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o-na [289]
3 years ago
11

The spontaneous reaction that occurs when the cell in the picture operates is as follows: 2Ag+ + Cd(s) à 2 Ag(s) + Cd2+ (A) Volt

age increases. (B) Voltage decreases but remains > zero. (C) Voltage becomes zero and remains at zero. (D) No change in voltage occurs. (E) Direction of voltage change cannot be predicted without additional information.
Chemistry
1 answer:
murzikaleks [220]3 years ago
6 0

The question is incomplete, the remaining part of the question is

Which of the above occurs for each of the following circumstances?

A 50-milliliter sample of a 2-molar Cd(NO3)2 solution is added to the left beaker.

Answer:

Voltage decreases but remains > zero.

Explanation:

Given the balanced redox reaction equation:

2Ag^+(aq) + Cd(s) ---------------> 2 Ag(s) + Cd^2+(aq)

Concentration affects the cell voltage according to Nernst equation. Change in concentration must lead to a change in cell Voltage.

As the concentration of the Cd(NO3)2 solution is increased, voltage decreases because of the increase in the concentration values but voltages remains above zero.

You might be interested in
A(n) _____ is equivalent to an electron. gamma ray omega particle alpha particle beta particle
ladessa [460]

Answer:

beta particle

Explanation:

Explanation:

Alpha beta and gamma radiations are the examples of ionizing radiations. When an atom is an excited state and having high energy, the atom is in unstable state. The excess of energy is released by the atom to get the stability. The released energy is in the form of radiations which may include alpha, beta, gamma, X-ray etc.

¹⁴₆C → ¹⁴₇N + ⁰₋₁e

The beta radiations are emitted in this reaction. The one electron is ejected and neutron is converted into proton.

Beta radiations:

Beta radiations are result from the beta decay in which electron is ejected. The neutron inside of the nucleus converted into the proton an thus emit the electron which is called β particle.

The mass of beta particle is smaller than the alpha particles.

They can travel in air in few meter distance.

These radiations can penetrate into the human skin.

The sheet of aluminium is used to block the beta radiations

3 0
3 years ago
List out the moles for each element in the following compound ZnBr2
natulia [17]

Answer:

If the colors in a chromatography are able to dissolve and travel up a paper wick, what kind of chemical property do the colors have when mixed with rubbing alcohol?

(You may need to search "Chemical Properties")

4 0
2 years ago
Calculate the number of atoms in 52g of He.​
MissTica

Answer:

78.3 × 10²³ atoms of helium are present in 52 g.

Explanation:

Given data:

Mass of He = 52 g

Number of atoms = ?

Solution:

First of all we will calculate the number of moles of He

Number of moles = mass /molar mass

Number of moles = 52 g/ 4 g/mol

Number of moles = 13 mol

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

1 mole = 6.022 × 10²³ atoms of helium

13 mol  × 6.022 × 10²³ atoms of helium  / 1 mole

78.3 × 10²³ atoms of helium

5 0
3 years ago
A. Which reactant is the limiting reagent?
Tasya [4]

Answer:

a. Zinc is the limiting reactant.

b. m_{ZnBr_2}^{by\ Zn}=162.61gZnBr_2

c. m_{Br_2}^{leftover}=6.6g

Explanation:

Hello there!

a. In this case, when zinc metal reacts with bromine, the following chemical reaction takes place:

Zn+Br_2\rightarrow ZnBr_2

Thus, since zinc and bromine react in a 1:1 mole ratio, we can compute their reacting moles to identify the limiting reactant:

n_{Zn}=47.2g*\frac{1mol}{65.38g} =0.722molZn\\\\n_{Br_2}=122g*\frac{1mol}{159.8g} =0.763molBr_2

Thus, since zinc has the fewest moles we infer it is the limiting reactant.

b. Here, we compute the grams of zinc bromide via both reactants:

m_{ZnBr_2}^{by\ Zn}=0.722molZn*\frac{1molZnBr_3}{1molZn} *\frac{225.22gZnBr_2}{1molZnBr_2} =162.61gZnBr_2\\\\m_{ZnBr_2}^{by\ Br_2}=0.763molBr_2*\frac{1molZnBr_3}{1molBr_2} *\frac{225.22gZnBr_2}{1molZnBr_2} =171.95gZnBr_2

That is why zinc is the limiting reactant, as it yields the fewest moles of zinc bromide product.

c. Here, since just 0.722 mol of bromine would react, we compute the corresponding mass:

m_{Br_2}^{reacted}=0.722molBr_2*\frac{159.8gBr_2}{1molBr_2} =115.4gBr_2

Thus, the leftover of bromine is:

m_{Br_2}^{leftover}=122g-115.4g\\\\m_{Br_2}^{leftover}=6.6g

Best regards!

8 0
3 years ago
5.
ANEK [815]
Knowledge of chemistry allows the public to make informed decisions
5 0
3 years ago
Read 2 more answers
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