Answer:
See the explanation
Explanation:
1) The Lewis structure for
has a central Carbon<em> </em>atom attached to Oxygen atoms.
In the
we will have a structure: O=C=O the <u>central atom</u> "carbon" we will have <u>2 sigma bonds and 2 pi bonds</u>, therefore, we have an <u>Sp hybridization</u>. For O we have <u>1 pi and 1 sigma bond</u>, therefore, we have an <u>Sp2 hybridization</u>.
2) These atoms are held together by <u>double bonds.</u>
<u></u>
Again in the structure of
: O=C=O we only have double bonds.
3. Carbon dioxide has a Carbon dioxide has a <u>Linear</u> electron geometry.
Due to the double bonds we have to have a linear structure because in this geometry the atoms will be further apart from each other.
4. The carbon atom is <u>Sp</u> hybridized.
We will have for carbon 2 pi bonds, so we will have an <u>Sp</u> hybridization.
5. Carbon dioxide has two Carbon dioxide has two C(p) - O(p) π bonds and two C(sp) - O(Sp2) σ bonds.
(See figures)
Figure 1: Carbon hybridization
Figure 2: Oxygen hybridization
Answer:
-1815.4 kJ/mol
Explanation:
Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:
∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)
This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.
In this case:
ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol
When heat energy is supplied to a material it can raise the temperature of mass of the material.
Specific heat is the amount of energy required by 1 g of material to raise the temperature by 1 °C.
equation is
H = mcΔt
H - heat energy
m - mass of material
c - specific heat of the material
Δt - change in temperature
substituting the values in the equation
120 J = 10 g x c x 5 °C
c = 2.4 Jg⁻¹°C⁻¹
The simplified solubility of glucose at 30°C is 1.25 g/g of water. Considering that the density of water at 30°C is 1 g/mL, the equivalent mass of 400 mL of water is also 400g.
The concentration of the solution in water is,
550 g/400g of water = 1.375 g glucose / g of water
Since the concentration is higher compared to the solubility of glucose at the specified temperature, it can be said that the solution is SATURATED.