To achieve complete combustion of one mole of butane (C₄H₁₀), 6.5 moles of oxygen (O₂) are necessary, as indicated by the balanced chemical equation for combustion.
To determine the number of moles of oxygen (O₂) required for the complete combustion of one mole of butane (C₄H₁₀), we must first write the balanced chemical equation for the combustion reaction:
C₄H₁₀ + O₂ → CO₂ + H₂O
After balancing the equation, we have:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
This balanced equation tells us that for every 2 moles of butane, we need 13 moles of oxygen for complete combustion, which means that for every mole of butane, we need 6.5 moles of oxygen.
Why water pollution is a global problem
Answer:
Pollutants are typically the cause of major water quality degradation around the world. Globally, the most prevalent water quality problem is eutrophication, a result of high-nutrient loads (mainly phosphorus and nitrogen), which substantially impairs beneficial uses of water.
Explanation:
Answer:
Water pollution is caused by industrial waste, temperature rising, deforestation and many other factors. All these facts are affecting water and they happen all over the world in many large bodies of water. Thus, it is a global issue,
Explanation:
I explained it in the answer loll.
Silicones can be oils or rubber-like materials depending on ________. Select one: A. the silicon-to-oxygen ratio B. the length of the chain and degree of cross-linking C. the percentage of carbon in the chain D. the oxidation state of silicon in the chain E. the percentage of sulfur in the chain
Answer: B-- the length of the chain and degree of cross-linking.
Explanation: Silicones can be oils or rubber-like materials depending on ___the length of the chain and degree of cross-linking.____
Silicone rubber is an elastomer which is composed of silicone containing silicon together with carbon, hydrogen, and oxygen. silicone rubber can be found in a wide variety of products because of its ability to withstand very high temperatures and pressures. eg cooking and food storage products; clothing apparels , automotive devices etc while
Silicone oil is a linear chain siloxane repeating units Si--0 with radical side group as such as methyl, phenyl etc bonded to it. They are oils because of they are viscose and have ability to repel. Silicone oils are mostly employed in medicine for surgical tools.
175 mal of Cl2 gas is held in a flexible vessel at STP. If the vessel is transported to the bottom of the impact basin Hellas planitia on the surface of Mars where is 1.16 kPa and the temperature is -5.0ºC. What is the new volume of Cl2 gas in liters
Answer: it’s 15.0 L
Explanation: Trust me
which reaction can be used to make 2,3 dichloropentane?
A) esterification
B) substitution
C) addition
D) etherification
Answer:
B) substitution
Explanation:
2,3 dichloropentane is a hydrocarbon that contains two chlorine atoms in its chain. It has a chemical formula: [tex]C_{5} H_{10} Cl_{2}[/tex]. In the compound, two hydrogen atoms has been substituted by two chlorine atoms.
A substitution reaction is a process by which an atom of a compound is being replaced by another atom to form a new compound due to a chemical reaction. This is one of the general properties of alkanes.
The fuel value of hamburger is approximately 3.3 kcal/g. If a man eats 0.75 pounds of hamburger for lunch and none of the energy is stored in his body, estimate the amount of water that would have to be lost in perspiration to keep his body temperature constant. The heat of vaporization of water may be taken as 2.41 kJ/g.
Answer:
[tex]m_{w} = 1755.323\,g[/tex]
Explanation:
The energy stored in the hamburger is:
[tex]Q = (0.75\,pd)\cdot \left(\frac{453\,g}{1\,pd} \right)\cdot \left(3.3\,\frac{kcal}{g} \right)\cdot \left(4.186\,\frac{kJ}{kcal} \right)[/tex]
[tex]Q = 4693.239\,kJ[/tex]
The amount of water perspired is derived from the following formula:
[tex]Q = m_{w}\cdot (c_{p}\cdot \Delta T + L_{v})[/tex]
[tex]m_{w} = \frac{Q}{c_{p}\cdot \Delta T + L_{v}}[/tex]
[tex]m_{w} = \frac{4693.239\,kJ}{\left(4.186\times 10^{-3}\,\frac{kJ}{g\cdot ^{\circ}C} \right)\cdot (100^{\circ}C-37^{\circ}C)+2.41\,\frac{kJ}{g} }[/tex]
[tex]m_{w} = 1755.323\,g[/tex]
The rust that appears on steel surfaces is iron(III) oxide. If the rust found spread over the surfaces of a steel bicycle frame contains a total of 9.62×1022 oxygen atoms, how many grams of rust are present on the bicycle frame?
