To determine the absolute age of rocks and fossils, geologists use _____.

Answer:

Approximately [tex]21\; \rm g[/tex].

Explanation:

[tex]\rm H_2SO_4[/tex] (a diprotic acid) reacts with [tex]\rm NaOH[/tex] (a monoprotic base) at a one-to-two ratio:

[tex]\rm 2\; NaOH\, (s) + H_2SO_4\, (aq) \to Na_2SO_4\; (aq) + 2\; H_2O\, (l)[/tex].

In other words, if [tex]n(\mathrm{NaOH})[/tex] and [tex]n(\mathrm{H_2SO_4})[/tex] represent the number of moles of the two compounds reacted, then:

[tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex].

Look up the relative atomic mass data on a modern periodic table:

Calculate the (molar) formula mass of [tex]\rm H_2SO_4[/tex] and [tex]\rm NaOH[/tex]:

[tex]M(\mathrm{H_2SO_4}) = 2 \times 1.008 + 32.06 + 4 \times 15.999 = 98.072\; \rm g \cdot mol^{-1}[/tex].

[tex]M(\mathrm{NaOH}) = 22.990 + 15.999 + 1.008 = 39.997\; \rm g \cdot mol^{-1}[/tex].

Calculate the number of moles of formula units in that [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:

[tex]\begin{aligned}n(\mathrm{NaOH}) &= \frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})} \\ &= \frac{33.8\; \rm g}{39.997\; \rm g \cdot mol^{-1}} \approx 0.845\; \rm mol\end{aligned}[/tex].

Apply the ratio [tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex] to find the (maximum) number of moles of [tex]\rm H_2SO_4[/tex] that would react with the [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:

[tex]\begin{aligned}n(\mathrm{H_2SO_4}) &= \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} \cdot n(\mathrm{NaOH})\\ &= \frac{1}{2} \times 0.845 \approx 0.4225\; \rm mol\end{aligned}[/tex].

Calculate the mass of that [tex]0.4225\; \rm mol[/tex] of [tex]\rm H_2SO_4[/tex]:

[tex]\begin{aligned}m(\mathrm{H_2SO_4}) &= n(\mathrm{H_2SO_4}) \cdot M(\mathrm{H_2SO_4})\\ &= 0.4225 \; \rm mol \times 98.072\; \rm g \cdot mol^{-1} \approx 41.435\; \rm g \end{aligned}[/tex].

When the maximum amount of [tex]\rm H_2SO_4[/tex] is reacted, the minimum would be in excess. Hence, the minimum mass of

[tex]62\; \rm g - 41.435\; \rm g \approx 21\; \rm g[/tex] (rounded to two significant figures.)

Answer:

water stored in a dam

Explanation:

when the water is in dam it is ready to move bit is not moving

Answer:

You would be able to see the ants clearly with the unique body parts.

Explanation:

Convex lens is also known as Converging lens. It helps in converging rays of light to become a principal focus which is usually very clear and visible to the eyes. The converging lens when used to view the ant makes the ant appear very visible and the individual is able to see all the unique body parts of the ant. This type of lens is used in individuals who are longsighted.

Complete Question

The complete question is shown on the first uploaded image

Answer:

Part A

    activation barrier for the reaction [tex]E_a = 84 .0 \ KJ/mol[/tex]

Part B

    The frequency plot is  [tex]A = 2.4*10^{13} s^{-1}[/tex]    

Explanation:

From the question we are told that

     at  [tex]T_1 = 300 \ K[/tex]   [tex]k_1 = 5.70 *10^{-2}[/tex]

and  at  [tex]T_2 = 310 \ K[/tex]   [tex]k_2 = 0.169[/tex]

The  Arrhenius plot is mathematically represented as

      [tex]ln [\frac{k_2}{k_1} ] = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2} ][/tex]

Where [tex]E_a[/tex] is the activation barrier for the reaction

         R is the gas constant with a value of  [tex]R = 8.314*10^{-3} KJ/mol \cdot K[/tex]

Substituting values

          [tex]ln [\frac{0.169}{6*10^-2{}} ] = \frac{E_a}{8.314*10^{-3}} [\frac{1}{300} - \frac{1}{310} ][/tex]

=>       [tex]E_a = 84 .0 \ KJ/mol[/tex]

The  Arrhenius plot can also be  mathematically represented as

      [tex]k = A * e^{-\frac{E_a}{RT} }[/tex]

Here we can use any value of k from the data table with there corresponding temperature let take  [tex]k_2 \ and \ T_2[/tex]

So substituting values

        [tex]0.169 = A e ^{- \frac{84.0}{8.314*10^{-3} * 310} }[/tex]

=>       [tex]A = 2.4*10^{13} s^{-1}[/tex]    

Answer: D. They would device an experiment that could test the two scientists conclusions.

