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1. What does the expression “metals eating each other” as used by the electrician refer to? Explain and give an example.

2. What are the possible electrolytes in the rusted panel?

3. Using the table for electrode potential differences (Figure 2) identify the possible composition of the screws responsible for the corrosion observed in the contact area with the copper wire.

During respiration, the inhalation of Oxygen (O) using the energy from Glucose (C6H12O6) is the first chain reaction.

Products formed during respiration are: Carbon Dioxide (CO2) and Water (H2O)

Ionic compounds contain both cations and anions in a ratio that results in no net electrical charge

Answer:

The maximum mass of water produced is  [tex]m__{H_2O}} =116 \ g[/tex]

Explanation:

From the question we are told that

    The mass of  sucrose is [tex]m_s = 200 \ g[/tex]

    The chemical formula for  sucrose is  [tex]C_{12} H_{22} O_{11}[/tex]

The chemical equation for the dissociation of sucrose  is

          [tex]C_{12} H_{22}O_{4} \to 12C + 11H_2O[/tex]

The number of moles of  sucrose can be evaluated as

        [tex]n = \frac{m}{Z}[/tex]

Where Z is the molar mass of sucrose which has a constant value of

      [tex]Z = 342 \ g/mol[/tex]

So  

    [tex]n = \frac{200}{342}[/tex]

   [tex]n =0.585[/tex]

From the chemical equation one mole of sucrose produces 11 moles of water so  0.585 moles of sucrose will produce x moles of  water

Therefore

          [tex]x = \frac{0.585 * 11}{1}[/tex]

         [tex]x = 6.433 \ moles[/tex]

Now the mass of water produced is mathematically represented as

         [tex]m__{H_2O}} = x * Z__{H_2O}[/tex]

 Where  [tex]Z__{H_2O}[/tex] is the molar mass of water with a constant values of  [tex]Z__{H_2O}} = 18 \ g/mol[/tex]

  So  

        [tex]m__{H_2O}} = 6.43* 18[/tex]

       [tex]m__{H_2O}} =116 \ g[/tex]

Answer:

True if by rifting you mean continental rift. If not idk.

Answer:

CₙH₂ₙ

Explanation:

Answer:

[tex]\huge \boxed{\mathrm{C_nH_{2n}}}[/tex]

Explanation:

The general molecular formula for cycloalkanes is [tex]\mathrm{C_nH_{2n}}[/tex].

Answer:

A. 4Ni(s) + 3O2(g) —> 2Ni2O3(s)

B. Synthesis reaction.

C. 18.31g of O2

Explanation:

A. The equation for the reaction.

Ni(s) + O2(g) —> Ni2O3(s)

The above equation can be balance as follow:

There are 2 atoms of O on the left side and 3 atoms on the right side. It can be balance by putting 2 in front Ni2O3 and 3 in front of O2 as shown below:

Ni(s) + 3O2(g) —> 2Ni2O3(s)

There are 4 atoms of Ni on the right side and 1 atom on the left it can be balance by putting 4 in front of Ni as shown below:

4Ni(s) + 3O2(g) —> 2Ni2O3(s)

Now the equation is balanced

B. From the equation, we can see that two reactants combined to form a single product. Therefore, the reaction is termed synthesis reaction.

C. We'll begin by calculating the mass of Ni and O2 that reacted from the balanced equation. This is illustrated below:

4Ni(s) + 3O2(g) —> 2Ni2O3(s)

Molar mass of Ni = 59g/mol

Mass of Ni from the balanced equation = 4 x 59 = 236g

Molar mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96g

Summary:

From the balanced equation above,

236g of Ni reacted with 96g of O2.

Now, we can obtain the mass of O2 needed to react with 45g of Ni as follow:

From the balanced equation above,

236g of Ni reacted with 96g of O2.

Therefore, 45g of Ni will react with = (45 x 96)/236 = 18.31g of O2.