Answer:
The grams of rust present in the bicycle frame is [tex]x = 8.50 g[/tex]
Explanation:
From the question we are told that
The number of oxygen atom contained is [tex]n = 9.62 *10^{22} \ atoms[/tex]
The molar mass of the compound is [tex]M_{Fe_3O_3} = 159.69 g[/tex]
At standard temperature and pressure the number of oxygen atom in one mole of iron(III) oxide is mathematically evaluated as
[tex]N_o = 3 * Ne[/tex]
Where Ne is the avogadro's constant with a value [tex]N_e = 6.023 *10^{23} \ atoms[/tex]
So
[tex]N_o= 3 * 6.023*10^{23} \ atoms[/tex]
[tex]N_o= 1.8069*10^{24} \ atoms[/tex]
So
[tex]1.8069*10^{24} \ atoms[/tex] is contained in [tex]M_{Fe_3O_3} = 159.69 g[/tex]
[tex]9.62 *10^{22} \ atoms[/tex] is contained in x
So
[tex]x = \frac{159.69 * 9.62 *10^{22}}{1.8069 *10^{24}}[/tex]
[tex]x = 8.50 g[/tex]
Final answer:
To calculate the grams of rust on the bicycle frame, we need to use stoichiometry. By converting the number of oxygen atoms to moles and then to grams using the molar mass of iron(III) oxide, we find that there are approximately 5.37x10^22 grams of rust on the bicycle frame.
Explanation:
To determine the grams of rust present on the bicycle frame, we need to calculate the molar mass of iron(III) oxide and then use stoichiometry to convert the number of oxygen atoms to grams of rust. The molar mass of Fe2O3 is 159.69 g/mol. Given that 1 mole of Fe2O3 contains 3 moles of oxygen atoms, we can calculate the moles of Fe2O3 using the given number of oxygen atoms and then convert it to grams using the molar mass.
9.62x10^22 oxygen atoms x (1 mol Fe2O3 / 3 mol O) x (159.69 g Fe2O3 / 1 mol Fe2O3) = 5.37x10^22 g Fe2O3
Therefore, there are approximately 5.37x10^22 grams of rust present on the bicycle frame.
Which 0.10 M solutions will have the greatest electrical conductivity?
Complete Question: Which solution will have the greatest electrical conductivity?
A. 0.50MHCl
B. 0.10MRbOH
C. 0.50MK3PO4
D. 2.0MC6H12O6
Answer:
Option C =>0.50M K3PO4.
Explanation:
Electrical conductivity has do do with the mobility of electrons or ion, that is to say for electrical conductivity to occur there should be the movement of ions or electrons.
So, let us take option (A) and option (B) first, that is 0.50MHCl and 0.10MRbOH respectively. Both HCl and RbOH are good conductors of electricity that is to say they are both strong electrolytes. But, between the two, (that is HCl and RbOH) HCl will be better conductor of electricity that RbOH.
Option (D) is incorrect totally, and option (C) 0.50M K3PO4 is a strong electrolytes too and it will be will be the CORRECT ANSWER in this question because it gives 4 ions( K3PO4 <======> 3K^+ + PO4^-3) unlike HCl that only gives 2 ions(that is HCl <=====> H^+ + Cl^-).
Option C. 0.50M K₃PO₄.
Let's understand the concept behind electrical conductivity:
Electrical conductivity tells us how well a material will allow electricity to travel through it. It is proportional to the product of mobility and carrier concentration.
Let's look at options one by one:
In option (A) and (B) it is given 0.50M HCl and 0.10M RbOH respectively. Both HCl and RbOH are good conductors of electricity thus they both are strong electrolytes. Out of these two HCl is more stronger electrolyte than RbOH. Option (D) 2.0M C₆H₁₂O₆ -the molecule of glucose is considered as a non-electrolyte, because of its inability to dissociate into ions. Option (C) 0.50M K₃PO₄ is a strong electrolyte than HCl since K₃PO₄ is an ionic compound so it dissociates completely in water to form potassium ion and phosphate ion.K₃PO₄ can dissociate into ions which can be represented as:K₃PO₄(aq) → 3K⁺(aq) + PO₄³⁻(aq)
So out the following options K₃PO₄ will have the greatest electrical conductivity since it is the strongest electrolyte out of the given options.
