• Briefly discuss the cause of errors in the measurements

Answer: 8 (feet)

Explanation:

24 inches = 2 feet

48 inches = 4 feet

12 inches = 1 foot

To find volume you do Base * Width * Height

2*4*1 = 8

Hope this helps!

The correct answer is  8 (feet).

How to calculate ?

Therefore, 2*4*1 = 8

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#SPJ2

Answer:

18130 mm

Explanation:

Now we have a lot of unit conversions to do in order to correctly answer this question. We shall do these conversions gradually.

First we convert the weight in ounce to grams.

If 1 ounce = 28.4g

12 ounces = 12×28.4 = 340.8 g

Next we convert the area of aluminum from ft2 to m2

1ft2= 0.0929 m2

75 ft2= 75 × 0.0929= 6.9675m2

Now density of aluminum= 2.70 gcm-3

Density= mass/volume

But volume= area× thickness

Density= mass/ area × thickness

Density × area × thickness= mass

Thickness= mass/ density × area

Thickness= 340.8g / 2.70gcm-3 × 6.9675m2

Thickness= 340.8/18.8

Thickness= 18.13 m

Since 1000 milimeters make 1 metre

Thickness= 18130 mm

Answer:

[tex]n_N=0.041molN[/tex]

Explanation:

Hello,

In this case, for this mole-mass relationship, we are able to compute the moles of nitrogen atoms by firstly obtaining the moles of the given compound, considering its molar mass that is 194 g/mol:

[tex]n_{C_8H_{10}O_2N_4}=2.0gC_8H_{10}O_2N_4*\frac{1molC_8H_{10}O_2N_4}{194gC_8H_{10}O_2N_4} =0.01molC_8H_{10}O_2N_4[/tex]

Then, by knowing that one mole of the given compound has four moles of nitrogen atoms, we apply the following relationship:

[tex]n_N=0.01molC_8H_{10}O_2N_4*\frac{4molN}{1molC_8H_{10}O_2N_4} \\\\n_N=0.041molN[/tex]

Best regards.

Answer:

Oxygen is the limiting reactant.

Explanation:

Based on the reaction:

C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O

1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of CO₂ and 11 moles of H₂O.

10.0g of sucrose (Molar mass: 342.3g /mol) are:

10.0g C₁₂H₂₂O₁₁ × (1mole / 342.3g) = 0.0292 moles of C₁₂H₂₂O₁₁

And moles of 10.0g of oxygen (Molar mass: 32g/mol) are:

10.0g O₂ × (1mole / 32g) = 0.3125 moles of O₂

For a complete reaction of 0.0292 moles of C₁₂H₂₂O₁₁ you need (knowing 12 moles of oxygen react per mole of sucrose):

0.0292 moles of C₁₂H₂₂O₁₁ × (12 moles O₂ / 1 mole C₁₂H₂₂O₁₁) = 0.3504 moles of O₂

As you have just 0.3125 moles of O₂, oxygen is the limiting reactant.

Answer:

A. ethane > butane > decane > dodecane > heptadecane

Explanation:

In fractionating column, crude oil is separated by means of fractional distillation due to the wide range of boiling point of the crude products such as ethane, propane, butane pentane etc.

The product with the least weight rises to top height while the product with highest weight will move down.

For the given hydrocarbon products, the ranking according to their molecular weight, starting with the lighter product to heavier product is

ethane (C2), butane (C4), decane(C10), dodecane (C12), heptadecane(C17).

Thus, the correct ranking, starting with the product that will rise highest is ethane > butane > decane > dodecane > heptadecane

Answer:

1177.88g of C3H8

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Next we shall determine the number of mole of C3H8 required to produce 80.3 moles of CO2. This is illustrated below:

From the balanced equation above,

1 mole of C3H8 reacted to produce 3 moles of CO2.

Therefore, Xmol of C3H8 will react to produce 80.3 moles of CO2 i.e

Xmol of C3H8 = 80.3/3

Xmol of C3H8 = 26.77 moles

Finally, we shall convert 26.77 moles of C3H8 to grams.

Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol

Mole of C3H8 = 26.77 moles

Mass of C3H8 =..?

Mass = mole x molar mass

Mass of C3H8 = 26.77 x 44

Mass of C3H8 = 1177.88g

Therefore, 1177.88g of C3H8 are needed for the reaction

Answer:

The given blanks can be filled with below, below, larger, larger, larger, larger and smaller.

Explanation:

In the periodic table, rubidium comes below the potassium, and iodine comes below bromine. Therefore, it can be said that the ion of rubidium is larger in comparison to potassium ion, and similarly the ion of iodine is larger in comparison to the ion of bromine.  

When the atomic radius is larger it signifies that the distance in between the ion of iodine and the ion of rubidium is larger in comparison to that between the ion of potassium and the ion of bromine. Thus, smaller energy is associated with the interaction between iodine and rubidium, and potassium bromide's lattice energy is more exothermic.  

Answer:

8.09x10⁻⁵M of Fe³⁺

Explanation:

Using Lambert-Beer law, the absorbance of a sample is proportional to its concentration.

In the problem, the Fe³⁺ is reacting with KSCN to produce Fe(SCN)₃ -The red complex-

The concentration of Fe³⁺ in the reference sample is:

4.80x10⁻⁴M Fe³⁺ × (5.0mL / 50.0mL) = 4.80x10⁻⁵M Fe³⁺

Because reference sample was diluted from 5.0mL to 50.0mL.

That means a solution of  4.80x10⁻⁵M Fe³⁺ gives an absorbance of 0.512

Now, as the sample of the lake gives an absorbance of 0.345, its concentration is:

0.345 × (4.80x10⁻⁵M Fe³⁺ / 0.512) = 3.23x10⁻⁵M.  

As the solution was diluted from 20.0mL to 50.0mL, the concentration of Fe³⁺ in Jordan lake is:

3.23x10⁻⁵M Fe³⁺ × (50.0mL / 20.0mL) = 8.09x10⁻⁵M of Fe³⁺

The concentration of  Fe³⁺ in Jordan Lake is = 8.09* 10⁻⁵ M  

According to Lambert-Beer law ;The rate of absorbance of a sample is directly proportional to concentration of the sample

The reaction that produces a red complex

Fe³⁺ + KScN ----> Fe ( SCN )₃  ( red complex )

First step:  Determine the Concentration of  Fe³⁺ in  reference sample

= 4.80x10⁻⁴ *  ( 5.0 / 50.0 ) = 4.80 * 10⁻⁵M  

reference sample was diluted from 5.0 mL to 50.0 mL

∴ Concentration of 4.80 * 10⁻⁵M  has an absorbance = 0.512

Given that Lake sample absorbance = 0.345

Next step : Determine the concentration of the lake sample

Concentration of lake sample :

= absorbance of lake sample * ( conc of reference sample / absorbance )

= 0.345 *  (  4.80* 10⁻⁵ / 0.512  )  = 3.23* 10⁻⁵M.

Final step : Determine the concentration of Fe³⁺ in Jordan  lake

=  3.23 * 10⁻⁵ *  ( 50.0mL / 20.0mL) = 8.09* 10⁻⁵ M  

Note :  Solution  was diluted from 20.0 mL to 50.0 mL

Hence we can conclude that The concentration of  Fe³⁺ in Jordan Lake is = 8.09* 10⁻⁵ M  .

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Answer:

1. H4C-NH-CH3

2. H3C-NH-C-CH3

H2C-CH3

1 +

H3C-CH2-N-CH3

CH3

N

3. H

4.

.

.

.

.

.

.

Answer:

C. Na₂SO₄(aq) + Sr(NO₃)₂(aq) → SrSO₄(s) + 2 NaNO₃(aq)

Explanation:

Based on the equation:

Na₂SO₄(aq) + Sr(NO₃)₂(aq) → SrSO₄(s) + NaNO₃(aq)

As you can see, sulfate ions (SO₄) are been replaced for nitrate ions (NO₃). That is a double replacement reaction and is a very important information because 2 NO₃ ions in Sr(NO₃)₂ are producing 1 NO₃ ion. To balance NO₃:

Na₂SO₄(aq) + Sr(NO₃)₂(aq) → SrSO₄(s) + 2 NaNO₃(aq)

1 SO₄ ion in Na₂SO₄ produce 1 SO₄ ion in SrSO₄. And Na and Sr metals are balanced yet. Thus, the balanced form of this chemical equation is:

Answer:

The weakest conjugate is HClO-.

Explanation:

As a general rule, the stronger the Bronsted-Lowry acid, the weaker its conjugate base, and vice versa.  

Acid 1: HClO is a strong acid, hence its conjugate base would be weak

Acid 2: H3PO4 is a weak acid, hence its conjugate base would be strong

Acid 3: hydrogen sulphide is also a moderately weak acid with a moderately strong conjugate base.

In order of increasing strengths:

HClO < H2S < H3PO4

answer

1)maximise your software efficiency

2)Prevention against viruses and malware

3)Early detection of problematic issues

4)prevent data loss

5)Speed up your computer

Answer:

[tex]M=0.213M[/tex]

Explanation:

Hello,

In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:

[tex]n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-[/tex]

[tex]n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-[/tex]

[tex]n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-[/tex]

We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:

[tex]n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-[/tex]

Finally, we compute the molarity:

[tex]M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M[/tex]

Regards.

Answer:

Explanation:

From the given information ;the objective is to determine how the instructor made the 0.25L elution buffer

0.25 L elution buffer = 250 mL elution butter

The breaking buffer that we use this week contains

10mM Tris    =   0.01 M

150mM NaCl  =   0.15 M

300mM imidazole.  = 0.3 M

The stock concentration  of Tris in 1M

Therefore ; by using the formula: [tex]M_1V_1 = M_2 V_2[/tex]; we can determine the volume in the preparation; so;

[tex]1*V_1 = 0.0 1 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0.0 1 \ M * 250 \ mL}{1 }[/tex]

[tex]V_1 = 2.5 \ mL[/tex]

In NaCl, The amount of stock concentration is 5 M

so; using the same formula; we have:

[tex]5*V_1 = 0.15 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0. 15 \ M * 250 \ mL}{5 }[/tex]

[tex]V_1 = 7.5 \ mL[/tex]

From Imidazole ; the amount of stock concentration is

[tex]1*V_1 = 0.3 \ M * 250 \ mL[/tex]

[tex]V_1 = \dfrac{0. 3 \ M * 250 \ mL}{1 }[/tex]

[tex]V_1 = 75 \ mL[/tex]

Thus; we can have a table as shown as :

Stock concentration        volume to be added        Final concentration

1 M of Tris                              2.5 mL                            10 mM

5 M of  NaCl                          7.5 mL                             150 mM

1 M of Imidazole                    75  mL                            300  mM

In conclusion. the addition of all the volume make up the 250 mL elution buffer that is equivalent to 0.25 L.

Answer:

[Ar]3d^84s^2

Explanation:

From the question given, we are asked to write the condensed form of electronic configuration of nickel, Ni.

To do this, we simply write the symbol of the noble gas element before Ni in a squared bracket followed by the remaining electrons to make up the atomic number of Ni.

This is illustrated below:

The atomic number of Ni is 28.

The noble gas before Ni is Argon, Ar.

Therefore, the condensed electronic configuration of Ni is written as:

Ni(28) => [Ar]3d^84s^2

Answer:

[Ar] 4s^23d^8

Explanation:

Answer:

B) It can have different compositions, but it has a set of characteristics that does not change.

Explanation:

On e d g e n u i t y

I believe the answer is d lmk if  im  wrong or right

Answer:

Ba(OH)2(aq)+H2SO4(aq) gives us 2BaH+H2O

Explanation:

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The correct option is reagent B

b

The  skeletal structure of major organic product A is shown on the third uploaded image

Explanation:

The mechanism of the reaction for A and  B  are shown on the second the second reaction and looking at this we can see that the reagent that  predominately gives 1,2 addition is reagent B  

Answer:

1= Volume

= Length x breath x height

= 13.90 x 2.9 x 0.081

=3.26511

2= Density = Mass ÷ volume

= 11.3 ÷ 3.26511

= 3.461 (3d.p)

idk the rest because you haven't shown a picture of the rest

Answer:

1. 3.3 cm³; 2. 3.5 g/cm³; 3. barium; 4. 4%

Explanation:

Experimental data:

Mass          = 11.3    g

Length      = 13.90 cm

Width        =  2.9    cm

Thickness = 0.081 cm

Calculations:

1. Volume of bar

V = lwh = 13.90 cm × 2.9 cm × 0.081 cm = 3.3 cm³

2. Experimental density

[tex]\text{Density} = \dfrac{m}{V} = \dfrac{\text{11.3 g}}{\text{3.27 cm}^{3}} = \textbf{3.5 cm}^{\mathbf{3}}[/tex]

3. Identity of metal

The three most likely metals are scandium (3.00 g/cm³), barium (3.59 g/cm³), and yttrium (4.47 g/cm³)

The metal is probably barium.

