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1 year ago

Which metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe? auau cucu mnmn agag

1 answer:

3 0

Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.

Commonly, sacrificial electrodes are employed to stop another metal from corroding or oxidising. A metal that is more reactive than the metal being shielded must serve as the sacrificial electrode. Magnesium, aluminium, and zinc are the three metals most frequently used in sacrificial anodes.

Manganese-Magnesium (Mn-Mg) electrode is more suited for on-shore pipelines where the electrolyte (soil or water) resistivity is higher since it has the highest negative electropotential of the three. In order to replenish any electrons that could have been lost during the oxidation of the shielded metal, the highly active metal offers its electrons.

Therefore, Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.

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The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.

Brilliant_brown [7]

Answer : The rate of reaction is,

$Rate=4.77\times 10^{-19}M/s$

The appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)$

The rate law expression will be:

$Rate=k[NO_2][O_3]$

Given:

Rate constant = $k=1.69\times 10^{-4}M^{-1}s^{-1}$

$[NO_2]$ = $1.77\times 10^{-8}M$

$[O_3]$ = $1.59\times 10^{-7}M$

$Rate=k[NO_2][O_3]$

$Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})$

$Rate=4.77\times 10^{-19}M/s$

The expression for rate of appearance of $NO_3$ :

$\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}$

As, $\text{Rate of reaction}=4.77\times 10^{-19}M/s$

So, $\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s$

Thus, the appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

7 0

3 years ago

If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a

Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0

3 years ago

A hot air balloon is filled to 1250 m3 at 27 C. At what temperature will the balloon be filled to 1600 m3 if the pressure remain

SSSSS [86.1K]

Answer:

384.2 K

Explanation:

First we convert 27 °C to K:

27 °C + 273.16 = 300.16 K

With the absolute temperature we can use Charles' law to solve this problem. This law states that at constant pressure:

T₁V₂=T₂V₁

Where in this case:

T₁ = 300.16 K

V₂ = 1600 m³

T₂ = ?

V₁ = 1250 m³

We input the data:

300.16 K * 1600 m³ = T₂ * 1250 m³

And solve for T₂:

T₂ = 384.2 K

7 0

2 years ago

A unicellular organism has how many cells? O two O many O one O three

Flauer [41]

The answer is many.

5 0

3 years ago

Read 2 more answers

The graphic shows two isomers of a chemical compound with molecular formula C5H11NO2. Which type of isomers are these? A. struct

-BARSIC- [3]

The two compounds shown indeed have tha same molecular formula, C5 H11 NO2. One of the molecules has a group NH2 and a group COOH, the other molecule has a NOO group, that makes that the two isomers have a completely different structure, with the atoms arranged in a completely different order. This kind of isomers fits in the definition of structural isomers, so the answer is structural isomers.

6 0

3 years ago

Read 2 more answers