Therefore, 18.31g of O2 is needed to react with 45g of Ni

Answer:

The percentage is   k  [tex]= 20.8[/tex]%

Explanation:

From the question we are told that

    The mass of the stibnite is  [tex]m_s = 5.86 \ g[/tex]

   The volume of   KBrO3(aq) is  [tex]V = 26.6 mL = 26.6 *10^{-3} \ L[/tex]

     The concentration  of   KBrO3(aq) is  [tex]C = 0.125 M[/tex]

Now the balanced ionic  equation for this reaction is

        [tex]BrO_3 ^{-}+ 3Sb^{3+} + 6H^{+} \to Br^{1-} + 3Sb^{5+} + 3H_2O[/tex]

The number of moles of   [tex]BrO_3 ^{-}[/tex] is  

     [tex]n = C *V[/tex]

substituting values

     [tex]n = 26.6*10^{-3} * 0.125[/tex]

     [tex]n = 0.003325 \ mols[/tex]

from the reaction we see that 1 mole of [tex]BrO_3 ^{-}[/tex]  reacts with 3 moles of  [tex]Sb^{3+}[/tex]

so 0.003325 moles will react with x moles of  [tex]Sb^{3+}[/tex]

Therefore

               [tex]x = \frac{0.003325 * 3}{1}[/tex]

              [tex]x = 0.009975 \ mols[/tex]

Now the molar mass of [tex]Sb^{3+}[/tex] is a constant with a values of  [tex]Z = 121.76 \ g/mol[/tex]

Generally the mass of  [tex]Sb^{3+}[/tex] is mathematically represented as

        [tex]m = x * Z[/tex]

substituting values

        [tex]m = 1.215 \ g[/tex]

The percentage of  Sb(antimony) in the overall mass of the stibnite is mathematically evaluated as

            k  [tex]= \frac{1.215}{5.85 } * 100[/tex]

           k  [tex]= 20.8[/tex]%

Answer:

Explanation:

Step 1: Data given

Mass of stibnite (Sb2S3) = 5.85 grams

The Sb3+(aq) is completely oxidized by 26.6 mL of a  0.125 M aqueous solution of KBrO3(aq).

Step 2: The balanced equation

BrO3-(aq)+ 3Sb^3+(aq) + 6 H+ → Br-(aq) + 3Sb^5+(aq) + 3H2O (l)

Step 3: Calculate moles KBrO3

Moles KBrO3 = molarity * volume

Moles KBrO3 = 0.125 M *0.0266 L

Moles KBrO3 = 0.003325 moles

Step 4: Calculate moles Bro3-

in 1 mol KBrO3 we have 1 mol K+ and 1 mol BrO3-

In 0.003325 moles KBrO3 we have 0.003325 moles BrO3-

Step 5: Calculate moles Sb

For 1 mol BrO3- we need 3 mol Sb^3+ to produce 1 mol Br- and 3 mol Sb^5+

For 0.029085 moles BrO3- we need 3*0.003325 = 0.009975 moles Sb

Step 6: Calculate mass Sb

Mass Sb = moles Sb * molar mass Sb

Mass Sb = 0.009975 moles * 121.76 g/mol

Mass Sb = 1.21 grams

Step 7: Calculate the percentage of Sb in the ore

% Sb = (mass Sb / total mass) * 100%

% Sb = (1.21 grams / 5.85 grams) * 100 %

% Sb = 20.76 %

Answer:

11

Explanation:

pH=-log(H+)

Answer:

It is an example of double displacement reaction.

4.8 g of NaCl is needed to react.

Explanation:

Balanced reaction: [tex]H_{2}SO_{4}(aq.)+2NaCl(s)\rightarrow 2HCl(g)+Na_{2}SO_{4}(aq.)[/tex]

Here, oxidation states of H, S, O, Na and Cl do not change during reaction. Hence it is not a redox reaction.

In this reaction, cations and anions of the reactants interchange their partners during reaction. Hence, it is an example of double displacement reaction.

As [tex]H_{2}SO_{4}(aq.)[/tex] remain in excess amount therefore NaCl (s) is the limiting reagent. Hence production of HCl entirely depends on amount of NaCl used.

