Answer:
ΔH°r = -483.64 kJ
Explanation:
Let's consider the following balanced equation.
2 H₂(g) + O₂(g) ⇒ 2 H₂O(g)
We can calculate the standard enthalpy change of the reaction (ΔH°r) using the following expression.
ΔH°r = ∑ΔH°f(p) × np - ∑ΔH°f(r) × nr
where
ΔH°f: standard heat of formation
n: moles
p: products
r: reactants
ΔH°r = ΔH°f(H₂O(g)) × 2 mol - ΔH°f(H₂(g)) × 2 mol - ΔH°f(O₂(g)) × 1 mol
ΔH°r = (-241.82 kJ/mol) × 2 mol - 0 kJ/mol × 2 mol - 0 kJ/mol × 1 mol
ΔH°r = -483.64 kJ
Explanation:
The dipoles in CO are in opposite directions so they cancel each other out, although CO₂ has polar bonds, it is a nonpolar molecule. Therefore, the only intermolecular forces are London dispersion forces. Water (H2O) has hydrogen bond present which is a polar bond which has a high intermolecular force.
Water which has high intermolecular force will require more energy that is a higher temperature to overcome these attractions and are pulled together tightly to form a solid at higher temperatures, so their freezing point is higher.
As the temperature of a liquid decreases, the average kinetic energy of the molecules decreases and they move more slowly.
CO with lower intermolecular forces will not solidify until the temperature is lowered further.
Answer:
It depends
(plum will spoil more quickly at warm temperatures)
Explanation:
Answer:
A) 29.9g
Explanation:
first find the weight of 1 staple.
then multiply with 225