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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.

Learn more about electrode here:

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For ethanol, propanol, and n-butanol the boiling points, surface tensions, and viscosities all increase. what is the reason for

Contact [7]

Moving from Ethanol through Propanol to Butanol the physical properties like boiling points, surface tension and viscosity increases because of the increases in intermolecular interactions between the molecules of given compounds.

Explanation:
Ethanol, propanol and butanol all have hydroxyl groups in common, means all have hydrogen bond intractions between their molecules. So, taking the hydrogen bonding interaction constant we are left with only the difference in the number of carbon atoms.
Butanol has the greatest physical properties than other two because it has four carbon atom chain. So, as we know the London Dispersion forces or Van der Waal forces increases with increase in molecular size and chain length of hydrocarbon.
Therefore, the strength of London forces is greater in butanol than other two while ethanol has the smallest chain comparatively hence, lowest physical properties.

3 0

2 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

2 years ago

The first strings of holiday lights were manufactured as series circuits. Which explains why this method is no longer used? A. I

user100 [1]

B because once the circuit is burned the lights will all go of

3 0

2 years ago

Match each of the following topical dosage forms with its correct definition.

Dmitry [639]

Answer:

cream - contains a higher proportion of oil than water

ointment - dr4g mixed in approximately equal proportions of oil and water

i don't know about the other two sorry

8 0

1 year ago

Given the equation representing a reversible reaction:which formula represents the h+ acceptor in the forward reaction?

AnnyKZ [126]

Question is incomplete. Complete question is attached below
.............................................................................................................................

Answer: Option A: HCO3-(aq.)

Reason:
From the reaction, it can be seen that following reaction occurs in forward direct

HCO3-(aq) + H2O(l) → H2CO3(aq) + OH-(aq)

In above forward reaction, HCO3- accepts proton from H2O to generate H2CO3. Thus, according to Lowry and Bronsted theory of acid-base, HCO3- is a base, while H2CO3 is a conjugate acid.

8 0

2 years ago