A galvanic cell consists of a Cu(s)|Cu2+(aq) half-cell and a Cd(s)|Cd2+(aq) half-cell connected by a salt bridge. Oxidation occurs in the cadmium half-cell. The cell can be represented in standard notation as
Answer:
[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]
Explanation:
A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.
The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.
As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.
[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]
We discussed the different types of intermolecular forces in this lesson, which can affect the boiling point of a substance.
1. Which of these has the highest boiling point?
A) Ar
B) Kr
C) Xe
D) Ne
2. Which substance has the highest boiling point?
A) CH4
B) He
C) HF
D) Cl2
Answer:
1, C, Xe 2, B,He
Explanation:
1, cause as u go down a group the boiling point increases.
2, boiling point of single element is greater than a compound
According to periodic trends in periodic table boiling point increases down the group and hence Xe has highest boiling point and more amount of heat is required to boil an element hence He has highest boiling point.
What is periodic table?
Periodic table is a tabular arrangement of elements in the form of a table. In the periodic table, elements are arranged according to the modern periodic law which states that the properties of elements are a periodic function of their atomic numbers.
It is called as periodic because properties repeat after regular intervals of atomic numbers . It is a tabular arrangement consisting of seven horizontal rows called periods and eighteen vertical columns called groups.
Elements present in the same group have same number of valence electrons and hence have similar properties while elements present in the same period show gradual variation in properties due to addition of one electron for each successive element in a period.
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Calculate the number of hydrogen atoms in a 110.0 sample of tetraborane(B4H10) . Be sure your answer has a unit symbol if necessary, and round it to 4 significant digits.
Answer:
[tex]1.242 \times 10^{25}\text{ atoms H}[/tex]
Explanation:
You must convert the mass of B₄H₁₀ to moles of B₄H₁₀, then to molecules of B₄H₁₀, and finally to atoms of H.
1. Moles of B₄H₁₀
[tex]\text{Moles of B$_{4}$H}_{10} = \text{110.0 g B$_{4}$H}_{10} \times \dfrac{\text{1 mol B$_{4}$H}_{10}}{\text{53.32 g B$_{4}$H}_{10}} = \text{2.063 mol B$_{4}$H}_{10}[/tex]
2. Molecules of B₄H₁₀
[tex]\text{No. of molecules} = \text{2.063 mol B$_{4}$H}_{10} \times \dfrac{6.022 \times 10^{23}\text{ molecules B$_{4}$H}_{10}}{\text{1 mol B$_{4}$H}_{10}}\\\\=1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10}[/tex]
3. Atoms of H
[tex]\text{Atoms of H} = 1.242 \times 10^{24}\text{ molecules B$_{4}$H}_{10} \times \dfrac{\text{10 atoms H}}{\text{1 molecule B$_{4}$H}_{10}}\\\\= \mathbf{1.242 \times 10^{25}}\textbf{ atoms H}[/tex]
An atom of 108Te has a mass of 107.929550 amu. Calculate the binding energy per MOLE in kJ. Use the values: mass of 1H atom
Answer:
The binding energy = 8.64972649×10¹⁰ kJ/mole
Explanation:
Given that:
An atom of 108Te has a mass of 107.929550 amu.
In a 108 Te atom, there are 52 protons and 56 neutrons
where;
mass of proton= 1.007825 amu
mass of neutron= 1.008665 amu
Similarly; The atomic number of Te = 52
the mass of 52 protons = 52 × 1.007825 amu
the mass of 52 protons = 52.4069 amu
the mass of 56 neutrons = 56 × 1.008665 amu
the mass of 56 neutrons = 56.48524 amu
The total mass can now be = the mass of 52 protons + the mass of 56 neutrons
The total mass = 52.4069 amu + 56.48524 amu
The total mass = 108.89214 amu
Recall : it is given that An atom of 108Te has a mass of 107.929550 amu.
