−791.4 kJ is the standard enthalpy change for the reaction.
Given,
SO2(g) → S(s) + O2(g)
ΔH° = +296.8 kJ
2SO2(g) + O2(g) → 2SO3(g)
ΔH° = −197.8 kJ
Modified equation:
S(s) + 2O2(g) → 2SO2(g)
ΔH° = −593.6 kJ, multiply by 2 and flip
2SO2(g) + O2(g) → 2SO3(g)
ΔH° = −197.8 kJ here no change
The 2SO2 will balance out by canceling it when the equations are added. Add the enthalpies for the final answer:
−593.6 + (−197.8) = −791.4 kJ
The change in enthalpy of a compound when one mole of the compound is generated from all components of the same constituent is known as the standard enthalpy of formation, also known as the standard heat of formation.
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Can I know the answer ？
According to the given statement Mass of one atom of gold in grams is 196.78 g.
What is a atom ?A chemical element is uniquely defined by its atoms, that are small bits of substance. An atom is made up of a core nucleus one and or more negatively charged electrons that orbit it. The positively charged, relatively hefty protons and neutrons that make it up the nucleus may be present.
Who discovered the atom?Up until the 1800s, this idea of tiny, indivisible pieces of matter remained. The contemporary atomic idea was really started by the eminent chemist John Dalton (1766–1844). But his atom was like a rock-solid ball.
Briefing:Molar mass is the weight in grams of one mole of an atom.
1 mole = 6.022×10²³ atoms
Molar mass of Gold (Au)
= mass of an atom × number of atoms
=3.207×10-²²×6.022×10²³
=196.78g.
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C2H4O2 IS THE EMPIRICAL FORMULA FOR GLUCOSE?
The empirical formula of the glucose molecule is written as [tex]CH_{2} O[/tex].
What is the empirical formula for glucose?Let us recall that glucose is the molecule that is response for the quick release of energy in the body. It is also the molecule that is first formed in the process of photosynthesis.
The term carbohydrate actually means a compound that is a hydrate of carbon and it is composed of glucose molecules. The empirical formula shows the relationship or the ratio of the atoms that could be found in a compound. We can see that we can obtain the ratio from the number of each of the atoms present.
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Acetone has a density of 0.7857 g/cm^3. What is the volume in mL of 5.52 g of acetone?
Let's see that the formula to find the volume using density and mass is:
[tex]V=\frac{m}{d}\begin{cases}V=\text{volume} \\ m=\text{mass} \\ d=\text{density}\end{cases},[/tex]In this case, the problem is asking about the volume in mL but remember that mL is the same that cm^3, so using the formula we're going to obtain:
[tex]V=\frac{5.52\text{ g}}{0.7857\text{ }\frac{g}{mL}}=7.02\text{ mL.}[/tex]The volume of 5.52 g of acetone is 7.02 mL.
How does temperature of land far from water compare to that of land near water?
The temperature of land far from water is hotter than land near water.
The temperature of land far from water is less stable compared to land near water.
The temperature is similar for land both near and far from water.
The temperature is either too hot or too cold near water.
The temperature of land far from water is hotter than land near water.
Why the temperature of land far away from water different from land near water?The lower heat capacity of land often allows them to cool the nearby water temperature so it takes less energy to change the temperature of land compared to water bodies. This means that land heats and cools more quickly as compared to water. This difference affects the climate of different areas on Earth. Large bodies of water like oceans, seas and large lakes can affect the climate of the nearby regions such as coastal regions. Water heats and cools more slowly than land regions. The coastal regions will stay cooler in summer season and warmer in winter season, which creates moderate climate on the coastal regions.
So we can conclude that the temperature of land far from water is different from land near water because of cooling effect of water bodies.
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how many moles of HCL are required to prepare 0.80L of a 0.5M HCL solution
Answer:
[tex]0.4\text{ mol}[/tex]Explanation:
Here, we want to get the number of moles
Mathematically:
[tex]Number\text{ of moles = molarity }\times\text{ volume}[/tex]We have that as:
[tex]Number\text{ of moles = 0.5 }\times\text{ 0.8 = 0.4 mol}[/tex]PLEASE HELP!