Explanation:

The results of the scientific study must be verified by peer scientists or members of the scientific community to proof whether the research has been conducted produce a valid evidence.

In the given situation, the two scientists had developed different conclusion for the same experiment. This may mean either of the two may have put up an incorrect conclusion.

The scientific community may address this issue by performing the experiment. Every scientific conclusion is based upon the results of the experimental approach.

Answer:d

Explanation:

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899        0.01714        0           0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714    0.01714 (end)

0.00185  | 0 → 0.01714    0.01714 (end)  

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

Answer:

[tex]\rm [Na^{+}]= \text{0.055 12 mol/L}[/tex]

[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \text{0.018 99 mol/L}[/tex]

Explanation:

The overall equation for the reaction is  

Na₂HAsO₄ + NaOH ⟶ Na₃AsO₄ + H₂O

1. Mass balance for Na

All the Na⁺ comes from the Na₂HAsO₄ and the NaOH.

The mass balance equation for Na is  

[tex]\rm c_{Na^{+}} = 2[Na^{+}]_{Na_{2}HAsO_{4}} + [Na^{+}]_{NaOH}[/tex]

At the moment of mixing and before the reaction started, the total volume had doubled, so the concentrations of each component were halved.

[Na₂HAsO₄] = ½ × 0.037 98 =0.018 99 mol·L⁻¹

[NaOH]         = ½ × 0.034 28 = 0.017 14 mol·L⁻¹

[tex]\rm c_{Na^{+}} = 2\times 0.01899 + 0.01714 = 0.03798 + 0.01714\\c_{Na^{+}}= \textbf{0.055 12 mol/L}[/tex]

2. Mass balance for arsenate species

All the arsenate species come from the Na₂HAsO₄.

The reactions involved are

HAsO₄²⁻+ OH⁻ ⇌ AsO₄³⁻ + H₂O

HAsO₄²⁻ + H₂O ⇌ H₂AsO₄⁻ + OH⁻

H₂AsO₄⁻ + H₂O ⇌ H₃AsO₄ + OH⁻

The mass balance equation for arsenate species is

[tex]\rm c_{\text{arsenate}} = [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}][/tex]

At the moment of mixing, the concentration of Na₂HAsO₄ had halved.

[Na₂HAsO₄] = ½ × 0.039 78 = 0.018 99 mol·L⁻¹

[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \textbf{0.018 99 mol/L}[/tex]

Answer:

in short weight

Explanation:

weight is mass x gravitational pull on an object

Answer:

Based on the given question, the dimensions of the plate is 15 cm in length, 12.5 cm in width, and 3.50 mm in thickness (0.350 cm). Now the volume of the plate will be,  

V = 15 cm × 12.5 cm × 0.350 cm = 65.62 cm³

A hole of diameter 2.50 cm is drilled through the center of the plate, at the height of 3.50 mm or 0.350 cm. Now the volume of the hole is π(r)²h,  

= 22/7 × (1.25 cm)² × 0.350 cm = 1.72 cm³

Thus, the volume of the plate will be determined by subtracting the volume of plate with the volume of hole, which will be,  

65.62 cm³ - 1.72 cm³ = 63.9 cm³

The density of the alloy is 8.80 g/cm³, therefore, the mass of the alloy can be determined by using the formula, mass = density * volume  

mass = 8.80 g/cm³ × 63.9 cm³ = 562.32 grams

Of the total alloy, 0.090 percent is Si, that is,  

(0.090/100) × 562.32 g = 0.506 grams of Si

The natural abundance of the element is not determined by mass but by the number of atoms it possess. For this Avogadro's number and atomic mass of Si is used. Now the number of atoms of Si present is,  

(0.506 g) (1 mol/28.0855 g) (6.023 × 10²³ atoms /mol) = 1.08 × 10²² Si atoms

Of these Si atoms, 3.10 percent are Si-30 so,  

= (3.10 / 100) × (1.08 × 10²² atoms) / 1000 = 3.34 × 10²⁰ atoms of Si-30. or 3.4 × 10²⁰ atoms

Answer:

Chemical reactions, kinetics, organic chemistry

Explanation:

You might study the chemical reaction, learn about the differences between products and reactants, about delta H and exothermic and endothermic reactions. You may also study Kinetics by studying the rates of reactions with certain chemicals in a maple's enzymatic processes.

