The balanced chemical equation for the neutralization of HCl with 
is:
Given weight of = 5g
Moles of = 
Volume of HCl solution = 
Assuming the density of the solution is 1.0 g/mL
Mass of HCl solution = 50 g
Overall mass of the solution = 50 g + 5 g = 55 g
To find the heat of neutralization, we calculate:
Q = m C ΔT
where m equals the mass of the solution = 55 g
C represents the specific heat capacity of the solution = 4.184
ΔT signifies the temperature change = 6.8 K = (6.8 - 273) C = -266.2
The enthalpy of neutralization per mole of
= 
Solution:
Washing Clothes & Dissolving Sugar
Clarification:
Consider each scenario:
1) For washing clothes, water is essential; without it, washing is ineffective.
2) Connecting brake pedals to brake pads requires solids, not liquids.
3) To deodorize a space, one would likely reach for an aerosol spray, which is a gas.
4) Sculpting involves solid tools and a solid medium.
5) Dissolving sugar necessitates a liquid to be effective!
6) While one might assert that paint is a liquid, it still might not fit the category; I would categorize this application as solid.
7) Gears employed in machinery are solid components!
The response to this inquiry involves energy release. The bonds holding molecule atoms act as energy reserves. One method of energy release occurs when these bonds are severed, allowing energy to disperse outward. This breaking leads to smaller molecules rather than the creation of a larger one.
Answer:
In all listed reactions, ΔH°rxn does not correspond to the ΔH°f of the resulting product.
Explanation:
The standard enthalpy of formation (ΔH°f) signifies the enthalpy change that occurs when 1 mole of a product is created from its basic elements in their standard states.
1/2 O₂(g) + H₂O(g) ⟶ H₂O₂(g)
ΔH°rxn does not equal ΔH°f of the product, since H₂O(g) is a compound rather than an element.
Na⁺(g) + F⁻(g) ⟶ NaF(s)
ΔH°rxn is not the same as ΔH°f of the product because Na and F are not in their standard states (Na(s); F₂(g)).
K(g) + 1/2 Cl₂(g) ⟶ KCl(s)
ΔH°rxn is not equal to ΔH°f of the product due to K being outside its standard state (K(s)).
O₂(g) + 2 N₂(g) ⟶ 2 N₂O(g)
ΔH°rxn does not match ΔH°f of the product as 2 moles of N₂O are produced.
In none of the above cases does ΔHrxn match ΔHf of the product.
<span>13.6
grams of mercury are present in a barometer that holds
per milliliter.
13.6*6.5 = 88.4</span>