In a monomolecular reaction A->B , at t =250C, the initial concentration decrease at 25% in t =52 min. Calculate:
a) the constant rate;
b) the time after the initial concentration decrease with 75%;
c) the initial reaction rate, if the initial concentration of the reactant is 2.5 mol/L·s
1) The calculate rate constant is 9.22 * 10^-5 s-1
2) In the period of is 15033 s 75% is used up
3) From the calculation, there is an initial 9.22 * 10^-5 M.
How can we find reaction rate?In chemistry, the rate of reaction would give the idea that the reaction is proceeding quickly or slowly. If a reaction has a large rate of reaction then it tends to move on to completion.
From the question, it is clear that there is 25% in t =52 min.
Initial concentration [A]o = [A]o
Final concentration = [A]o - 0.25 [A]o = 0.75 [A]o
Time taken = 52 min or 3120 s
From the formula that can be applied to a first order reaction;
ln[A] = -kt + ln[A]0
k = -(ln[A] - ln[A]0)/t
k = - (ln0.75 [A]o/A]0)/3120
k = 9.22 * 10^-5 s-1
b)
Then we have to find the time after the initial concentration decrease with 75%
[A] = [A]o - 0.75 [A]o = 0.25 [A]o
ln[A] = -kt + ln[A]0
t = -(ln[A] - ln[A]0)/k
t = - (ln0.25 [A]o/A]0)/9.22 * 10^-5
t = 15033 s
c) Given that in this case, the initial concentration of the reactant is 2.5 mol/L·s
k = - (ln0.25 (2.5)/ln(2.5))/15033
k = 9.22 * 10^-5 M
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Which is an example of plasmas in nature?
Answer: nbhhvyyvub
Explanation:
What is the volume, in mL, of a solution that is 0.421 M K2SO3 and was prepared by dissolving 16.7 g of solid potassium sulfite in water?
Which of the following changes are chemical changes?
Answer:
flammability, toxicity etc..
Explanation:
are chemical changes
At equilibrium, a reaction vessel contains 4.50 atm of Br₂ and 1.10 atm of NBr₃. According to the reaction: 2 NBr₃ (g) ⇌ N₂ (g) + 3 Br₂ (g) Kp = 4.8
Determine the equilibrium partial pressure of N₂.
Equilibrium constant is the only concept that is to be used here to calculate partial pressure of nitrogen. The partial pressure of Nitrogen comes out to be 1.17atm
What is equilibrium constant?Equilibrium constant is a rate constant at equilibrium shows the values of reaction with respect to the atmospheric pressure and concentration
[tex]K_{c}[/tex] is equilibrium constant with respect to concentration.
[tex]K_{p}[/tex] is equilibrium constant with respect to atmospheric pressure.
Mathematically,
[tex]K_{p}=\frac{[N_{2}][Br_{2} ] }{[NBr_{3}] }[/tex]
Substituting values
(4.8×1.10atm)÷ 4.50atm=[tex]{[N_{2}][/tex]
1.17atm=[tex]{[N_{2}][/tex]
Thus the equilibrium partial pressure of N₂ is 1.17atm.
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How many possible combinations are there for the values of l and ml when n = 3?
There are three possible values for the l and there are nine possible values for ml.
What are quantum numbers?The term quantum numbers has to do with the description that shows the most probable position of the electron in an atom. We know that the orbital is a region in space, where there is a high probability of finding the electrons.
Given that the values of l must be in the range of 0 to n - l and the values of ml must be in the rang of -l to + l. we then can use this to know the number of the possible combinations of l and ml when n = 3.
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A 0.325g of KHP required 20.15ml of NaOH for neutralization. Calculate the molarity of NaOH
The molarity of NaOH solution that required 20.15mL is 0.0004032M.
How to calculate molarity?Molarity is the concentration of a substance in solution, expressed as the number moles of solute per litre of solution.