Molar mass of HCl = 36.46 g/mol

So, 3.0 g of HCl = [tex]\frac{3.0}{36.46}[/tex] mol of HCl = 0.082 mol of HCl

According to balanced equation-

2 moles of HCl are produced from 2 moles of NaCl

So, 0.082 moles of HCl are produced from 0.082 moles of NaCl

Molar mass of NaCl = 58.44 g/mol

So, mass of 0.082 moles of NaCl = [tex](0.082\times 58.44)[/tex] g = 4.8 g

Hence 4.8 g of NaCl is needed to react.

Explanation:

[tex]CH_2O[/tex] is an energy rich molecule, which when reacts with oxygen produces carbon dioxide and energy. In aerobic reaction oxygen is consumed. IN this biochemical reaction oxygen is consumed and CO2 + CH4 is produced. Thus the reaction is aerobic one.

Answer:

H2O / H+

Explanation:

In alkene hydration, water, being a very weak acid, with insufficient proton concentration, does not have the capacity to initiate the electrophilic addition reaction. An acid must be added to the medium for the reaction to take place.

The mechanism proceeds with the formation of a carbocation after adding the proton to the double bond. Alkene hydration is Markovnikov, that is, the proton is added to the less substituted carbon of the alkene.

The  mass of solid BaSO4 formed - 7.64 grams. and the pH of the mixed solution - 13.13

We can calculate the mass of the Ba(OH)2 by calculating the moles each solution contains

n(Ba(OH)2) = 1.00M x 0.05L

= 0.05 moles

0.494M x 0.0864L

= 0.04268 moles

According to balanced reaction 1 mole Ba(OH)2 react with 1 mole H2SO4 molar ratio between Ba(OH)2 to H2SO4 is 1:1 therefore to react with 0.05 mole Ba(OH)2 required H2SO4 = 0.05 mole but H2SO4 therefore H2SO4 is limiting reactant

Ba(OH)2 + H2SO4 ----> BaSO4 + 2 H2O  

0.04268<--- 0.04268  

= 0.0172 mole

0.0172:(0.05 + 0.0664)

= 0.1478 M

Ba(OH)2 is strong base therefore dissociate completely and 1 mole of Ba(OH)2 form to mole of OH- ion therefore concentration of OH- is double than Ba(OH)2

14 - -log[0.1478]

= 14 - 0.83

= 13.13 pH

0.0328 x ( 137 + 32 + 16 x 4 )

= 7.64 grams.

Thus, The  mass of solid BaSO4 formed - 7.64 grams. and the pH of the mixed solution - 13.13

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Answer:

Cu2+ (aq) + e- -------->Cu+ (aq)​E° = 0.15 V anode

Br2 (l) + 2e- -------------> 2Br- (aq) ​E° = 1.08 V cathode

Explanation:

The half equation having a more positive reduction potential indicates the reduction half equation while the half equation having the less positive reduction potential indicates the oxidation half equation.

The overall redox reaction equation is;

2Cu^+(aq) + Br2(g) ----> 2Cu^2+(aq) + 2Br^-(aq)

E°cell= E°cathode -E°anode

E°cathode= 1.08 V

E°anode= 0.15 V

E°cell= 1.08V-0.15V

E°cell= 0.93V

From;

∆G=-nFE°cell

n= 2

F=96500C

E°cell= 0.93V

∆G= -(2×96500×0.93)

∆G= - 179.49 J

From ∆G= -RTlnK

∆G= - 179.49 J

R=8.314Jmol-K-1

T= 25° +273= 298K

K= ???

lnK= ∆G/-RT

lnK=-( - 179.49/(8.314×298))

lnK= 0.0724

K= e^0.0724

Kc= 1.075

Answer:

4.83% of acetic acid in the vinegar

Explanation:

The neutralization reaction of acetic acid (CH₃COOH) with sodium hydroxide (NaOH) is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

Assuming the sodium hydroxide solution has a concentration of 0.500M, moles used in the neutralization reaction are:

0.0185L × (0.500mol / L) = 0.00925 moles of NaOH = 0.00925 moles CH₃COOH

Because 1 mole of acid reacts per mole of NaOH

Using molar mass of acetic acid (60g/mol):

0.00925moles CH₃COOH ₓ (60g / mol) = 0.555g of CH₃COOH

As mass of vinegar sample is 11.5g (d = 1g/mL), percentage of acetic acid by mass is:

0.555g CH₃COOH / 11.5g vinegar × 100 = 4.83% of acetic acid in the vinegar

Answer:

- 12.1

Explanation:

M(C3H8O3) = 3*12.0 + 8*1.0 + 3*16.0 = 92 g/mol

210.0 g C3H8O3 * 1 mol C3H8O3/92 g C3H8O3 = 210/92 mol C3H8O3

350.g = 0.350 kg H2O

Molality = mol soluty/ kg solvent = (210/92 mol C3H8O3) /(0.350 kg H2O) = =6. 522 molal

ΔT =i* Kf* m

(T2 - T1) = i* Kf* m

(T2 - 0°C) = 1*(-1.86°C/molal *6.522 molal)

T2= - 12.1°C

The change in freezing point and the new freeze point are -12.1°C and -12.1°C.

Given the following data:

We know that the temperature at which water freezes is 0°C.

To determine the change in freezing point and the new freeze point:

First of all, we would determine the molar mass of glycerol:

Molar mass of glycerol ([tex]C_3H_8O_3[/tex]) = [tex]12 \times (1 \times 8)\times (16 \times 3)[/tex]

Molar mass of glycerol ([tex]C_3H_8O_3[/tex]) = [tex]12 \times (8)\times (48)[/tex]

Molar mass of glycerol ([tex]C_3H_8O_3[/tex]) = 92 g/mol.

Next, we would find the number of moles of glycerol that is required:

[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{210.0}{92}[/tex]

Number of moles = 2.28 moles

To find the molality concentration:

[tex]Molality = \frac{moles\;of \;solute}{mass\;of \;solvent} \\\\Molality = \frac{2.28}{0.35}[/tex]

Molality = 6.51 molal.

Mathematically, the freezing point elevation of a liquid is given by the formula:

[tex]\Delta T = K_f m[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]\Delta T = -1.86 \times 6.51[/tex]

Change in temperature = -12.1°C.

Now, we can determine the new freeze point:

[tex]\Delta T = T_n - T_f\\\\T_n = \Delta T + T_f\\\\T_n = -12.1 + 0[/tex]

New freeze point = -12.1°C.

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After 3 half-life, 12 of the 14 of the C-14 = 18 of the C-14 would remain.

Answer:

Br2 (l) + 2e- ---------> 2Br- (aq) E° = 1.08 V cathode

Cu2+ (aq) + e- --------->Cu+ (aq) E° = 0.15 V anode

Explanation:

We have to first state the fact that the reaction having the most positive reduction potential occurs at the cathode in any spontaneous electrochemical cell. The half reaction with the less positive electrode potential usually occurs at the anode.

The overall reaction equation is;

2Cu2+ (aq) + Br2 (l) ----->2Cu+ (aq) + 2Br- (aq)

E°cell= E°cathode - E°anode

E°cathode= 1.08 V

E°anode= 0.15V

E°cell = 1.08-0.15 = 0.93 V

But

∆G°= -nFE°cell

n= 2, F=96500C, E°cell= 0.93V

∆G° = -(2× 96500× 0.93)

∆G= -179490 J

But;

∆G = -RTlnK

R=8.314 JK-1

T= 25+273= 298K

Kc= the unknown

∆G° = -179490 J

Substituting values and making lnK the subject of the formula

lnK= ∆G/-RT

lnK= -( -179490/8.314 × 298)

lnK= 72.45

K= e^72.45

K= 2.91×10^31

Answer: 7.505

×

10

22

atoms. (rounded to third place of decimal)

Explanation:

Answer:

Percent yield for chemical reaction = 60%

Explanation:

Given:

Theoretical yield of chemical reaction = 20 gram

Actual yield of chemical reaction = 12 gram

Find:

Percent yield for chemical reaction = ?