Therefore, the mass defect will be = 108.89214 amu - 107.929550 amu
the mass defect = 0.96259amu
where 1 amu = 1.66× 10⁻²⁷ kg
Therefore, 0.96259amu = (0.96259 × 1.66× 10⁻²⁷) kg
= 1.5978994 × 10⁻²⁷kg
The binding energy = mass defect × (speed of light)²
where;
speed of light c = 2.99792 × 10⁸ m/s
The binding energy = 1.5978994 × 10⁻²⁷kg × 2.99792 × 10⁸ m/s
The binding energy = 1.43611597 × 10⁻¹⁰ J
The binding energy = 1.43611597 × 10⁻¹³ kJ/atom
since 1 mole = 6.023 × 10²³ atom (avogadro's constant)
Then;
The binding energy = ( 1.43611597 × 10⁻¹³ )× (6.023 × 10²³) kJ/mole
The binding energy = 8.64972649×10¹⁰ kJ/mole
A simplified version of photosynthesis can be represented as carbon dioxide combining with water to form glucose and oxygen: 6CO2+6H20 C6H12O6+6O2 In this reaction, ________ is oxidized. 1.Carbon dioxide 2.Hydrogen 3.Carbon 4.Oxygen
Answer:
2, hydrogen
Explanation:
i think
Answer:
Answer is not hydrogen
Explanation:
did the test and got it wrong
1.46 mol sample of nitrogen gas at a temperature of 14.0 °C is found to occupy a volume of 23.7 liters. The pressure of this gas sample is mm Hg
Explanation:
Since there are no gas constant with the unit of mmHg, you just have to use the gas constant for atm, and then convert that into mmHg
What is the oxidizing agent in the redox reaction represented by the following cell notation? Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
Answer:
Silver.
Explanation:
Hello,
In this case, for the redox reaction:
[tex]Ni^0(s)+Ag^+(aq)\rightarrow Ni^{2+}+Ag^0(s)[/tex]
We can see the nickel is being oxidized as its oxidation state increases from 0 to 2+ whereas the oxidation state of silver decreases from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.
Best regards.
The oxidizing agent in the redox reaction represented by the following cell notation is Silver.
Calculation of the oxidizing agent:The redox reaction is
Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
here the nickel is being oxidized since its oxidation state rises from 0 to 2+ while on the other hand, the oxidation state of silver is reduced from +1 to 0, it means that the oxidizing agent is silver and the reducing agent is nickel.
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A student completed the experiment but found that the total amount of material recovered weighed more than the original sample. What is the most likely source of error and how may it be corrected?
Answer:
This is due to the water moisture present in the recovered sample.
Explanation:
The total amount of material recovered isn’t meant to weigh more than the original sample. However when this happens then it means there is the presence of water moisture in the recovered sample.
The recovered samples however needs to be heated to make it dry and eliminate the water moisture through evaporation.
Which ONE of these cations has the same number of unpaired electrons as Fe2+ ? A) Ni2+ B) Fe3+ C) Cr2+ D) Mn2+ E) Co2+
Answer:
Explanation:
Fe2+ Has 4 unpaired electrons.
By method of elimination;
Option A: Ni2+ has two unpaired electrons. so this option is wrong.
Option B: There are 5 unpaired electrons in the Fe3+ ion. so this option is wrong.
Option C: There are 4 unpaired electrons in the Cr2+ ion. so this option is correct.
Option D: There are 5 unpaired electrons in the Mn2+ ion. so this option is wrong.
Option E: There are 3 unpaired electrons in the Co2+ ion. so this option is wrong.
Which one of the following compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide?
a. butanal
b. 2-methylbutanal
c. 3-methylpentanal
d. 2, 2-dimethylbutanal
The compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide is 2, 2-dimethylbutanal.
What is aldol reaction?The Aldol Reaction occurs when the enolate of an aldehyde or ketone combines with the carbonyl of another molecule at the aplha-carbon under basic or acidic circumstances to produce beta-hydroxy aldehyde or ketone.
For the formation of enolate ion, compound should contain alpha hydrogen in it and among the given compound only 2, 2-dimethylbutanal will not have alpha hydrogens.
Butanal, 2-methylbutanal and 3-methylpentanal will have aplha hydrogens in it so that it takes part in the aldol reaction.Hence 2, 2-dimethylbutanal does not undergo an aldol addition reaction.
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Describe how you would prepare 500ml of 40% (w/v) aqueous iodine solution.
[Atomic mass of iodine =127g/mol].
Answer:
- Weight 333.3 grams of iodine.
- Measure 500 mL of water.
- Vigorously mix the resulting solution.