Molarity of a solution can be calculated by dividing the number of moles in the solution by the volume as follows:
Molarity = no of moles ÷ volume
According to this question, 0.325g of KHP required 20.15ml of NaOH for neutralization. The number of moles in this mass can be calculated as follows:
no of moles = 0.325g ÷ 40g/mol
no of moles = 0.008125mol
Molarity = 0.008125mol ÷ 20.15mL
Molarity = 0.0004032M
Therefore, 0.0004032M is the molarity of the solution.
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The average college student uses 320 pounds of paper per year. How many
ounces of paper does an average college student use per day?
1 pound = 16 ounces 1 year= 365 days
*
Based on the conversion factor, the number of ounces of paper an average college student uses per day is 14 ounces of paper per day
What is the conversion factor from pounds to ounces?A conversion factor is a value that is used to make the conversion from one unit of measurement to another.
The conversion factor from pounds to ounces is given below as follows:
1 pound = 16 ounces
Also, the conversion factor from years to days is given below as follows:
1 year = 365 days
Hence, the ounces of paper an average college student uses per day are calculated as follows;
320 pounds = 320 * 16 ounces
320 pounds = 5120 ounces
Ounces of paper per day = 5120 / 365
Ounces of paper per day = 14 ounces per day
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If 335g water at gains 102.3 J of heat, how much does the temperature of the water change? The specific heat of water is 4.184 J/g*C
Temperature is a measure of how hot a substance or radiation is expressed numerically.
There are three different types of temperature scales:
those that depend only on macroscopic properties and thermodynamic principles, like Kelvin's original definition;
those that depend on practical empirical properties of particles rather than theoretical principles;
and those that are defined by the average translational kinetic energy per freely moving microscopic particle, like an atom, molecule, or electron, in a body, like the SI scale.
How much heat is gained or lost by a sample can be calculated using the equation q = mcΔT, where m is the mass of the sample, c is the specific heat, and T is the temperature change (q).
Therefore,
q = m*c*ΔT
102.3 = 335 * 4.184 * ΔT
ΔT = 0.073 °c
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You have 0.834 moles of Potassium. The NIH recommends 2,600mg of potassium as a daily intake recommendation. Convert the moles to mg.
Answer:
0.834 mole = 13293.32 mg
2,600 mg = 0.16 mole
Explanation:
1 mole of K = 39.098 g
=> 0.34 mole has (0.34)(39.098) = 13.29332g
13.29332 g x 10^3 = 13293.32 mg
so 0.834 mole of K = 13293.32 mg
mole of 2,600mg = (0.834)(2,600)/(13293.32) = 0.16311952168 or 0.16 mole
Convert 0.05090Kg/mol to dg/mmol
Answer: .509
Explanation: Im pretty sure
Explain how you would calculate the total change in bond energy for the reaction H2+Cl2->2HCI. How would you know if the reaction was endothermic or exothermic?
To calculate the total change in binding energy we must apply the following equation:
[tex]\Delta H=\Delta H_{f(\text{Reagents)}}-=\Delta H_{f(\text{Products)}}[/tex]This equation tells us that the energy change will be equal to the sum of the potential energy of the product bonds minus the sum of the potential energy of the reagent bonds.
Now, let's calculate each term separately.
[tex]\begin{gathered} \Delta H_{f(\text{Reagents)}}=n_{H2}\times\text{Energy bond H-H + }n_{Cl2}\text{Energy bond Cl-Cl} \\ \Delta H_{f(\text{Reagents)}}=1molH_2\times432\frac{kJ}{mol}+1molCl_2\times239\frac{kJ}{mol} \\ \Delta H_{f(\text{Reagents)}}=671kJ \end{gathered}[/tex][tex]\begin{gathered} \Delta H_{f(\text{Products)}}=n_{\text{HCl}}\times EnergyBond\text{ H-Cl} \\ \Delta H_{f(\text{Products)}}=2molHCl\times427\frac{kJ}{mol} \\ \Delta H_{f(\text{Products)}}=854kJ \end{gathered}[/tex]So, the change in bond energy will be:
[tex]\begin{gathered} \Delta H=\Delta H_{f(\text{Reagents)}}-=\Delta H_{f(\text{Products)}} \\ \Delta H=671kJ-854kJ=-183kJ \end{gathered}[/tex]We have a negative value in the result, when this happens it means that the reaction is exothermic, that is to say, that it releases heat and the energy of the products is greater than that of the reagents.