Computation:

[tex]Percent\ yield \ for\ chemical\ reaction = [\frac{Actual yield}{Theoretical yield} ]100\\\\Percent\ yield \ for\ chemical\ reaction = [\frac{12}{20} ]100\\\\Percent\ yield \ for\ chemical\ reaction = [0.6 ]100\\\\Percent\ yield \ for\ chemical\ reaction = 60[/tex]

Percent yield for chemical reaction = 60%

The given question is incomplete, the complete question is:

Given the balanced equation representing a redox reaction:

2Al + 3Cu2+ --> 2Al3+ + 3Cu

Which statement is true about this reaction?

(1) Each Al loses 2e- and each Cu2+ gains 3e-

(2) Each Al loses 3e- and each Cu2+ gains 2e-

(3) Each Al3+ gains 2e- and each Cu loses 3e-

(4) Each Al3+ gains 3e- and each Cu loses 2e-

Answer:

The correct answer is option 2, that is, in the given reaction each of Al loses three electrons and each of the Cu^2+ ion acquires two electrons.

Explanation:

To understand the concept, let us suppose that when an atom A loses an electron, it turns into an atom A^1+, and becomes A^2+ when it loses two electrons. On the other hand, when an atom A gains an electron, it turns into A^1- and becomes A^2- when it acquires two electrons.  

Therefore, in the given case, 2Al + 3Cu^2+ --> 2Al^3+ + 3Cu, each of the Al atom turns into Al^3+ when it loses three electrons, and when each of the Cu^2+ ions acquires two electrons, it turns into Cu (Cu^2+ ^2- = Cu).  

Answer:

.

Explanation:

.

Answer:

1. C+ ---- O-

2. O+ ---- Cl-

3. O+ ----- F-

4. C+ ----- N-

5. Cl- ----- C+

6. S- ----- H+

7. S+ ----- Cl -

Explanation:

Electronegativity determines the polarity . There may be two atoms in a bond with high electronegativity, in such cases the positive charge is given to atom with comparatively lower electronegativity. Electronegativity determines the easiness with which an atom attract electrons in a chemical bond. A polar bond is formed when the difference in the electronegativity of two combining atoms is between 0.4 and 1.7. The correct direction is

1. C+ ---- O-

2. O+ ---- Cl-

3. O+ ----- F-

4. C+ ----- N-

5. Cl- ----- C+

6. S- ----- H+

7. S+ ----- Cl -

Answer:

The description including its given problem is outlined in the following section on the clarification.

Explanation:

The given values are:

RBCC = 0.12584 nm

RFCC = 0.12894 nm

The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:

√3 ABCC = 4RBCC

⇒  ABCC = [tex]\frac{4RBCC}{\sqrt{3} }[/tex]

⇒             = [tex]\frac{4\times 0.12584}{\sqrt{3}}[/tex]

⇒             = [tex]0.29062 \ nm[/tex]

Likewise AFCC as well as RFCC are interconnected by  

√2AFCC = 4RFCC

⇒  AFCC = [tex]\frac{4RFCC}{\sqrt{2}}[/tex]

⇒             = [tex]\frac{4\times 0.12894}{\sqrt{2} }[/tex]

⇒             = [tex]0.36470 \ nm[/tex]

Now,

The Change in Percent Volume,

= [tex]\frac{V \ final-V \ initial}{V \ initial}\times 100 \ percent[/tex]

= [tex]\frac{(VFCC)unit \ cell-(VBCC)unit \ cell}{(VBCC)unit \ cell}\times 100 \ percent[/tex]

= [tex]\frac{(aFCC)^3-(aBCC)^3}{(aBCC)^3}\times 100 \ percent[/tex]

= [tex]\frac{(0.36470)^3-(0.29062)^3}{(0.29062)^3}\times 100 \ percent[/tex]

