Answer:
B. ΔHreaction = ΔH°f reactants- ΔH°f products
Explanation:
Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction.
Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).
For example, in ΔH°f of H₂O, the equation is:
H₂(g) + 1/2O₂(g) → H₂O(g)
The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).
Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:
NaOH + HCl → H₂O + NaCl. ΔHr
The ΔH°f for each substance in the reaction is:
NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH (1)
HCl: 1/2H₂ + 1/2Cl₂ → HCl (2)
H₂O: H₂ + 1/2O₂ → H₂O (3)
NaCl: Na + 1/2Cl₂ → NaCl (4)
The algebraic sum of (3) + (4) is -(ΔH°f reactants):
H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants
This reaction - {(1)+(2)} ΔH°f products
NaOH + HCl → H₂O + NaCl.
ΔHr = ΔH°f reactants- ΔH°f products
In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:
B. ΔHreaction = ΔH°f reactants- ΔH°f productsThere are 154,000 mg of sugar in a
container of orange juice at Mindy's
house. She only wants to have 11 g of
sugar from the orange juice per day.
How many days can she drink the
orange juice before it is gone?
Explanation:
1g = 1000mg
154000mg = 154g
No of days she can drink = 154÷ 11 = 14days
A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg?
PLEASE HELP, will mark brainliest!!!
Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
Explanation:
mass of nitrogen = 37.8 g
mass of oxygen = (100-37.8) g = 62.2 g
Using the equation given by Raoult's law, we get:
[tex]p_A=\chi_A\times P_T[/tex]
[tex]p_{O_2}[/tex] = partial pressure of [tex]O_2[/tex] = ?
[tex]\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}[/tex]
[tex]P_{T}[/tex] = total pressure of mixture = 525 mmHg
[tex]{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles[/tex]
[tex]{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles[/tex]
Total moles = 1.94 + 1.35 = 3.29 moles
[tex]\chi_{O_2}=\frac{1.94}{3.29}=0.59[/tex]
[tex]p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg[/tex]
Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.
A gaseous mixture of O₂ and N₂ contains 37.8% nitrogen by mass, that is, in 100 g of the mixture, there are 37.8 g of N₂. The mass of O₂ in 100 g of the mixture is:
[tex]mO_2 = 100 g - 37.8 g = 62.2 g[/tex]
We will convert both masses to moles using their molar masses.
[tex]N_2: 37.8 g \times 1 mol/28.00 g = 1.35 mol\\\\O_2: 62.2 g \times 1 mol/32.00 g = 1.94 mol[/tex]
The mole fraction of O₂ is:
[tex]\chi(O_2) = \frac{nO_2}{nN_2+nO_2} = \frac{1.94mol}{1.35mol+1.94mol} = 0.590[/tex]
Given the total pressure (P) is 525 mmHg, we can calculate the partial pressure of oxygen using the following expression.
[tex]pO_2 = P \times \chi(O_2) = 525 mmHg \times 0.590 = 310 mmHg[/tex]
A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.
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