When we have a positive value the reaction will be endothermic, this means that it needs energy.
on a solid mixture, X. The inferences made are recorded in Table 2. Complete Table 2 by filling in the observations based on the inferences made. TABLE 2: TESTS ON MIXTURE X Test Observation Inferences a) Distilled water was added to a portion of X and the resulting mixture stirred and filtered. (The residue was set aside for use later.) The filtrate was divided into 3 equal portions and tests (b) to (d) done on separate portions. Cl- ions are present. b) Dilute nitric acid followed by a few drops of silver nitrate solution was added. Ammonium hydroxide solution was added to the resulting mixture. (3 marks)
Answer:
a) Upon adding silver nitrate, a white precipitate is observed
b) Upon adding ammonium hydroxide, the white precipitate dissolves to give a clear, colorless solution
Explanation:
Here, we want to state the observations when testing for chloride ions
From what we have:
a) When silver nitrate is added, a white precipitate is formed
This is as a result of the following chemical reaction:
[tex]Ag\placeholder{⬚}_{(aq)}^+\text{ + Cl}_{(aq)}^-\text{ }\rightarrow\text{ AgCl}_{(s)}[/tex]The AgCl is the white precipitate formed
b) Upon the addition of the ammonium hydroxide solution, a colorless and clear solution is observed showing that the white precipitate has dissolved
Find the volume in liters of the 0.505 molar NaOH solution needed to react with 40 milliliters of the 0.505 molar H2SO4 solution.
A//: 80 mililiters
Please, it's urgent, I need it ASAP
The volume of NaOH required is 80ml.
Sulfuric acid is a dibasic acid and NaOH is monoacidic base.
M1V1 = M2V2
where, M1 = initial concentration,
V1 = initial volume,
M2 = concentration after mixing
V2 = total final volume.
This formula is used for calculating the final volume or molarity after mixing two solutions.
Given the question,
M1 = 0.505M
V1 =?
M2 = 0.505M
V2 = 40 ml
M1V1 = 2×M2V2
0.505 × V2 = 2 × 0.505 × 40
V2 = 2×0.505×40
0.505
V2 = 80ml
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How many grams of Br are in 395 g CaBr2 ?
In order to answer this question we will need to use the molar mass of Calcium bromide (CaBr2), which is 199.89 g/mol, now we can find how much of this mass is only Bromide, so what is the percentage of Bromide that makes up this whole compound, to do so we will also use the molar mass of Bromide (Br2) which is 159.8 g/mol
199.89 g/mol = 100% of the compound
159.8 g/mol = x %
x = roughly 80% (is like 79.9%)
So, regardless of the mass of CaBr2, bromide needs to be 80% of its mass, now solving our question
395 g = 100%
x grams = 80%
x = 316 grams is the mass of Bromide in the compound
What is made of two or more substances that are together in the same place but are not chemically combined
When two or more substances are together in the same place but are not chemically combined they form a mixture.
What is a mixture?The word mixture is used in chemistry to denote a combination of two or more elements that conserve their original properties, which is fundamental to achieving the separation of such elements in the future.
Therefore, with this data, we can see that a mixture is a mix or combination of two or more elements that still conserve their physical and chemical properties, thereby allowing their separation through different techniques.
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In the following equation, how many grams of O2 are needed to react with 24.0 g of NH3?
4 NH3(g)+302(g) → 2N2(g)+6H₂O(1)
a) 45.1
b) 22.6
c) 33.9
d) 45.0
e) 135.5
33.6 g grams of O2 are needed to react with 24.0 g of NH3.
The molecular mass of NH3 = 14 + 3 = 17g
Given mass of NH3 = 24g
Firstly, we will calculate the number of moles.
Moles is defined as the ratio of given mass of substance to the molecular mass of substance.