= [tex]97.62 \ percent (approximately)[/tex]

Note: percent = %

Answer:

Animal cells don't have a dividing cell wall like plant cells do

Explanation:

Plants cells use photosynthesis from the sun, which requires them to have chloroplast filled with chlorophyll to complete this function; animal cells do not have chloroplasts

Answer:

0.209M

Explanation:

M1V1=M2V2

(28.5 mL)(0.183M)=(25.0mL)(M)

M2= 0.209M

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Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

0.25 mol

Explanation:

Moles= No.of molecules / Avogadros number

Answer:

0.25

Explanation:

Answer:

1.23 g/mL

-18.2%

Explanation:

We need to find the average, which is just the sum of the numbers divided by the number of numbers. Here, the sum will be 1.24 + 1.21 + 1.23 = 3.68 g/mL. There are 3 numbers, so divide 3.68 by 3: 3.68 / 3 ≈ 1.2266...

However, we need to round this and take into account significant figures. Each trial gave a number with 3 significant figures, so we round our number off to three: 1.22666... ≈ 1.23. So, circle the first number under the X column.

We now need to find the percent error, which is RE (%). To calculate this, we take the measured value (1.23 in this case) and subtract the exact value (1.50 here) from it, and then divide that by the exact value:

(1.23 - 1.50) / 1.50 ≈ -0.1822...

Again, we need to round to 3 significant figures, which would make it:

-0.1822... ≈ -0.182

Thus, circle the last number under the RE (%) column.

Answer:

- 622.5kJ

Explanation:

1)      2NH3 + 3N2O ----> 4N2 + 3H2O    ΔH° 1= - 1010kJ

-  3 (N2O  + 3H2     --->  N2H4 + H2O    ΔH° 2 = - 317 kJ)

2)      2NH3 + 3N2O ----> 4N2 + 3H2O        ΔH° 1= - 1010kJ

     - 3N2O   - 9 H2     ----> - 3N2H4 - 3H2O      - 3ΔH° 2 = 3*317 kJ

2NH3  + 3N2H4 --->4N2+9H2,     ΔH° 1 -  3*ΔH° 2

2NH3  + 3N2H4  --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

3) -(2NH3 +1/2O2 ---> N2H4 + H2O,     ΔH°3 = -143 kJ)

   -2NH3 - 1/2O2 ---> - N2H4 - H2O,    - ΔH°3 = 143 kJ

 2NH3  + 3N2H4 --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

 -2NH3 - 1/2O2              ---> - N2H4 - H2O,             - ΔH°3 = 143 kJ

 4N2H4 + H2O  ---> 4N2  + 9H2 + 1/2O2,    ΔH° 1 -  3*ΔH° 2 - ΔH°3

4)

H2 + 1/2O2 ---> H2O,    ΔH° 4 = - 286 kJ

9*(H2 + 1/2O2 ---> H2O,    ΔH° 4 = - 286 kJ)

9H2+ 9/2 O2 ---> 9H2O,  9*ΔH° 4 = 9*(- 286) kJ

4N2H4 + H2O  ---> 4N2  + 9H2 + 1/2O2,    ΔH° 1 -  3*ΔH° 2 - ΔH°3

9H2+ 9/2 O2    ---> 9H2O,                             9*ΔH° 4 = 9*(- 286) kJ

4N2H4 +4O2 --->4N2+8H2O,         ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

5)

1/4*(4N2H4 +4O2 --->4N2+8H2O,    ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

N2H4 +O2 --->N2+2H2O,        1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)

6)

N2H4 +O2 --->N2+2,        

1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)=

=1/4(-1010kJ - 3*(-317kJ) - (-143kJ) + 9*(-286kJ))= - 622.5 kJ

Explanation:

Molarity = mol/liter

Solution 240 mL = NaCl 9.0 g

Solution 1000 mL = NaCl 9.0/240 × 1000 = ..... g

mol NaCl = g/MW

MW NaCl = ?....

...........

Molarity = ......... Molar