The molar enthalpy of
mole of a liquid.
is the heat required to vaporize one

thermochemical equation

combustion

released

vaporization

fusion

absorbed

heat

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:

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.

Answer:

P is the limiting reagent

Explanation:

P = phosphorus  = 30.97g/mol

Cl2 = Chlorine = 70.91g/mol

PCl3 = Phosphorus Trichloride = 137.33g/mol

P + Cl2 = PCl3

Left Side

P = 1

Cl = 2

Right Side

P = 1

Cl = 3

So equation needs to be balanced first

2P + 3Cl = 2PCl3

Left Side

P = 2

Cl = 6

Right Side

P = 2

Cl = 6

That's better.

Ok so we have 12.39g of P so we have 0.4 moles of it

We then have 52.54g of Cl so we have 0.74 moles of it

For every P we need 1.5 Cl so we have an excess of Cl

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:

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:

The reaction produces 201.4 g of hydrogen gas and 12.2 kg of Nickel Phosphate.

Explanation:

English Translation

9.7 Kg of a 70% nickel mineral react with 8L of a 60% phosphoric acid solution and with a density of 1.36g / ml.

Solution

The problem doesn't seen to be complete as it doesn't ask a question in the end. But, we will just calculate the amount of each product expected to cover the grounds.

The balanced chemical reaction between Nickel and Phosphoric acid is given as

3Ni + 2H₃PO₄ → 3H₂ + Ni₃(PO₄)₂

We need to first obtain the limiting reagent, that is, the reagent that is used up during the reaction and is in short supply. This reagent determines the amount of products that will be formed.

Mass of nickel that is present at the start = 70% of 9.7 kg = 6.79 kg

Mass of Phosphoric acid present at the start of the reaction = 60% of (8000 mL × 1.36 g/mL) = 6528 g = 6.528 kg

Converting both of these to number of moles

Number of moles = (mass)/(Molar mass)

For nickel,

Mass = 6.79 kg = 6790 g

Molar mass = 58.6934 g/mol

Number of moles at the start = (6790/58.6934) = 115.7 moles

For Phosphoric acid

Mass = 6528 g

Molar mass = 97.994 g/mol

Number of moles = (6528/97.994) = 66.6 moles

3 moles of Ni reacts with 2 moles of H₃PO₄

From the number of moles present initially, shows that Phosphoric acid is in limited supply and is the limiting reagent.

From the stoichiometric balance of the reaction

2 moles of H₃PO₄ gives 3 moles of H₂

66.6 moles of H₃PO₄ will give (66.6×3/2) of H₂, that is, 99.9 moles of H₂.

Mass of H₂ liberated from the reaction = (Number of moles) × (molar mass) = 99.9 × 2.016 = 201.3984 g = 201.4 g

2 moles of H₃PO₄ gives 1 mole of Ni₃(PO₄)₂

66.6 moles of H₃PO₄ will give (66.6×1/2) of Ni₃(PO₄)₂, that is, 33.3 moles of Ni₃(PO₄)₂.

Mass of Ni₃(PO₄)₂ produced from the reaction = (Number of moles) × (molar mass) = 33.3 × 366.02 = 12,188.466 g = 12.2 kg

Hope this Helps!!!

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.

The correct answer is C. Bohr's model

Explanation:

Bohr's model of the atom developed in 1913 proposed each atom contained a nucleus with protons and neutrons. Also, there were electrons that orbited the nucleus. About this, Niels Bohr proposed the orbits of electrons were similar to those of planets around the sun; however, these did not occur due to gravity but to attraction forces. This model integrated new accurate ideas about the atom. However, this model was still inaccurate because particles in an atom are electrically charged and electrons do not orbit in fixed stable orbits and cannot be compared to the movement of planets around a star.  

Answer:

Bohr's model

Explanation:

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:

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: 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:

C3H6O

Explanation:

Step 1:

Data obtained from the question include the following:

Mass of the compound = 0.225g

Mass of CO2 = 0.512g

Mass of H2O = 0.209g

Step 2:

Determination of the masses of carbon, hydrogen and oxygen present in the compound.

This is illustrated below:

For Carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 0.512 = 0.1396g

For Hydrogen, H:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 0.209 = 0.0232g

For Oxygen, O:

Mass of O = 0.225 – (0.1396 + 0.0232)

Mass of O = 0.0622g

Step 3:

Determination of the empirical formula for caprioc acid.