Moles = given mass/ molecular mass
Number of moles of NH3 = 24/17
= 1.4 mole
Chemical reaction4 NH3(g)+302(g) → 2N2(g)+6H₂O(1)
4 moles of NH3 require 3 moles of O₂ to react.
1 moles of NH3 require 3/4 moles of O₂ to react.
1.4 moles of NH3 require 1.05 moles of O₂ to react.
Now we calculate the grams of O₂.
1.05 × 32 = 33.6 g.
Thus, we concluded that the 33.6 g grams of O2 are needed to react with 24.0 g of NH3.
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can atoms of different elements have the same atomic number?
No, Atoms of different element can not have same atomic number because only same type of atoms combine to form element. Atoms belonging to different element can have different atomic number.
What is element?Element generally consist of atoms or we can atoms combine to form element. Atoms of an element is always same, means all the properties of all atoms of one type of element is same. Two or more than two atoms with different physical or chemical properties can not combine together to form an element.
So we can say that atomic number of all atoms that is constituting an element is same. It can never be different.
Thus Atoms of different element can not have same atomic number.
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How should the electrode of a pH meter be preserved? Explain your answer.
We can care for the pH meter electrode by;
1) Making sure that the electrode remains moist
2) The electrode should be stored in a 4M solution of KCl
3) Electrodes should not come in contact with deionized water.
What is pH?The term pH has to do with the negative logarithm of the hydrogen ion concentration. Now we know that the pH of a solution tells us the amount of the hydrogen ions or the hydroxyl ions that is present in the solution. The pH mete is the instrument that we could use to be able to measure the pH of the solution.
The pH scale runs between 0 - 14. The points on the scale that have been labeled from 0 - 6 tells us that the solution is an acidic solution and contains more hydrogen ions. If the solution has a pH of 7, then it is neutral and contains equal concentration of hydrogen and hydroxyl ions. A solution of pH 8 - 14 is basic and contains more hydroxyl than hydrogen ions.
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Calculate the number of mol corresponding to 18.8 g Na2SO4.
Answer: 0.132 moles of Na2SO4
Explanation:
Multiply the grams of Na2SO4 by the 1/molar mass of Na2SO4, which is the sum of all the elements 2(Na)+S+4(O)= 142.04.
Below is the word equation for the reaction between sodium hydroxide and hydrochloric acid. Identify the reactant(s) in the reaction. Sodium hydroxide + Hydrochloric acid → Sodium chloride + Water
I a chemical reaction, we have the reactants in the left part,followed by an arrow and then the products.
So, the reactants are the two that are in the left part of the equation:
Sodium hydroxide and Hydrochloric acid
Scientists are looking for ways to help increase the amount of calcium carbonate in oceans so that coral reefs can be healthy ecosystems again. Why is the amount of calcium carbonate in the ocean decreasing? (1 point)
Responses
Increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.
Decreased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.
Increased amounts of carbon dioxide cause reactions to happen in the water that increase the pH and decrease the amount of calcium carbonate.
Decreased amounts of carbon dioxide cause reactions to happen in the water that increase the pH and decrease the amount of calcium carbonate.
Scientists are looking for ways to help increase the amount of calcium carbonate in oceans so that coral reefs can be healthy ecosystems again mount of calcium carbonate in the ocean decreasing because increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate
Calcium carbonate is a compound CaCO₃ found in nature as calcite and aragonite and in plant ashes, bones and shells and used especially in making lime and portland cement and as a gastric antacid and as ocean acidification increases available carbonate ion bond with excess hydrogen resulting in fewer carbonate ion available for calcifying organisms to build and maintain their shells, skeletons, and other calcium carbonate structures
Ocean acidification describe the lowering of sweater pH and carbonate saturation that result from increasing atmospheric CO₂ concentration and that's why increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.
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Answer: Its C I did my research and that's what makes sense to me. because increased amount of carbon dioxide decreases the PH and the amount of calcium carbonate.
