calculate the moles of 25.2 g Na2S2O8

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: Ethanol is the odd one out.

Explanation:

A polar compound is defined as the compound which is formed when there is a difference of electronegativities between the atoms. It is also defined as the bond which is formed due to the unequal sharing of electrons between the atoms.

Non-polar compound is defined as the compound which is formed when there is no difference of electronegativities between the atoms or the polarities cancel out.

Hexane [tex](C_6H_{14}), Oil (mixture of hydrocarbons) and carbon tetrachloride [tex](CCl_4)[/tex] all are non polar whereas ethanol is polar due to electronegative difference between hydrogen and oxygen.

Answer:

C

Explanation:

It looks pretty reasonable to me

Answer: the exciation of molecules is brount by absorption of energy  in spectroscpy

Explanation:

Answer:

43.964

Explanation:

i think i used a calculator so let me know if its wrong

Answer:

39.49 kg

Explanation:

:)

Answer:

Mass of Ga = 0.73694 gram

Explanation:

Given:

Current = 0.850 A

Time = 60 minutes

Find:

Amount of gas deposit.

Computation:

Total charge = Current × Time in second

Total charge = 0.850 × 60 × 60

Total charge = 3,060 C

Mole of electron = Total charge / Faraday constant         [Faraday constant = 96,485.3329]

Mole of electron = 3,060 / 96,485.3329

Mole of electron = 0.0317146

Moles of Ga = 1/3 [Mole of electron]

Moles of Ga = 1/3 [0.0317146]

Moles of Ga = 0.01057

Mass of Ga = molar mass × Moles of Ga

Mass of Ga = 69.72 × 0.01057

Mass of Ga = 0.73694 gram

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:

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:

[tex]m=14,105.71 g Fe[/tex]

Explanation:

Hello,

In this case, the first step is to compute the volume of the block considering the length, height and width:

[tex]V=L \times W \times H =12.5 in\times 3.50 in \times 2.50 in =109.375 in^3[/tex]

Then, we compute the volume in cubic centimetres:

[tex]V=109.375in^3\times \frac{16.3871 cm^3}{1in^3} =1792.34cm^3[/tex]

Finally, as the density is given by:

[tex]\rho =\frac{m}{V}[/tex]

We solve for the mass:

[tex]m=\rho \times V= 7.87\frac{g}{cm^3} \times 1792.34 cm^3\\\\m=14,105.71 g Fe[/tex]

Best regards.

Answer:

A. The pH value of rainwater is acidic about 4.4

B. The molar concentrations of the Hydronium ions are more than that of the hydroxide ions. That is why the rainwater is acidic with a pH of less than 7

Explanation:

A. Methyl orange is an acid indicator that is used to detect acidic solutions which have pH values that fall within the range of about 4.4 to 7.  The distinct yellow colour change that was shown by the methyl orange as it was added to the water shows that the pH value is acidic, with a value above 4.4. (it has to be like this before methyl orange changes to  yellow colour)

B. The Hydronium ( H30+) ion concentrations and the hydroxide (OH-) ion concentrations are used to measure the pH values of substances.

We can tell that the Hydronium ( H30+) ion concentrations are more than the hydroxide (OH-) ion concentrations in the sample of rainwater tested. This can be detected from the colour change that both the methyl orange and the litmus paper gave. The indicators showed that the rainwater solution was indeed acidic. Hence, the pH value will be less than 7, but greater than 4.4.

Answer:

B. the molecules in liquid are loosely packed and scattered thus, they cannot be arranged

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:

93.15 %

Explanation:

We have to start with the chemical reaction:

[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~NaCl[/tex]

Now, we can balance the reaction:

[tex]CaCl_2~+~Na_2CO_3~->~CaCO_3~+~2NaCl[/tex]

Our initial data are the 15.71 g of [tex]CaCl_2[/tex], so we have to do the following steps:

1) Convert from grams to moles of [tex]CaCl_2[/tex] using the molar mass (110.98 g/mol).

2) Convert from moles of [tex]CaCl_2[/tex] to moles of [tex]CaCO_3[/tex] using the molar ratio. ( 1 mol [tex]CaCl_2[/tex]= 1 mol of [tex]CaCO_3[/tex]).

