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KiRa [710]
4 years ago
15

Sn =0.14 (E0) Sn+2 = -0.15 (E0) Which form of tin is the stronger reducing agent?

Chemistry
2 answers:
Gemiola [76]4 years ago
7 0

<u>Answer:</u> The stronger reducing agent will be Sn^{2+}

<u>Explanation:</u>

Reducing agent is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

We are given:

E_o(Sn)=0.14V\\E_o(Sn^{2+})=-0.15V

The electrode potential values of given substances determine the tendency to gain electrons. Higher value of positive electrode potential means that it will gain more and hence, will be a stronger oxidizing agent.

If the specie has negative electrode potential, it means that it will easily loose electrons and hence, will be considered as a stronger reducing agent.

Thus, the stronger reducing agent among the given species will be Sn^{2+}

dimaraw [331]4 years ago
3 0

Sn is the only form of tin that can act as a reducing agent.

<em>Reduction</em>: Sn²⁺ + 2e⁻ ⟶ Sn; E° = -0.15 V

The Sn²⁺ is being <em>reduced</em>, so it is acting as an <em>oxidizing agent</em>.

<em>Oxidation</em>: Sn ⟶ Sn²⁺ + 2e⁻; E° = +0.15 V

The Sn is being <em>oxidized</em>, so it is acting as a reducing agent.

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Which choice describes the function of the pancreas gland?
gayaneshka [121]

Answer:

A. It secretes hormones that regulate blood sugar levels.

Explanation:

The pancreas is the organ that helps in the process of digestion of the food. The location of the pancreas is in the abdomen. It helps in the process of digestion of the food and regulating the blood sugar level in the body. It helps in regulating and controlling the glucose level in the body during the process of digestion. Pancreas produces enzymes that are useful in the process of digestion.

6 0
3 years ago
How many electrons in an atom can have each of the following quantum number or sublevel designations?
storchak [24]

10 electrons in an atom can have each of the 3d quantum number or sublevel designations.

<h3>What are four quantum numbers?</h3>

Quantum numbers are used to characterize the values of conserved quantities in a quantum system's dynamics in both quantum physics and chemistry.

1) The "n" sign refers to the principal quantum number. It shows how many shells there are.

2) The "l" sign indicates the azimuthal quantum number. It gives the form of the orbitals and their angular momentum. 0 to n-1 make up its range.

3) "m_{l}" stands for the magnetic quantum number. It serves as a cue for spatial orientation. The range of it is -l to +l.

4) Spin quantum number, represented by the symbol "m_{s}". It shows how the electron is spinning.

Learn more about quantum numbers here:

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4 0
2 years ago
A slice of pizza has 500 kcal. If we could burn the pizza and use all the heat to warm a 50-L container of cold water, what woul
gayaneshka [121]

Answer:

10°C  

Explanation:

Heat gain by water = Heat lost by the slice of pizza

Thus,    

m_{water}\times C_{water}\times \Delta T=Q

<u>For water:  </u>

Volume = 50.0 L

Density of water= 1 kg/L  

So, mass of the water:  

Mass\ of\ water=Density \times {Volume\ of\ water}  

Mass\ of\ water=1 kg/L \times {50.0\ L}  

Mass of water  = 50 kg

Specific heat of water = 1 kcal/kg°C  

ΔT = ?

For slice of pizza:  

Q = 500 kcal

So,

50\times 1\times \Delta T=500  

ΔT = 10°C  

Increase in temperature = 10°C  

3 0
3 years ago
Check all the statements that are correct. Conservation of mass means:
Llana [10]

Conservation of mass means:

  • We must start and end with the same amount of matter
  • Matter cannot be created

<em>In a closed / isolated system, the total mass of the substance before the reaction will be equal to the total mass of the reaction product. </em>

<h3>Further explanation </h3>

In general, the reaction takes place in the open air. If the reaction results in the form of gas as in combustion, the mass of the reaction results will be smaller than the original mass. Or if the reaction binds to a larger than the environment then the reaction results will produce a greater mass than the original mass.

For example, in a paper combustion reaction the mass of combustion ash is smaller than the mass of paper that is burned because there is a gas released into the air

Conservation of mass applies to a closed system, where the masses before and after the reaction are the same

Let see the answer choices :

  • We can start a reaction with different types of elements than we have at the end of the reaction.

This statement is true, the elements in the reaction will also be found in the product, but in the form of other compounds

  • Matter cannot be created

Matter cannot be created nor destroyed by chemical reactions

This statement is true, because matter only changes or is rearranged

  • Matter can be destroyed in a chemical reaction

Incorrect statement

<h3>Learn more </h3>

Complete and balance the molecular equation

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Lavoisier transforms the field of Chemistry

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Keywords : matter,  Conservation of mass, chemical reaction,closed system

8 0
3 years ago
Read 2 more answers
5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0
3 years ago
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