This can be obtain as follow:

C = 0.1396g

H = 0.0232g

O = 0.0622g

Divide by their molar mass

C = 0.1396/12 = 0.0116

H = 0.0232/1 = 0.0232

O = 0.0622/16 = 0.0039

Divide by the smallest

C = 0.0116/0.0039 = 3

H = 0.0232/0.0039 = 6

O = 0.0039/0.0039 = 1

Therefore, the empirical formula for caprioc acid is C3H6O

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:

Kc = [Pb²⁺]³.[PO₄³⁻]²

Explanation:

Let's consider the following reaction at equilibrium.

Pb₃(PO₄)₂(s) ⇄ 3 Pb²⁺(aq) + 2 PO₄³⁻(aq)

The concentration equilibrium constant is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.

Kc = [Pb²⁺]³.[PO₄³⁻]²

This equilibrium constant is known as the solubility product of Pb₃(PO₄)₂.

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

Learn more: https://brainly.com/question/9352088

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.

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:

When hypercapnia processes occur, where the concentration of carbon dioxide gas increases in the blood, the protonization of the blood increases, this means that the H + ions increase in concentration, thus generating metabolic acidosis.

This metabolic acidosis is regulated by various systems, but the respiratory system collaborates by generating hyperventilation, to increase blood oxygen pressures, decrease CO2 emissions, and indirectly decrease acidity.

Explanation:

This method of regulating the body is crucial, since the proteins in our body will not be altered if they do not happen.

The enzymes, the red globules, and many more fundamental things for life ARE PROTEINS, that in front of acidic media these modify their structure by denaturing themselves and ceasing to fulfill their functions. This is the reason why it seeks to neutralize the blood pH when it comes to an increase in CO2.

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:

Explanation:

O B. Hypothesizing how pieces of ancient pottery were used

Answer:

The correct answer is 12.43 Liters.

Explanation:

Based on the given question, the volume V₁ occupied by the sample of carbon dioxide gas is 14.2 liters at temperature (T₁) 65 degree C or 65+273 K = 338 K.  

The gas is cooled at a temperature (T₂) 23 degree C or 273+23 K = 296 K

The volume of the gas (V₂) after cooling can be determined by using the formula,  

V₁/T₁ = V₂/T₂

14.2/338 = V₂/296

0.0420 = V₂/296

V₂ = 0.0420 * 296  

V₂ = 12.43 Liters.  

Answer:

E. Gain of electrons

Explanation:

A reduction reaction is one part of the two concurrent reactions that take place in a redox (reduction-oxidation) reaction.

During reduction, an atom gains electrons from a donor atom, and it's oxidation number becomes smaller.

Option A is wrong because reduction does not increase oxidation state nor are neutrons involved

Option B is wrong because reduction is not a nuclear reaction (does not involve the nucleons)

Option C is wrong because reduction leads to reduction in oxidation state

Option D is wrong leads to a reduction in oxidation state when electrons are gained

Option E is correct because reduction involves gain of electrons

Answer:

I am not sure, but I think this is the answer 72.987 g/mol

Answer:

CONVERT IT:

50°F is equal to 10°C

Answer:

10 degrees Celsius

Explanation:

(50°F − 32) × 5/9 = 10°C

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:

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:

c

Explanation:

Answer:

409.0 kg of sodium sulfate decahydrate will produce 4.49×10⁵ kJ

of heat energy.

Explanation:

CHECK THE COMPLETE QUESTION BELOW

To make use of an ionic hydrate for storing solar energy, you place 409.0 kg of sodium sulfate decahydrate on your house roof. Assuming complete reaction and 100% efficiency of heat transfer, how much heat (in kJ) is released to your house at night? Note that sodium sulfate decahydrate will transfer 354 kJ/mol

EXPLANATION

Here we were asked to calculate the amount of heat will be generated by 409.0 kg of sodium sulfate decahydrate at night assuming there Isa complete reaction and 100% efficiency of heat transfer in the process

The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S) is needed here, so it must be firstly calculated.

The molecular weight of sodium sulfate decahydrate (H₂₀Na₂O₁₄S)

( 1*20) + (22.98*2) + (16*14)+ (32*14)= 322.186 g/mol.

Thus 409.0 kg of H₂₀Na₂O₁₄S will have a value which is equivalent to = (409000g)/(322.186 g/mol.)

=1269.453mol of H₂₀Na₂O₁₄S.

But it was stated in the the question that per mole of H₂₀Na₂O₁₄S will transfer 354 kJ heat.

Therefore, 1269.453mol will transfer 1269.453× 354 kJ = 4.49×10⁵ kJ of heat.

Hence, 409.0 kg of sodium sulfate decahydrate will produce

4.49×10⁵ kJ of heat energy.

Answer:

in short weight

Explanation:

weight is mass x gravitational pull on an object