3) Convert from moles of [tex]CaCO_3[/tex] to grams of [tex]CaCO_3[/tex] using the molar mass. (100 g/mol).

[tex]15.71~g~CaCl_2\frac{1~mol~CaCl_2}{110.98~g~CaCl_2}\frac{1~mol~CaCO_3}{1~mol~CaCl_2}\frac{100~g~CaCO_3}{1~mol~CaCO_3}=14.16~g~CaCO_3[/tex]

Finally, we can calculate the yield percent:

[tex]%~=~\frac{13.19~g~CaCO_3}{14.16~g~CaCO_3}*100=93.15~%[/tex]

I hope it helps!

The percentage yield obtained when excess sodium carbonate, Na₂CO₃, is added to a solution containing 15.71 g CaCl₂ is 93.2%

Molar mass of CaCl₂ = 40 + (35.5×2) = 111 g/mol

Mass of CaCl₂ from the balanced equation = 1 × 111 = 111 g

Molar mass of CaCO₃ = 40 + 12 + (16×3) = 100 g/mol

Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g

SUMMARY

From the balanced equation above,

111 g of CaCl₂ reacted to produce 100 g of CaCO₃

From the balanced equation above,

111 g of CaCl₂ reacted to produce 100 g of CaCO₃.

Therefore,

15.71 g of CaCl₂ will react to produce = [tex]\frac{15.71 * 100}{111} \\\\[/tex] = 14.15 g of CaCO₃.

Thus, the theoretical yield of of CaCO₃ is 14.15 g

Actual yield of CaCO₃ = 13.19 g

Theoretical yield of CaCO₃ = 14.15 g

[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{13.19}{14.15} * 100\\\\[/tex]

Therefore, the percentage yield of the reaction is 93.2%

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Answer:

[tex]1.66~V[/tex]

Explanation:

We have to start with the half-reactions for both ions:

[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76

[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80

If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:

[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76

[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80

If we want to calculate ºE we have to add the two values, so:

ºE=0.76+0.80 = 1.56 V

Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:

[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]

On this case, Q is equal to:

[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]

Because the total reaction is:

[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]

So, the value of "Q" is:

[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]

Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:

[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]

I hope it helps!

Answer:

[CH₂Cl₂] = 7.07x10⁻² M

[CH₄] = 0.319 M

[CCl₄] = 0.164 M  

Explanation:

The equilibrium reaction is the following:

2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)  

The equilibrium constant of the above reaction is:

[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]

When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:

[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]

[tex]C_{CH_{4}} = 0.328 M[/tex]      

Now, the concentrations at the equilibrium are:

2CH₂Cl₂(g)   ⇄   CH₄(g)  +  CCl₄(g)

5.35x10⁻² - 2x   0.328 + x   0.173 + x    

[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]

[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]

Solving the above equation for x:  

x₁ = 0.076 and x₂ = -0.0086

Hence, the concentration of the three gases once equilibrium has been reestablished is:

[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M

[CH₄] = 0.328 + (-0.0086) = 0.319 M

[CCl₄] = 0.173 + (-0.0086) = 0.164 M  

We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.

I hope it helps you!

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: The solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.

Explanation:

The given data is as follows.

Volume of sample water = 46 ml

Temperature = [tex]21^oC[/tex]

After vaporization, washes and then drying the weight of mineral X = 0.87 g

This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.

          1 ml of water = [tex]\frac{0.87 g}{46.0 ml}[/tex]

                                = [tex]1.891 \times 10^{-2}[/tex] g/ml

Therefore, we can conclude that solubility of X in water is [tex]1.891 \times 10^{-2}[/tex] g/ml.

Answer:

Numbers 4,3

Explanation:

Ionic bond is between nonmental and metals

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.

Explanation:

Alkali metal refers to group1 elements.

Alkali earth metal refers to group 2 elements.

Non metals refers to elements in grouos 4 to group 7.

Halogen refers to group 17 elements

Transition Metal refers to group 3 to group 12 elements

Noble gases refer to elements in group 18.

