Answer:
5H3AsO3(aq) + 2MnO4-(aq) + 6H+(aq) → 5H3AsO4(aq) + 2Mn2+(aq) + 3H2O(aq)
Explanation:
Every net balanced ionic equation is composed of a union of two half equations;
The oxidation half equation (indicating electron loss) and the reduction half equation (indicating electron gain). Remember that redox reactions is a process in which electrons are lost and gained by chemical species simultaneously. One specie looses electrons in the oxidation half equation while the other specie gains electrons in the reduction half equation.
The balanced redox reaction equation shows the overall redox process and shows at a glance the total number of elect tribe lost or gained in the redox process. The overall redox reaction equation for the titration described in the question is;
5H3AsO3(aq) + 2MnO4-(aq) + 6H+(aq) → 5H3AsO4(aq) + 2Mn2+(aq) + 3H2O(aq)
Which elements cannot have more than an octet of electrons? Select all that apply
C
S
O
N
Br
Answer:
{ Carbon, Oxygen, Nitrogen }
Explanation:
Elements can only have more than an octet of electrons if they demonstrate an expanded octet. This is if they belong to groups in or beyond the third group. Why? Well these elements have d - orbitals that they can rely on to expand the number of electrons that could otherwise be limited. * Here we are focusing on main group elements, P - block elements more specifically. *
Carbon belongs to the 2 group, and thus doesn't have an empty d - orbital. Thus, it can't have more than an octet of electrons. Sulfur belongs to group 3, hence has an empty d - orbital, and can have more than an octet of electrons. Oxygen belongs to the 2 group, and thus doesn't have an empty d - orbital, so it can't have more than an octet of electrons. Same goes for Nitrogen. Bromine belongs to group 4, thus has empty d - orbitals, and can expand further than Sulfur can - it can have more than an octet of electrons.
Solution = { Carbon, Oxygen, Nitrogen }
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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