Explanation:
Hello,
In this case, since 500 mL of a 40% (w/v) aqueous solution iodine is required, we can compute the required mass of iodine by defining the given mass-volume percent:
[tex]\% w/v=\frac{m_{iodine}}{m_{solution}} *100%=\frac{m_{iodine}}{m_{water}+m_{iodine}} *100%[/tex]
In such a way, we need to find mass of iodine, which is computed as:
[tex]m_{iodine}=\frac{\%w/v*m_{water}}{100w/v-\%} \\\\m_{iodine}=\frac{40*500}{100-40}\\ \\m_{iodine}=333.3g\\[/tex]
Thereby, the procedure will be:
- Weight 333.3 grams of iodine.
- Measure 500 mL of water.
- Vigorously mix the resulting solution.
Best regards.
Which of the following is required for the flow of current in all systems?
a) the presence of ions
b) an electrical potential ofo
c) a closed circuit
d) a short circuit
Answer:
I would say c) a closed circuit.
Hope I was right.
Automotive air bags inflate when sodium azide decomposes explosively to its constituent elements: 2NaN3 (s) → 2Na (s) + 3N2 (g) How many grams of sodium azide are required to produce 30.5 g of nitroge
Answer:
NaN3 = 47.2 g
Explanation:
Given:
2 NaN3 ⇒ 2 Na + 3 N2
Find:
Amount of NaN3
Computation:
N2 moles = Product of N2 / molar mass of N2
N2 moles =30.5/28
N2 moles = 1.0893
2NaN3 makes 3(N2 )
So,
NaN3 moles = (2/3) moles of N2
NaN3 moles = ( 2/3) × 1.0893
NaN3 moles = = 0.7262
NaN3 mass = 0.7262 x 65
NaN3 = 47.2 g
Answer:
NaN3 = 47.2 g
Explanation:
Given:
2 NaN3 ⇒ 2 Na + 3 N2
Find:
Amount of NaN3
Computation:
N2 moles = Product of N2 / molar mass of N2
N2 moles =30.5/28
N2 moles = 1.0893
2NaN3 makes 3(N2 )
So,
NaN3 moles = (2/3) moles of N2
NaN3 moles = ( 2/3) × 1.0893
NaN3 moles = = 0.7262
NaN3 mass = 0.7262 x 65
NaN3 = 47.2 g
Explanation:
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed? Explain how you solved for your answers.
Balanced Equation: 2NH3 + 3CuO → 3Cu + N2 + 3H2O
I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.
After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.
A chemical reaction that has the general formula of nA → (A)n is best classified as a ____ reaction. A. synthesis B. polymerization C. decomposition D. oxidation E. replacement
Answer:
B.
Explanation:
A chemical reaction that has the general formula of nA → (A)n is best classified as a polymerization reaction.
Answer:
B. Polymerization
Explanation:
I'm just smart
Convert 59800 kilograms to pounds
Answer:
131836.43 pounds
Explanation:
one kilogram is 2.20462 pounds. multiply 2.20462 by 59800
Answer: 131836.43
Formula: Multiply the mass value by 2.205
59800×2.205=131836.43
Match the tools with the advantages they offer to astronomers.
photography
space telescope
radio telescope
optical telescope
w
detects electromagnetic frequencies outside
the visible spectrum that reaches Earth
captures images that can be shared and
compared by scientists
obtains a magnified and clear view of a part
of the sky to observe celestial objects
allows access to images taken from outside
Earth's atmosphere
Answer:
- photography (captures images that can be shared and
compared by scientists)
- space telescope (allows access to images taken from outside
Earth's atmosphere)
- radio telescope (detects electromagnetic frequencies outside
the visible spectrum that reaches Earth)
- optical telescope (obtains a magnified and clear view of a part
of the sky to observe celestial objects)
Explanation:
Photography is used to capture still images based on the principle that some compounds react in the presence of optical energy.
Space telescope is a type of observatory telescope positioned in outer space to observe distant planets, galaxies and other astronomical objects. Space telescopes reduces the interference from ultraviolet frequencies, X-rays and gamma rays; as well as light pollution which ground-based observatories encounter.
A radio telescope is a specialized antenna and radio receiver used to receive radio waves from astronomical radio sources in the sky. Radio telescope is used to study radio frequencies emitted by astronomical objects, that fall outside the visible light spectrum.
