The density of human fat is 
. The mass of the pure fat is 11.1 lbs.
First, convert the mass from pounds to grams as follows:
1 lb=453.592 g
Thus,
Density is defined as mass per unit volume, meaning volume can be calculated as:
By substituting the values,
Consequently, the volume gained by the individual will be 
.
The percentage of calcium carbonate that reacted is 2.5%. The reaction in question allows us to determine the equilibrium Kp: Kp = the partial pressure of carbon dioxide, since the other components are solids. We'll apply the ICE table to the provided equilibrium. At the start, we have 0.2 for calcium carbonate with no initial moles of other substances. As the reaction progresses, we set the changes to be -x for calcium carbonate, +x for carbon dioxide, and +x for the other product, leading us to an equilibrium of 0.2-x for calcium carbonate while both other products are at x. Using Kp = Kc(RT)ⁿ, where n represents the mole difference of gaseous products and reactants, we find n to equal 1 for this reaction. With R as the gas constant (8.314 J/mol K) and the temperature at 800 °C (1073 K), we substitute the values accordingly. Upon calculation, we find x = 0.005, which indicates the amount of calcium carbonate that dissociated or reacted, leading us to the reacted percentage.
Answer:
The glycerol solution has a molality of 2.960×10^-2 mol/kg.
Explanation:
Calculating the moles of glycerol involves the formula: Moles = Molarity × Volume of solution = 2.950×10^-2 M × 1 L = 2.950×10^-2 moles.
To find the mass of water, use: Mass = Density × Volume = 0.9982 g/mL × 998.7 mL = 996.90 g, which converts to 0.9969 kg.
The formula for molality is: Molality = Moles of solute/Mass of solvent (in kg) = 2.950×10^-2/0.9969 = 2.960×10^-2 mol/kg.
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.
