)] + ΔHfo[CO2(g)] } - { ΔHfo[Na2CO3(s)] }
to solve for the enthalpy of formation of Na2O(s):
ΔHfo[Na2O(s)] = ΔHoreaction - ΔHfo[CO2(g)] + ΔHfo[Na2CO3(s)]
Since the coefficients are all 1,
ΔHfo[Na2O(s)] = 321.5 - (-393.509) + (-1130.68)
ΔHfo[Na2O(s)] = -415.671 kJ/mol ≈ -415.7 kJ/mol