An optical telescope is used to gather and focuses light, from a far distant object. Optical telescope is used within the visible light spectrum of the electromagnetic spectrum, to create a magnified image, for direct view, or to make a photograph, or to collect data through electronic image sensors.
Answer:
- photography (captures images that can be shared and
compared by scientists)
- space telescope (allows access to images taken from outside
Earth's atmosphere)
- radio telescope (detects electromagnetic frequencies outside
the visible spectrum that reaches Earth)
- optical telescope (obtains a magnified and clear view of a part
of the sky to observe celestial objects)
Explanation:
Calculate the molar hydrogen ion concentration of each of the following biological solutions given the pH:
(a) gastric juice, pH= 1.80
(b) urine, pH 4.75 56.
Answer: A
Explanation: Calculate the molar hydrogen ion concentration of each of the following biological solutions given the pH, Urine pH= 4.90
g . Calculate the molar concentration for each of the following solutions. (a) 1.50 g NaCl in 100.0 mL of solution (b) 1.50 g K2Cr2O7 in 100.0 mL of solution (c) 5.55 g CaCl2 in 125 mL of solution (d) 5.55 g Na2SO4 in 125 mL of solution
Answer:
(a) [tex]M=0.257M[/tex]
(b) [tex]M=0.0510M[/tex]
(c) [tex]M =0.500M[/tex]
(d) [tex]M= 0.391M[/tex]
Explanation:
Hello,
In this case, since the molarity or molar concentration of a solution is computed by dividing the moles of solute by the volume of solution in liters, we proceed as follows:
(a) The molar mass of sodium chloride is 58.45 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:
[tex]M=\frac{1.50gNaCl}{0.100L} *\frac{1molNaCl}{58.45gNaCl} =0.257M[/tex]
(b) The molar of potassium dichromate is 294.2 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:
[tex]M=\frac{1.50gK_2Cr_2O_7}{0.100L} *\frac{1molK_2Cr_2O_7}{294.2gK_2Cr_2O_7} =0.0510M[/tex]
(c) The molar of calcium chloride is 111 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:
[tex]M=\frac{5.55gCaCl_2}{0.100L} *\frac{1molCaCl_2}{111gCaCl_2} =0.500M[/tex]
(d) The molar of sodium sulfate is 142 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:
[tex]M=\frac{5.55gNa_2SO_4}{0.100L} *\frac{1molNa_2SO_4}{142gNa_2SO_4} = 0.391M[/tex]
Best regards.
Which one of these four atoms has the most neutrons?
O A.
79
Se
34
B.
80
Br
35
OC.
73
Ge
32
84
Kr
36
O D.
Answer:
kr
Explanation:
it is radioactive so it loss electron due to which it jas most neutons
A quantity of liquid methanol, CH 3OH, is introduced into a rigid 3.00-L vessel, the vessel is sealed, and the temperature is raised to 500K. At this temperature, the methanol vaporizes and decomposes according to the reaction CH 3OH(g) CO(g) + 2 H 2(g), K c= 6.90×10 –2. If the concentration of H 2 in the equilibrium mixture is 0.426M, what mass of methanol was initially introduced into the vessel?
Answer:
74.3g of methanol were introduced into the vessel
Explanation:
In the equilibrium:
CH₃OH(g) ⇄ CO(g) + 2H₂(g)
Kc is defined as the ratio between concentrations in equilibrium of :
Kc = 6.90x10⁻² = [CO] [H₂]² / [CH₃OH]
Some methanol added to the vessel will react producing H₂ and CO. And equilibrium concentrations must be:
[CH₃OH] = ? - X
[CO] = X
[H₂] = 2X
Where ? is the initial concentration of methanol
As [H₂] = 2X = 0.426M; X = 0.213M
[CH₃OH] = ? - 0.213M
[CO] = 0.213M
[H₂] = 0.426M
Replacing in Kc to solve equilibrium concentration of methanol:
6.90x10⁻² = [0.213] [0.426]² / [CH₃OH]
[CH₃OH] = 0.560
As:
[CH₃OH] = ? - 0.213M = 0.560M
? = 0.773M
0.7733M was the initial concentration of methanol. As volume of vessel is 3.00L, moles of methanol are:
3.00L * (0.773 mol / L) = 2.319 moles methanol.