Another thing that you might learn about is organic chemistry. The glucose molecules, carbohydrates, lipids, nucleic acids, all have a structure based on the Carbon atom. You can learn about the specific structures of some chemicals that are involved in photosynthesis and simple hydrocarbons that are involved in photosynthetic/bio-synthetic pathways.

There's probably a lot more - but these are the most basic things I could think of.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  ammonia

Explanation:

The mixture contains two base compound which are

           ammonia,

and     diethylamine

Now the addition of HCl which is  a strong acid in step 1  will cause the protonation of  the  two base compound , which makes the soluble hence resulting in them being extracted to the aqueous layer as represented in below

       [tex]NH_3 + HCl\to NH_4 ^{+} + Cl^-[/tex]

and

     [tex](CH 3CH 2) 2NH + HCl \to (CH 3CH 2) 2NH_2^{+} + Cl[/tex]

Answer:

99.24 gm of nitrogen .

Explanation:

molecular weight of ammonia = 17 , molecular weight of nitrogen = 28.

2 NH₃(g) + 3 CuO(s) → 1N₂(g) + 3 Cu(s) + 3 H₂O(g)

2 x 17 gm                      28 gm

( 34 gm )

34 gm of ammonia forms 28 gms of nitrogen

1 gm of ammonia   forms 28 / 34 gms of nitrogen

120.51 gn of ammonia forms 28 x 120.51 / 34 gms of nitrogen

28 x 120.51 / 34 gms

= 99.24 gms of nitrogen will be formed .

Answer:

Explanation:

in your case ,

Meaured value = 112.3

actual value = 119.5

Absolute error= measured value - actual value

Percent error = [measured value - actual value  / actual value ] x 100

Hope this help you to find the answer

Answer:

gravitational and quantum ARE NOT, but electric and magnetic ARE.    there is a similar question to this but it's the exact opposite, so don't get confused

Answer:

4 HBr + O2  →  + 2H2O + 2Br2

Explanation:

Based on the following reaction mechanism:

HBr + O2 → HOOBr (slow)

HOOBr + HBr → 2HOBr (fast)

HOBr + HBr → H2O + Br2 (fast)

The equation for the overall reaction is the sum of the three reactions in which intermediaries of reaction (HOBr and HOOBr are canceled). That is 1 + 2 + 2*(3):

HBr + O2 + HOOBr + HBr + 2HOBr + 2HBr → HOOBr + 2HOBr + 2H2O + 2Br2

Beeing this reaction the equation of the overall reaction.

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 SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

                     mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

Answer:

c

Explanation:

Answer:

Mass is the same but it weights 64 Newtons

Explanation:

First of Mass is the same in any sort of gravity. Now let's calculate weight

W = MG

where W = Weight

M = Mass

G = Gravity

W = (40kg)(1.6)

W = 64

Sorry for the spelling mistakes, hope this helps

Answer:6.61kilo

Explanation: fdfv

Explanation:

A Bronsted-Lowry base is a substance that accepts a proton in the form of a hydrogen (H) atom.

On the other hand;

Bronsted-Lowry acid is the substance that donates the proton.

HF (aq) + SO32- ⇌ F- + HSO3-

In the forward reaction;

Bronsted-Lowry acid : HF

Bronsted-Lowry base: SO32-

In the backward reaction;

Bronsted-Lowry acid : HSO3-

Bronsted-Lowry base: F-

The conjugate base of HF is F-

The conjugate acid of SO32- is HSO3-

Answer:

Explanation:

The given chemical reaction is:

[tex]2H_2O_{(l)} \to^{I^-}} 2H_2O_{(l)}+O_2_{(g)}[/tex]

From above equation  [tex]I^-[/tex] serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.

Given that :

the volume of potassium iodide [tex]V_{KI} = 15 \ ml[/tex]

the molarity of potassium [tex]M_{KI} = 0.1 \ M[/tex]

the volume of distilled water [tex]V_W = 15 \ mL[/tex]

The volume of 3% [tex]H_2O_2 \ \ V_{H_2O_2} = 5 \ mL[/tex]

Molarity of AgNO₃ [tex]M_{AgNO_3} = 0.1 \ M[/tex]

Let take an integral look with the reaction between KI and AgNO₃; we have

[tex]KI + AgNO_3 \to KNO_3 + AgI[/tex]

At the end point; the moles of KI will definitely be equal to the moles of AgNO₃

So;

[tex]M_{KI}V_{KI}= M_{AgNO_3}V_{AgNO_3} \\ \\ V_{AgNO_3} = \dfrac{M_{KI}V_{KI}}{M_{AgNO_3}} \\ \\ \\ V_{AgNO_3} = \dfrac{ 0.1*15}{0.1}[/tex]

[tex]V_{AgNO_3} = 15 \ ml[/tex]

Thus; the volume of 0.1 M AgNO₃  needed to reach the end point is 15 mL

Answer:

ΔG = -2.17 kJ/mol

Explanation:

ΔG of a reaction at any moment could be obtained thus:

ΔG = ΔG° + RT ln Q

Where ΔG° is standard change in free energy of a particular reaction (7.53kJ/mol for the reaction of the problem, R is gas constant (8.314×10⁻³kJ/molK), T is absolute temperature (298K) and Q is reaction quotient of the reaction.

For the reaction:

dihydroxyacetone phosphate ⇄ glyceraldehyde−3−phosphate

Q is defined as:

Q = [glyceraldehyde−3−phosphate] / [dihydroxyacetone phosphate]

Replacing values in ΔG formula:

ΔG = 7.53kJ/mol + 8.314×10⁻³kJ/molK × 298.15K ln [0.00200M] / [0.100M]

ΔG = -2.17 kJ/mol

Answer:

Q = 12.38

Explanation:

The Nernst equation is given as; Ecell = E°cell - (2.303RT/nF) log Q  ;where Q is the reaction quotient.

The reaction quotient, Q  in a reaction, is the product of the concentrations of the products divided by the product of the concentrations of the reactants.

In an electrochemical cell, Q is the ratio of the concentration of the electrolyte at the anode to that of the electrolyte at the cathode.

Q = [anode]/[cathode]

therefore , Q = 0.052/0.0042 = 12.38

Answer:

Explanation:

CHECK THE ATTACHMENT FOR THE COMPLETE QUESTION AND THE DETAILED EXPLANATION

NOTE:

Equatorial atoms are referred to atoms that are attached to carbons in the cyclohexane ring which is found at the equator of the ring.

Axial atoms are atoms that exist in a bond which is parallel to the axis of the ring in cyclohexane

Answer:

M=0.816M

Explanation:

Hello,

In this case, we should consider the following reaction:

[tex]AgNO_3+KCl\rightarrow KNO_3+AgCl[/tex]

Thus, by knowing the 1:1 molar ratio of silver nitrate and potassium chloride, we can easily compute the moles of silver nitrate precipitating the 1570 mg of potassium chloride considering its molar mass of 74.5513 g/mol:

[tex]n_{AgNO_3}=1570mgKCl*\frac{1gKCl}{1000mgKCl} *\frac{1molKCl}{74.5513gKCl}*\frac{1molAgNO_3}{1molKCl} \\\\n_{AgNO_3}=0.021molAgNO_3[/tex]

Then, by using the volume of silver nitrate in liters (0.0258 L), we can directly compute the molarity:

[tex]M=\frac{0.021molAgNO_3}{0.0258L}\\ \\M=0.816M[/tex]

Regards.

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:

0.053moles

Explanation:

Hello,

To calculate the number of moles of gas remaining in his after he exhale, we'll have to use Avogadro's law which states that the volume of a given mass of gas is directly proportional to its number of moles provided that temperature and pressure are kept constant. Mathematically,

V = kN, k = V / N

V1 / N1 = V2 / N2= V3 / N3 = Vx / Nx

V1 = 1.7L

N1 = 0.070mol

V2 = 1.3L

N2 = ?

From the above equation,

V1 / N1 = V2 / N2

Make N2 the subject of formula

N2 = (N1 × V2) / V1

N2 = (0.07 × 1.3) / 1.7

N2 = 0.053mol

The number of moles of gas in his lungs when he exhale is 0.053 moles

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.

Answer:

is capable of combining with oxygen to form iron oxide

Answer:

CONVERT IT:

50°F is equal to 10°C

Answer:

10 degrees Celsius

Explanation:

(50°F − 32) × 5/9 = 10°C