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sattari [20]
3 years ago
12

An iron object alloyed with cobalt rusts more quickly than a pure iron object. However, an iron object alloyed with manganese ru

sts less quickly than a pure iron object under the same conditions. This is true because: __________
a. cobalt is a stronger reducing agent than iron
b. iron is a stronger reducing agent than manganese
c. cobalt exhibits more metallic character than either iron or manganese in the iron-manganese alloy
Chemistry
1 answer:
HACTEHA [7]3 years ago
5 0

Rusting is an electrochemical reaction. Iron rusts faster when alloyed with cobalt than when alloyed with manganese because, in the iron-manganese alloy, manganese is rendered the anode and iron is rendered the cathode

An alloy is a combination of two metals. There are various reasons for producing alloys such as greater tensile strength, corrosion resistance and improved aesthetic appearance.

When iron is alloyed with cobalt, the iron rusts faster than pure iron because iron is rendered the anode and cobalt is rendered the cathode. When the iron is alloyed with manganese, it rusts more slowly than pure iron because in the iron-manganese alloy, manganese is rendered the anode and iron is rendered the cathode.

Missing parts;

An iron object alloyed with cobalt rusts more quickly than a pure iron object. However, an iron object alloyed with manganese rusts less quickly than a pure iron object under the same conditions. This is true because

(1) cobalt is a stronger reducing agent than iron

(2) iron is a stronger reducing agent than manganese

(3) cobalt exhibits more metallic character than either iron or manganese

(4) in the iron-manganese alloy, manganese is rendered the anode and iron is rendered the cathode

(5) in the iron-cobalt alloy, cobalt is rendered the anode and iron is rendered the cathode

Learn more: brainly.com/question/14281129

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viktelen [127]
Lewis Structure is drawn in following steps,

1) Calculate Number of Valence Electrons:
    
# of Valence electrons in Mg  =  2
# of Valence electrons in I      =  7
# of Valence electrons in I      =  7
                                               ---------
Total Valence electrons          =  16

2) Draw Mg as a central atom surround it by two atoms of Iodine.

3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.

Total Valence electrons           16
- Four electrons                    -   4
                                              ----------
                                                  12

4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.

The final lewis structure formed is as follow, 

4 0
3 years ago
Which of these elements has the greatest average atomic mass?4
Arte-miy333 [17]
Hi the answer is actually B
7 0
3 years ago
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The initial pressure of a gas is 1.58 atm and occupies 1.76 L of space at constant temperature. This gas is compressed so that i
jok3333 [9.3K]

Answer:

Final volume = 0.39 L

Explanation:

V1P1 = V2P2

V2= V1P1/P2

V2= (1.76×1.58)/7.08

V2 = 0.39 L

7 0
3 years ago
Ethanol melts at -114 degree C. The enthalpy of fusion
Brut [27]

Answer: The heat required is 6.88 kJ.

Explanation:

The conversions involved in this process are :

(1):ethanol(s)(-135^0C)\rightarrow ethanol(s)(-114^0C)\\\\(2):ethanol(s)(-114^0C)\rightarrow ethanol(l)(-114^0C)\\\\(3):ethanol(l)(-114^0C)\rightarrow ethanol(l)(-50^0C)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change = ?

m = mass of ethanol = 25.0 g

c_{p,s} = specific heat of solid ethanol= 0.97 J/gK

c_{p,l} = specific heat of liquid ethanol = 2.31 J/gK

n = number of moles of ethanol = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}}=\frac{25.0g}{46g/mole}=0.543mole

\Delta H_{fusion} = enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole

T_{final}-T_{initial}=\Delta T = change in temperature

The value of change in temperature always same in Kelvin and degree Celsius.

Now put all the given values in the above expression, we get

\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]

\Delta H=6885.93J=6.88kJ     (1 KJ = 1000 J)

Therefore, the heat required is 6.88 kJ

3 0
3 years ago
Describe, in detail, to a freshman undergraduate how to make 1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /m
mario62 [17]

Answer:

Explanation:

The objective here is to prepare  1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media.

Given that :

the Stock concentration of Amp: 100 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml since 100 mg/ml = 100000  μg/ml

However, using formula C₁V₁=C₂V₂ (Ampicilin),

where:

C₁ = 100000 μg/ml,

V₁=?,

C₂= 50  μg/ml,

V₂=1000 ml

100000 μg/ml × V₁ = 50  μg/ml × 1000 ml

V₁ =  50  μg/ml × 1000 ml/100000 μg/ml

V₁ =   0.5 ml

Given that:

the Stock concentration of Kan: 25 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml , 25 mg/ml = 25000 μg/ml

Now by using formula C₁V₁=C₂V₂ (Kanamycin),

C₁ = 25000 μg/ml,

V₁=?,

C₂= 100  μg/ml,

V₂=1000 ml

25000 μg/ml × V₁ = 100  μg/ml × 1000 ml

V₁ =  100  μg/ml × 1000 ml/25000 μg/ml

V₁ =   4 ml

Thus; in 1 lite of Lb+ Kan+Amp preparation;

0.5 ml of Amp & 4 ml of kanamycin is used for their stock preparation.

Finally;

Sterilization step should be carried out in flask (Clean dry glass wares) for media in an autoclave, the container size should be twice the volume of media which is prepared.

5 0
3 years ago
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