Answer: Slowly move at least 3 m away from the side mirror. Observe your image as you ... Compare the images formed in different mirrors. ... but if i stand 3m away, even though there is no light, it reflects the other material it can ... is transparent so you cannot see yourself much in the mirror, that is my observation.
Explanation:
Answer:
90.3 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ
We can find the standard enthalpy of formation for NO using the following expression.
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))
ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol
ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol
ΔH°f(NO(g)) = 90.3 kJ/mol
Answer:
Explanation:
In CF4 and NF3, the valence electron groups on the central C and N atoms have a tetrahedral arrangement. The shapes of the molecules are determined by the number of bonding and nonbonding of electrons: since CF4 has four bonded atom(s) and zero lone pair(s) of electrons, the shape is tetrahedral.
Answer:
ΔH° = -186.2 kJ
Explanation:
Hello,
This case in which the Hess method is applied to compute the required chemical reaction. Thus, we should arrange the given first two reactions as:
(1) it is changed as:
SnCl2(s) --> Sn(s) + Cl2(g)...... ΔH° = 325.1 kJ
That is why the enthalpy of reaction sign is inverted.
(2) remains the same:
Sn(s) + 2Cl2(g) --> SnCl4(l)......ΔH° = -511.3 kJ
Therefore, by adding them, we obtain the requested chemical reaction:
(3) SnCl2(s) + Cl2(g) --> SnCl4(l)
For which the enthalpy change is:
ΔH° = 325.1 kJ - 511.3 kJ
ΔH° = -186.2 kJ
Best regards.