Using molar mass of methanol (32.04g/mol), initial mass of methanol added was:
2.319 moles * (32.04g / mol) =
74.3g of methanol were introduced into the vesselIf 2.9g of water is heated from 23.9C to 98.9C, how much heat (in calories) was added to the water?
Answer:
Explanation:
we know that
ΔH=m C ΔT
where ΔH is the change in enthalpy (j)
m is the mass of the given substance which is water in this case
ΔT IS the change in temperature and c is the specific heat constant
we know that given mass=2.9 g
ΔT=T2-T1 =98.9 °C-23.9°C=75°C
specific heat constant for water is 4.18 j/g°C
therefore ΔH=2.9 g*4.18 j/g°C*75°C
ΔH=909.15 j
Calculate the mass of naphthalene required to react stoichiometrically with the moles of 2-bromo-2-methylpropane that you have placed in the flask. Dispense that quantity of naphthalene into the weighing dish. There is some imprecision in delivering the powder onto the weigh boat, but a mass within 0.040 g of the calculated stoichiometric quantity will be sufficient for this experiment. Note that you can use a combination of the 1 g and the 0.1 g buttons to add the required mass. If you overshoot the goal, you can discard the weighing boat and get a new one. RECord the sample mass dispensed below. mass of naphthalene (g)
The reaction of 2-bromo-2-methylpropane and naphthalene is stoichiometrically in a mole ratio of 1:1.
Notice that in the question, the moles of 2-bromo-2-methylpropane placed in the flask wasn't mentioned. if I assume that it was 1.24 moles then;
1 mole of naphthalene reacts with 1 mole of 2-bromo-2-methylpropane
x moles of naphthalene reacts with 1.24 mole of 2-bromo-2-methylpropane
x = 1 * 1.24/1
= 1.24 moles of naphthalene
Molar mass of naphthalene = 128.2 g/mol
Mass of naphthalene = 128.2 g/mol * 1.24 moles of naphthalene
Mass of naphthalene =159 g
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2NH3 → N2 + 3H2 If 2.22 moles of ammonia (NH3) decomposes according to the reaction shown, how many moles of hydrogen (H2) are formed? A) 2.22 moles of H2 B) 1.11 moles of H2 C) 3.33 moles of H2 D) 6.66 moles of H2
Answer:
C
Explanation:
According to the mole ratio, using 2NH3 will give you 3H2. Which means in order to find the moles of H2 you would only need to divide 2 and multiply 3 to get the amount of moles of H2 produced.
Answer:
I think it's C
Explanation:
Please, tell me if I'm incorrect.
How to prepared sodium chloride solution in the laboratory.
hope it will help you
.
.
.
hope it will helps
Explain why only the lone pairs on the central atom are taken into consideration when predicting molecular shape
Answer:
Lone pairs cause more repulsion than bond pairs
Explanation:
A lone pair takes up more space around the central atom than bond pairs of electrons. This is because, a lone pair is attracted to only one nucleus while bond pairs are attracted to two nuclei.
Hence the repulsion between lone pairs is far greater than the repulsion between bond pairs or repulsion between a lone pair and a bond pair. The presence of a lone pair therefore distorts a molecule away from the ideal shape predicted on the basis of the valence shell electron pair repulsion theory.
Lone pairs are found to decrease the observed bond angles in a molecule.
Compare and contrast the quantum mechanical model and Niels Bohr of the atom
Answer:
the quantum mechanical model VS. Neils Bohr of the atom
Explanation:
The Bohr model and quantum model are models that explain the structure of an atom. ... The key difference between Bohr and quantum model is that Bohr model states that electrons behave as particles whereas quantum model explains that the electron has both particle and wave behavior.
What happens to the electropositive character of elements on moving from left to right along a period in a periodic table?
Answer:
On moving from left to right in a period the electropositive character decreases as the tendency to lose electrons decreases.
Explanation:
How are Math, Physics, Chemistry, and Biology all related?
Answer:
- you have to do maths in all 3
- atoms make up everything even parts of a cel and theyre studied in chem and physics
- chemistry is used in biology by finding out what different substances are eg cytoplasm in a cell
Calculate the enthalpy change (∆H) for the reaction- N2(g) + 3 F2(g) –––> 2 NF3(g) given the following bond enthalpies: N≡N 945 kJ/mol F–F 155 kJ/mol N–F 283 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Explanation: