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Effectus [21]
2 years ago
6

Consider the following voltaic cell. Identify the anode and cathode, and indicate the direction of Na ion and NO3- ion flow from

the salt bridge. A. Sn is the anode and Fe is the cathode; Na ions flow into half-cell compartment (A) and NO3- ions flow into half-cell compartment (B). B. Fe is the anode and Sn is the cathode; NO3- ions flow into half-cell compartment (A) and Na ions flow into half-cell compartment (B). C. Sn is the anode and Fe is the cathode; NO3- ions flow into half-cell compartment (A) and Na ions flow into half-cell compartment (B). D. Fe is the anode and Sn is the cathode; Na ions flow into half-cell compartment (A) and NO3- ions flow into half-cell compartment (B).
Chemistry
1 answer:
levacccp [35]2 years ago
4 0

Answer:

Fe is the anode and Sn is the cathode; NO3- ions flow into half-cell compartment (A) and Na ions flow into half-cell compartment (B).

Explanation:

The purpose of a salt bridge in a voltaic cell is not to move electrons from the electrolyte, it's main function is the maintenance of charge balance between the half cells .The electrons flow from the anode to the cathode.

The iron half cell which has a higher reduction potential serves as anode. Iron atoms give up electrons to form positive ions according to the equation;

Fe(s)-----> Fe^2+(aq) + 2e

Hence there will be an excess of positive ions in the anode (compartment A). NO3^- ions from the salt bridge will migrate towards this compartment to ensure charge balance.

Similarly, in compartment B(cathode) Sn^2+ ions will accept two electrons according to the reaction;

Sn^2+(aq) + 2e ------> Sn(s)

This reaction causes a depletion of positive ions in compartment B, hence Na^+ from the salt bridge move towards this compartment to ensure charge balance.

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If a car travels 1255 miles how many kilometers is that i mile = 1.609km
ANTONII [103]

Answer:

2,019 km

Explanation:

Step 1: Given data

Distance traveled by the car (D): 1,255 mi

Step 2: Convert the distance traveled by the car to kilometers

To convert one unit into another, we use a conversion factor. In this case, the appropriate conversion factor between miles and kilometers is 1 mile = 1.609 km. The distance traveled by the car, in kilometers, is:

D = 1,255 mi × (1.609 km/1 mi) = 2,019 km

3 0
2 years ago
At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a
djverab [1.8K]

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2

<em>Where P is the pressure of each compound in equilibrium.</em>

If initial pressure of H2S is 3.00atm, concentrations in equilibrium are:

H2S = 3.00 atm - 2X

H2 = 2X

S2: = X

Replacing:

3.2x10^{-7} = (2X)^2X / (3-2X)^2

3.2x10^{-7} = 4X^3 / 9- 6X+4X^2

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Solving for X:

X = 0.008945 atm

As in equilibrium, pressure of S2 is X, <em>pressure is 0.008945 atm</em>

7 0
2 years ago
In the formula X2O5, the symbol X could represent an element in Group
topjm [15]

Answer: (3) 15

Explanation: We criss-cross down the oxidation numbers to get the subscripts for the correct formulas. That means the X has an oxidation number of 5. The element with the + oxidation number is always written first so it is +5. Of the groups names, only group 15 has +5 as an oxidation number.

6 0
2 years ago
Which statements describe how stars are born? Check all that apply. Stars are born in clouds of gas and dust called nebulas. The
prohojiy [21]

Answer: 1.Stars are born in clouds of gas and dust called nebulas.

2.The gas and dust are pulled together by gravity.

3.Heat and pressure cause nuclear fusion, which signals the birth of a star.

Explanation:

6 0
2 years ago
Read 2 more answers
What is the net ionic equation for 2Sb(OH)3 (s) + 3Na2S (aq) = Sb2S3 + 6NaOH
tatuchka [14]

Answer:

2Sb^(+3) (aq) + 3S^(-2) (aq) = Sb_2•S_3

Explanation:

First of all, let us balance the equation to give;

2Sb(OH)3 (s) + 3Na2S (aq) = Sb2S3 + 3NaOH

Now, we can observe the presence of positive Sodium ions (Na+) and negative hydroxyl ions (OH-) on both left and right sides of the equation.

Now, the two ions will cancel out. These ions are not really involved in the overall reaction and thus do not require being written in the overall equation. Hence, the overall net ionic reaction can now be written as:

2Sb^(+3) (aq) + 3S^(-2) (aq) = Sb_2•S_3

6 0
2 years ago
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