4. Percent difference

[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{ True - Calculated}\lvert}{ \text{True}} \times 100 \,\%\\\\& = & \dfrac{\lvert 3.59 - 3.5\lvert}{3.59} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.1\lvert}{3.59} \times 100 \, \%\\ \\& = & 0.04 \times 100 \, \%\\& = & \mathbf{4 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{4 \, \%} }$}[/tex]

Answer: The Gibbs free energy of the reaction is -114629.4 J

Explanation:

To calculate the Gibbs free energy of the reaction, we use the equation:

[tex]\Delta G^o=-RT\ln K_{eq}[/tex]

where,

[tex]\Delta G^o[/tex] = Gibbs free energy of the reaction = ?

R = Gas constant = [tex]8.314 J/K.mol[/tex]

T = temperature of the reaction = 298 K

[tex]K_{eq}[/tex] = equilibrium constant of the reaction = [tex]1.24\times 10^{20}[/tex]

Putting values in above equation, we get:

[tex]\Delta G^o=-(8.314J/mol.K\times 298K\times \ln (1.24\times 10^{20}))\\\\\Delta G^o=-114629.4J[/tex]

Hence, the Gibbs free energy of the reaction is -114629.4 J

Answer:

2 Hydrogen One oxygen

Explanation:

Complete Question

The complete question is shown on the first uploaded image

Answer:

The mass is [tex]m = 0.0349 \ g[/tex]

Explanation:

From the question we are told that

   The Henry's Law constant is  [tex]k = 2.4 *10^{10} M/atm[/tex]

   The volume of water is [tex]V = 525 \ ml = 0.525 \ L[/tex]

   The partial pressure is  [tex]P = 1.90 \ atm[/tex]

   The temperature is [tex]T = 25 ^oC[/tex]

Henry's  law is mathematically represented as

       [tex]C = P * k[/tex]

Where C is  the concentration of sulfur hexafluoride(SP)

substituting value

        [tex]C = 1.90 * 2.4*10^{-4}[/tex]

        [tex]C = 4.56*10^{-4} \ M[/tex]

The number of moles of  SP is mathematically represented as

        [tex]n = C * V[/tex]

substituting value

       [tex]n = 0.525 * 4.56*10^{-4}[/tex]

       [tex]n = 2.39 *10^{-4} \ moles[/tex]

The mass of SP that dissolved is

          [tex]m = n * Z[/tex]

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

           [tex]Z = 146 g/mole[/tex]

So

         [tex]m = 2.394*10^{-4} * 146[/tex]

         [tex]m = 0.0349 \ g[/tex]

Answer:

Explanation:

From the question, osmotic pressure exerted by a solution is equal to the MOLARITY multiplied by the absolute TEMPERATURE and the GAS CONSTANT r.

Let P = osmotic pressure,

C = molarity, then

T = absolute temperature

r=gas constant

The Osmotic pressure Equation exerted by a solution [tex]P=C*T*r[/tex]

[tex]P=CTr[/tex]

Then it was required in the question to write an equation that will let you calculate the molarity c of this solution, and this equation should contain ONLY symbols

C= molarity of the solution

P=osmotic pressure

r = gas constant

T= absolute temperature

[tex]C=P/(rT)[/tex]

The equation that will let us calculate the molarity c of this solution = [tex]C=P/(rT)[/tex]

Answer:

It is the portion of internal energy that can be transferred from one substance to another.

Thermal energy is the portion of internal energy that can be transferred from one substance to another.

Thermal energy is the energy an object posses which is as a result of particles movement within it.

It is also the internal energy system in a state of thermodynamic equilibrium which is as a result of its temperature. Thermal energy cannot be concert to useful work easily.

Therefore, thermal energy is the portion of internal energy that can be transferred from one substance to another.

Learn more about thermal energy from the link below.

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Answer:

The percentage of the void space out of the total volume occupied is 93.11%

Explanation:

A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.

To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom

Let's calculate this one after the other.

