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25 days ago

According to reference table adv-10, which reaction will take place spontaneously?

1 answer:

5 0

Missing table!! list out the elements with their symbols starting with capital letters!!!

http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

Ni2+ +Pb(s) → Ni(s) + Pb2+
The oxidation potential of Pb(s) --> Pb2+(aq) is 0.126 V
The reduction potential for Ni2+(aq) --> Ni(s) is -0.25 V

Summing these gives a potential of -0.124 V (NOT SPONTANEOUS, AS THE SIGN IS NEGATIVE)

Au3+ + Al(s) → Au(s) + Al3+ Au3+(aq) --> Au(s) +1.5 V Al --> Al3+ +1.66 VV= 3.16 (SPONTANEOUS, POSITIVE SIGN)Sr2+ + Sn(s) → Sr(s) + Sn2+

Sr2+(aq) + 2 e– Sr(s) V= -2.89V
Sn --> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)

Fe2+ + Cu(s) → Fe(s) + Cu2+
Fe2+(aq) + 2 e– Fe(s) V= -0.44 V
Cu --> Cu2+ V = -0.337V

V= -0.777V (no spontaneous)

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A sealed vessel contains 0.200 mol of oxygen gas, 0.100 mol of nitrogen gas, and 0.200 mol of argon gas. The total pressure of t

Tems11 [2400]

Respuesta:

Explicación:

Utilizamos la relación de moles para calcular la presión parcial. El número total de moles es 0.2 + 0.2 + 0.1 = 0.5 moles

Ahora, sabemos que la fracción molar del gas argón es 0.2/0.5

La presión parcial se determina así. Para calcular esto, simplemente multiplicamos el número de moles por la presión total.

0.2/0.5 * 5 = 1.0/0.5 = 2.00atm

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18 days ago

The symbol for xenon (Xe) would be a part of the noble gas notation for the element antimony. cesium. radium. uranium.

castortr0y [2743]

Noble gas notation serves as a condensed form of indicating electron configurations. This notation employs the symbol for the preceding noble gas in the electron configuration of an element. For antimony, the noble gas prior is Kr, which means Xe is not used in its electron configuration. Similarly, for radium, the prior noble gas is Rn, whereas, for uranium, it is also Rn. However, for cesium, the preceding noble gas is Xe, thus it is utilized in the noble gas notation for Sb, specifically written as: Cs: [Xe] 6s.

Answer: cesium

8 0

1 month ago

Read 2 more answers

Write the balanced molecular and net ionic equation for the reaction that occurs when the contents of the two beakers are added

alisha [2704]

1) reacting hydrochloric acid with nickel:

Balanced molecular equation: Ni(s) + 2HCl(aq) → NiCl₂(aq) + H₂(g).

Ionic equation: Ni(s) + 2H⁺(aq) + 2Cl⁻(aq) → Ni²⁺(aq) + 2Cl⁻(aq) + H₂(g).

Net ionic equation: Ni(s) + 2H⁺(aq) → Ni²⁺(aq) + H₂(g).

In this reaction, nickel undergoes oxidation, changing from an oxidation state of 0 to +2, while hydrogen is reduced from +1 to 0 (H₂).

2) reacting sulfuric acid with iron:

Balanced molecular equation: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g).

Ionic equation: Fe(s) + 2H⁺(aq) + SO₄²⁻(aq) → Fe²⁺(aq) + SO₄²⁻(aq) + H₂(g).

Net ionic equation: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g).

In this scenario, iron is oxidized from an oxidation state of 0 to +2, while hydrogen experiences reduction from +1 to 0 (H₂).

3) hydrobromic acid reacting with magnesium:

Balanced molecular equation: Mg(s) + 2HBr(aq) → MgBr₂(aq) + H₂(g).

Ionic equation: Mg(s) + 2H⁺(aq) + 2Br⁻(aq) → Mg²⁺(aq) + 2Br⁻(aq) + H₂(g).

Net ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).

This reaction sees magnesium oxidized from 0 to +2, and hydrogen reduced from +1 to 0 (H₂).

4) acetic acid reacting with zinc:

Balanced molecular equation: Zn(s) + 2CH₃COOH(aq) → (CH₃COO)₂Zn(aq) + H₂(g).

Ionic equation: Zn(s) + 2H⁺(aq) + 2CH₃COO⁻(aq) → Zn²⁺(aq) + 2CH₃COO⁻(aq) + H₂(g).

Net ionic equation: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).

Here, zinc gets oxidized from 0 to +2 (Zn²⁺), while hydrogen is reduced from +1 to 0 (H₂).

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1 month ago

In a group assignment, students are required to fill 10 beakers with 0.720 M CaCl2. If the molar mass of CaCl2 is 110.98 g/mol a

alisha [2704]

The result is 200 g. Given that the molar mass of CaCl2 is 110.98 g/mol, this indicates that there are 110.98 g in 1 L of a 1 M solution. Let's calculate the amount of CaCl2 in 0.720 M. Using the proportion 110.98 g: 1 M = x: 0.720 M, we find x to be 79.90 g. Therefore, in 1 L of a 0.720 M solution, there is 79.90 g. Next, we need to create ten beakers with 250 mL each, totaling 10 * 250 mL = 2500 mL or 2.5 L. Then, using the equation 79.90 g: 1 L = x: 2.5 L, we calculate x = 79.90 g * 2.5 L: 1 L, resulting in x = 199.75 g, approximately 200 g.

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9 days ago

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If you pump air into cycle tyre a slight warming effect is noticed at valve stem why

castortr0y [2743]

The slight warm feeling noticed at the valve stem when air is pumped into the tire is likely due to the kinetic energy generated by the friction from the pump and the resultant increase in gas pressure within the tire.

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27 days ago