To obtain the group number from the electronic configuration, we calculate the total number of electrons in the principal quantum number (coefficient of the letters).

a. [Ne]3s1

Principal quantum number = 3

Number of electrons present = 1

This element belongs to group 1. It is an Alkali Metal.

b. [Ne]3s23p3

Principal quantum number = 3

Number of electrons present = 2 + 3 = 5

This element belongs to group 15 (5A). It is a Non metal

c. [Ar]4s23d104p5

Principal quantum number = 4

Number of electrons present = 2 + 5 = 7

This element belongs to group 17 (7A). It is an Halogen.

d. [Kr]5s24d1

This configuration belongs to the element yttrium and has an incomplete d shell. Hence it is a transition metal.

e. [Kr]5s24d105p6

Principal quantum number = 5

Number of electrons present = 2 + 6 = 8

This element belongs to group 18 (8A). It is a Noble gas.

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%

Compounds formed from non-metals consist of molecules. The atoms in a molecule are joined together by covalent bonds. These bonds form when atoms share pairs of electrons.

Complete Question

The diagram for this question is shown on the second uploaded image

Answer:

The organic product obtained is  shown  on the first uploaded image

Explanation:

The process that lead to this product formation is known as oxidative cleavage   which is a reaction that involves the cleavage of a carbon to carbon bond at the same time this carbon which formed the carbon bond are oxidized i.e oxygen is been added to them

Answer:

water stored in a dam

Explanation:

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

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 correct answer is 4.502 g per ml.

Explanation:

Based on the given question, the sum of the mass of mesitylene and barium together is 148.792 grams. The mass of barium given is 77.240 grams. Therefore, the mass of mesitylene will be,  

Mass of mesitylene = Total mass - Mass of barium

= 148.792 - 77.240

= 71.552 grams

The density of mesitylene is 0.86370 g per ml. To calculate the volume of mesitylene, the formula to be used is,  

Volume = mass / density. Now, putting the values we get,  

Volume = 71.552 / 0.86370 = 82.8436 ml.  

As the total volume is 100 ml, therefore, the volume of Ba will be,  

Volume of Ba = 100-82.8436 = 17.1564 ml

The density of Ba at 20 degree C can be calculated by using the formula,  

Density = mass / volume. Now putting the values we get,  

Density = 77.240 g / 17.1564 ml  

= 4.502 g per ml

Answer:

The concentration of the sample is 3.564x10⁻³M.

Explanation:

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

The general graph of the absorbance of the standards with different concentrations is:

Y = 75.9X + 0.0045

R² = 0.9946

Where Y is the absorbance of the sample and X its concentration in mole/L.

If a solution has an absorbance of 0.275:

0.275 = 75.9X + 0.0045

0.2705 = 75.9X

3.564x10⁻³M = X → The concentration of the sample.

Answer:

Amide

Explanation:

O=NH2 is the Amide group versus NH2, which is the amine group.

Answer:

Butamide

Explanation:

C3H7-C(=O)-NH2 IUPAC NAME

C4H9NO

     H   H   H

H - C - C - C - C = O

     H   H   H   N - H

                      H

But amide

Amide because R-CO-NH2 ie C(=O)-NH2

But because 4 Cabon

Answer:

[tex]Q=-3.11x10^5kJ[/tex]

Explanation:

Hello,

In this case, for the given reaction, we are given the standard enthalpy of reaction per mole of ammonia that is -92.6 kJ, it means, that forming one mole of ammonia will release 92.6 kJ of energy. In such a way, for the formation of 5.71x10⁴ g of ammonia, the following amount of heat will be released:

[tex]Q=5.71x10^4gNH_3*\frac{1molNH_3}{17gNH_3}*-92.6\frac{kJ}{molNH_3}\\ \\Q=-3.11x10^5kJ[/tex]

Best regards.

The amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ

We'll begin by calculating the number of mole in 5.71×10⁴ g of NH₃

Mass of NH₃ = 5.71×10⁴ g

Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mole = mass / molar mass

Mole of NH₃ = 5.71×10⁴ / 17

N₂(g) + 3H₂(g) —> 2NH₃(g) ΔH°rxn = −92.6 kJ

Since reaction took place at standard conditions, it means:

1 moles of NH₃ required −92.6 kJ

Therefore,

3358.82 moles of NH₃ will require = 3358.82 × –92.6 = –311026.732 KJ

Thus, the amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ

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