For the hydrogen, formula for the volume will be

[tex]V_{hydrogen[/tex] = 2 × 4/3 × π × [tex]r_{H}^{3}[/tex]

where [tex]r_{H}^{3}[/tex] = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 × [tex]10^{-12}[/tex] meters

Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31

we multiply this by the avogadro's number = 6.02 * 10^23

= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3

We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen

Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3

adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)

Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water

= (18g/mol)/(1 g/ml) = 18 ml/mol

Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml

Now, the percentage of void = volume of void/total volume * 100%

= 16.76/18 * 100% = 93.11%

Answer:

The brain

Explanation:

With all those instructions the body recqures to respond to it must be so

Hope it helps

Answer:

7.5 atm

Explanation:

Initial pressure P1 = 1.0 ATM

Initial volume V1= 196 L

Final pressure P2= the unknown

Final volume V2= 26000ml or 26 L

From Boyle's law we have;

P1V1= P2V2

P2= P1V1/V2

P2= 1.0 × 196/26

P2 = 7.5 atm

Therefore, as the air is compressed, the pressure increases to 7.5 atm.

Answer:

[tex]\Delta H=-11897J[/tex]

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

[tex]\Delta H=\Delta U+V\Delta P[/tex]

Whereas the change in the internal energy is computed by:

[tex]\Delta U=nCv\Delta T[/tex]

So we compute the initial and final temperatures for one mole of the ideal gas:

[tex]T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}[/tex]

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

[tex]\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J[/tex]

Then, the volume-pressure product in Joules:

[tex]V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J[/tex]

Finally, the change in the enthalpy for the process:

[tex]\Delta H=-7140J-4757J\\\\\Delta H=-11897J[/tex]

Best regards.

The change in enthalpy is 70.42J

Data;

The change of enthalpy is calculated as

[tex]\delta H = \delta V + \delta nRT\\\delta n = 0\\\delta H = \delta U \\[/tex]

The volume change is negligible

The change in enthalpy here is equal to change in internal energy over ΔE

[tex]\delta H = \delta U = nCv\delta T\\\delta H = \frac{3}{2}(nR\delta T)\\\delta H = \frac{3}{2}\{\delta PV)\\ \delta H = \frac{3}{2}[(10.90-1.24)*4.86] \\\delta H = 70.42J[/tex]

The change in enthalpy is 70.42J

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Answer:

Concentration of Copper (II) Sulfate in the original sample in mol/L = 0.0035 M

Concentration of Copper (II) Sulfate in the original sample in g/L = 0.56 g/L

Explanation:

Complete Question

Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)

Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.

Solution

Noting that the precipitate is Copper as it is the only solid by-product of this reaction.

89 mg of Copper is produced from this reaction.

We convert this into number of moles for further stoichiometric calculations

Mass of Copper = 89 mg = 0.089 g

Molar mass of Copper = 63.546 amu

Number of moles of Copper produced from the reaction = (0.089/63.546) = 0.0014005602 = 0.001401 mole

From the stoichiometric balance of the reaction,

1 mole of Copper is produced from 1 mole of Copper (II) Sulfate

0.001401 mole of Copper will be produced similarly from 0.001401 mole of Copper (II) Sulfate.

Number of moles of Copper (II) Sulfate in the original sample = 0.001401 mole

Concentration of Copper (II) Sulfate in the original sample in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = 0.001401 mole

Volume in L = (400/1000) = 0.4 L

Concentration of Copper (II) Sulfate in the original sample in mol/L = (0.001401/0.4) = 0.0035025 mol/L = 0.0035 mol/L to 2 s.f.

Concentration in g/L = (Concentration in mol/L) × (Molar Mass)

Concentration in mol/L = 0.0035025 M

Molar mass of Copper (II) Sulfate = 159.609 g/mol

Concentration of Copper (II) Sulfate in the original sample in g/L = 0.0035025 × 159.609 = 0.559 g/L = 0.56 g/L to 2 s.f

Hope this Helps!!!!

The concentration of the original copper solution is 0.035 M.

The equation of the reaction is;

Fe(s) + CuSO4(aq) -------> FeSO4(aq) + Cu(s)

Number of moles of copper obtained = 89 × 10^-3g/63.5 = 0.0014 moles

Since the reaction is 1:1, the number of moles of copper sulfate that reacted is c.

From the question, we are told that the volume of solution is 400.mL or 0.04L.

Hence, the concentration of the solution is; number of moles /volume

=  0.0014 moles/0.04L = 0.035 M

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Missing parts;

Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.