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3 years ago

5

Is corrosion garvanic or electrolytic?Why?

2 answers:

sukhopar [10]3 years ago

5 0

Answer:

<h2>While galvanic corrosion is driven by the difference in corrosion potential between two metals , electrolytic corrosion is driven by the external sources of EMF </h2>

prisoha [69]3 years ago

3 0

<h2><u><em>While galvanic corrosion is driven by the difference in corrosion potential between two metals, electrolytic corrosion is driven by the external sources of EMF.</em></u></h2>

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3. Theoretically how many grams of magnesium is required to produce to 5.0 g of

FinnZ [79.3K]

Answer:

3grams

Explanation:

The reaction for the production of Magnesium dioxide will be

Mg + O2 → MgO

we have 5g of MgO (molar mass 40g)

no of moles of MgO = 5/40 = 0.125

Using unitary method we have

1 mole of Mg require 1 mole of MgO

0.125 Mole of MgO = 0.125mole of Mg

n = given mass /molar mass

0.125 = mass / molar mass

mass = 0.125* 24 = 3grams

8 0

3 years ago

The smallest unit into which a compound can be divided and still be that same compound is a(n) ___.

mina [271]

The molecule of that compound! Hope this helps!

8 0

3 years ago

The compound trimethylamine, (CH3)3N, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibr

GREYUIT [131]

Answer:

The $K_b$ expression for the weak base equilibrium is:

$K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

Explanation:

$(CH_3)_3N(aq)+H_2O(l)\rightlefharpoons (CH_3)_3NH^++OH^-(aq)$

The expression of the equilibrium constant of base $K_c$ can be given as:

$K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}$

] $K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

As we know, water is pure solvent, we can put $[H_2O]=1$

$K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

So, the the $K_b$ expression for the weak base equilibrium is:

$K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

6 0

3 years ago

Oil (specific gravity of 0.80 and a viscosity of 0.000042 lbf/ft2) at a temperature of 80 F flows through two separate pipes 10

Elenna [48]

Answer:

The velocity of flow in 10in pipe is 4.16 ft/s.

Explanation:

Given that

Specific gravity = 0.8

Viscosity =0.00042 $lbf/ft^2$

For pipe 1

$V_1=6 ft/s,d_1=12\ in$

For pipe 1

$V_2,d_1=10\ in$

If we assume that flow in the both pipe is laminar

For laminar flow through circular pipe

$\dfrac{\Delta P}{L}=\dfrac{32V\mu }{d^2}$

So same pressure drop we can say that

$\dfrac{V_1 }{d^2_1}=\dfrac{V_2}{d^2_2}$

$\dfrac{6}{12^2}=\dfrac{V_2}{10^2}$

$V_2=4.16 ft/s$

So the velocity of flow in 10in pipe is 4.16 ft/s.

6 0

4 years ago

I need your help please...

iVinArrow [24]

Answer:

its c and A

Explanation:

6 0

3 years ago

Read 2 more answers