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1 year ago

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Chemistry

1 answer:

Cerrena [4.2K]1 year ago

3 0

Answer:

1. Galvanic oxidation. Example is the corrosion of aluminium wires when in contact with copper wires under wet conditions.

2. Rainwater or Damp/moist air

3. Chromium-plated steel screws or stainless steel screws or galvanized steel screws

Explanation:

1. Galvanic oxidation or corrosion occurs when two different metals with different electrode potentials are brought into contact with each other by means of an electrolyte (usually a aqueous solution), such that a redox reaction occurs leading to one metal with the more negative electrode potential (the anode) becoming oxidized, while the other less negative potential (the cathode) is reduced.

In order for galvanic corrosion to occur, three elements are required.

i. Two metals with different corrosion potentials (anode and cathode)

ii. Direct metal-to-metal electrical contact

iii. A conductive electrolyte solution (e.g. water) must connect the two metals on a regular basis.

For example oxidation (corrosion) of aluminium wires when in contact with copper wire under wet conditions.

2. The most likely electrolyte will be rainwater containing dissoved solutes (if the panel is in an exposed part of the house) or damp/moist air.

3. From the table, the most likely screw will be chromium-plated steel screws or stainless steel (made of iron and nickel) screws or galvanized steel (zinc-plated) screws.

All these possible screw components have a more negative electrode potential than copper. Thus they will serve as the anode in a galvanic oxidation with copper.

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Use the information in the square to answer the questions about copper. A purple box has C u at the center and 29 above. Below i

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Answer:

29 protons 29 electrons

34 neutrons or (65-29)=36 neutrons in its nucleus.

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2 years ago

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2. Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:

Nesterboy [21]

Answer

increase in temperature

decrease in pressure

continuous removal of PH3
adding more of P into the system

Explanation:

In the reaction P4(g)+6H2(g) ⇌ 4PH3(g);

The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:

heat+A⇌BΔH=+

Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.

According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.

4 0

1 year ago

Oxyacetylene torches produce such high temperature that they are often used to weld and cut metal. When 1.53 g of acetylene (C2H

hodyreva [135]

Answer: $4403kJ/mole$

Explanation:

First we have to calculate the heat absorbed by bomb calorimeter

Formula used :

$q_b=c_b\times (T_{final}-T_{initial})$

$q_b$ = heat absorbed by calorimeter = ?

$c_b$ = specific heat of = 10.69 kJ/K

$T_{final}$ = final temperature = $44.688^oC=(273+44.688)K=317.688K$

$T_{initial}$ = initial temperature = $20.486^oC=(273+20.486)K=293.486K$

$q_b=10.69kJ/K\times (317.688-293.486)=258.7kJ$

As heat absorbed by calorimeter is equal to the heat released by acetylene during combustion.

Thus 1.53 gram of acetylene releases heat of combustion = 258.7kJ

So, 26.04 g/mole of acetylene releases heat of combustion $\frac{258.7}{1.53}\times 26.04=4403kJ/mole$

Therefore, the heat of combustion of acetylene is, $4403kJ/mole$

7 0

2 years ago

495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F

I am Lyosha [343]

Answer:

Total pressure of the flask is 2.8999 atm.

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

= 0.495 L (1 cm³ = 1 mL = 0.001 L)

Volume of nitrogen (N2) gas = 877 cm3

= 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

= 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature = 25 °C

T = (25°C + 273.15) K

= 298.15 K

Pressure = 114.7 kPa

= 114.700 Pa

Pressure (torr) = 114,700 / 101325

= 1.132 atm

Formula:

PV=nRT (ideal gas equation)

P = pressure

V = volume

R (gas constnt)= 0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

For Oxygen:

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

= 0.560 / 24.47

= 0.0229 moles

For Nitrogen:

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

= 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

= 0.0229 moles + 0.0406 moles

n = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L

P = 1.554 / 0.536

P = 2.8999 atm

Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.

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1 year ago

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A tiny sample of an aqueous solution of two substances R and P is sketched below as if it was under an imaginary microscope so p

PilotLPTM [1.2K]

Answer:

R = 6, P = 6

Explanation:

Answer

Number of R molecules = 6

Number of P molecules = 6

Explanation

R(aq) <-------> P(aq)

K = [P]/[R]

at equillibrium

[ P ] = 10+x

[R] = 2-x

10 + x /(2-x) = 1

10 + x = 2- x

2x = -8

x = -4

Therefore,

[ R] = 2 - ( -4) = 6

[ P ] = 10 + ( -4) = 6

7 0

1 year ago