Following the Experiment 3 procedure, you combine 50.0 mL H2O, 50.0 mL of 2.2 M HCl(aq), and 4.039 g NaOH(s) (molar mass of NaOH = 40.00 g/mol) in a Styrofoam calorimeter, recording a maximum solution temperature change of 22.9 oC and a final solution mass of 104.153 g (csolution = 4.184 J/g.oC). Calculate the molar change in enthalpy of reaction
The molar change in enthalpy of reaction is 99.6 kJ/mol.
the formula for the specific heat capacity expressed as :
Q = mcΔT
where,
m , mass = 104.153 g
c, specific heat = 4.184 J/g °C
dt, change in temperature = 22.9 °C
Q = 104.153 × 4.184 × 22.9
Q = 9969.7 J = 9.96 kJ
now, the molar change in enthalpy is give by:
Q = ΔH / n
n is no. of moles
morality of HCl = 2.2 M
v = 50 mL = 0.05 L
n = 0.05 × 2.2
= 0.11 moles
no. of moles of NaOH = mass / molar mass
= 4.039 / 40
= 0.100
NaOH is limiting reactant.
using the formula we get:
Q = ΔH / n
ΔH = Q / n
ΔH = 9.96 kJ/ 0.10
ΔH, change in enthalpy = 99.6 kJ/mol
Thus, combine 50.0 mL H₂O, 50.0 mL of 2.2 M HCl(aq), and 4.039 g NaOH(s) (molar mass of NaOH = 40.00 g/mol) in a Styrofoam calorimeter, recording a maximum solution temperature change of 22.9 °C and a final solution mass of 104.153 g (c solution = 4.184 J/g°C). the molar change in enthalpy of reaction is 99.6 kJ/mol.
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Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). if 2.46 g of carbon dioxide is produced from the reaction of 2.71 g of ethane and 16.7 g of oxygen gas, calculate the percent yield of carbon dioxide. Round to 3 sig figs
Taking into account definition of percent yield, the percent yield of carbon dioxide is 30.90%
Reaction stoichiometryIn first place, the balanced reaction is:
2 CH₃CH₃ + 7 O₂ → 4 CO₂ + 6 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
CH₃CH₃: 2 molesO₂: 7 molesCO₂: 4 molesH₂O: 6 molesThe molar mass of the compounds is:
CH₃CH₃: 30 g/moleO₂: 32 g/moleCO₂: 44 g/moleH₂O: 18 g/moleThen, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
CH₃CH₃: 2 moles ×30 g/mole= 60 gramsO₂: 7 moles ×32 g/mole= 224 gramsCO₂: 4 moles ×44 g/mole= 176 gramsH₂O: 6 moles ×18 g/mole= 108 gramsLimiting reagentThe limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
Limiting reagent in this caseTo determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 224 grams of O₂ reacts with 60 grams of CH₃CH₃, 16.7 grams of O₂ reacts with how much mass of CH₃CH₃?
mass of CH₃CH₃= (16.7 grams of O₂× 60 grams of CH₃CH₃)÷ 224 grams of O₂
mass of CH₃CH₃= 4.47 grams
But 4.47 grams of CH₃CH₃ are not available, 2.71 grams are available. Since you have less mass than you need to react with 16.7 grams of O₂, CH₃CH₃ will be the limiting reagent.
Percent yieldThe percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.
The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
percent yield= (actual yield÷ theoretical yield)×100%
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
Theoretical yield of CO₂Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 60 grams of CH₃CH₃ form 176 grams of CO₂, 2.71 grams of CH₃CH₃ form how much mass of CO₂?
mass of CO₂= (2.71 grams of CH₃CH₃× 176 grams of CO₂)÷ 60 grams of CH₃CH₃
mass of CO₂= 7.96 grams
Then, the theoretical yield of CO₂ is 7.96 grams.
Percent yield for the reaction in this caseIn this case, you know:
actual yield= 2.46 gramstheorical yield= 7.96 gramsReplacing in the definition of percent yields:
percent yield= (2.46 grams÷ 7.96 grams)×100%
Solving:
percent yield= 30.90%
Finally, the percent yield for the reaction is 30.90%.
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What is the oxidation number or Br in NaBrO?