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Match the following: Molecule Test 9. Carbohydrates A. Biuret test 10. Starch and glycogen B. Paper test 11. Lipids C. I2KI test 12. Proteins D. Benedict test 13. Nucleic acids E. None of the above
Answer:
Carbohydrates------Benedict test
Starch and glycogen --------I2KI test
Proteins--------Biuret test
Lipids-------Paper test
Nucleic acids------None of the above
Explanation:
Benedict's Test is can be used to detect simple carbohydrates. The Benedict's test can detect reducing sugars (monosaccharide's and some disaccharides), having free ketone or aldehyde functional groups.
The Biuret Test shows the presence of peptide bonds, which are the basis for proteins. These bonds makes the blue Biuret reagent turn purple. The rest is carried out by adding an equal amount of NaOH to a solution of the food, mix carefully and add a few drops of 1% CuSO4, without shaking the mixture.
Lipids form a translucent stain on paper while starch/glycogen turns I2/KI solution blue-black
When you hear the word solution, you tend to think of a liquid. However, a solution can be a homogeneous mixture of any phase. These two questions are about a solid solution that has properties of a metal. When some atoms of the solvent are replaced by solute atoms of a similar size, the solution is called a heterogeneous alloy. an intermetallic compound. a substitutional alloy. an interstitial alloy.
Both questions are answered below:
Substitutional alloysInterstitial alloysExplanation:
Some atoms of the solvent can be replaced by atoms of relatively similar sizes, allowing for atom exchange or substitution, Whenever these atoms can be substituted within the matrix, it forms a substitutional alloy. Examples include bronze and brass.
Alternatively, some atoms are much smaller in size and cannot successfully be exchanged for the other. Instead, they get trapped within the matrix. these are called interstitial alloys, e.g steel.
Arthropods do not have internal bones. Instead, an arthropod's body is supported by:
a) two or three segments.
b) an exoskeleton.
c) a set of antennae.
Answer:an exoskeleton
Explanation:jus did it
Answer:
B
Explanation:
For the following reaction, 7.53 grams of benzene (C6H6) are allowed to react with 8.33 grams of oxygen gas. benzene (C6H6) (l) + oxygen (g) carbon monoxide (g) + water (g) What is the maximum amount of carbon monoxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams
Answer:
The maximum amount of CO2 that can be formed is 9.15 grams CO2
O2 is the limiting reactant
There will remain 4.82 grams of benzene
Explanation:
Step 1: Data given
Mass of benzene = 7.53 grams
Mass of oxygen gas = 8.33 grams
Molar mass of benzene = 78.11 g/mol
Molar mass oxygen gas = 32.00 g/mol
Step 2: The balanced equation
2C6H6 + 15O2 → 12CO2 + 6H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles C6H6 = 7.53 grams / 78.11 g/mol
Moles C6H6 = 0.0964 moles
Moles O2 = 8.33 grams / 32.00 g/mol
Moles O2 = 0.2603 moles
Step 4: Calculate the limiting reactant
For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O
O2 is the limiting reactant. It will completely be consumed ( 0.2603 moles). Benzene is in excess. there will react 2/15 * 0.2603 = 0.0347 moles
There will remain 0.0964 - 0.0347 = 0.0617 moles benzene
This is 0.0617 moles * 78.11 g/mol = 4.82 grams benzene
Step 5: Calculate moles CO2
For 2 moles benzene we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O
For 0.2603 moles O2 we'll have 12/15 * 0.2603 = 0.208 moles CO2
Step 6: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.208 moles * 44.01 g/mol
Mass CO2 = 9.15 grams
Lactic acid C3H6O3 is found in sour milk where it is produced by the action of lactobacilli in lactose or the sugar in milk. The pH of a 0.045 M solution of lactic acid was determined using a pH probe and found to be 2.63. a. Calculate the equilibrium constant for this acid. b. Had you not been given the pH of the acid and you had to measure it yourself, how would the method in part 2 be applied to the determination of Ka? Would you expect an improvement in the accuracy of your result with the application of the method of this experiment? Explain why or why not.
Answer:
see explanation below
Explanation:
In this case, we need to write the overall reaction:
HC₃H₅O₃ + H₂O <-------> C₃H₅O₃⁻ + H₃O⁺
The lactic acid is a weak acid, so, when it dissociates in it's ions, part of the acid is dissociated. This depends of it's Ka to know which quantity was dissociated.
To calculate Ka, let's write an ICE chart first:
HC₃H₅O₃ + H₂O <-------> C₃H₅O₃⁻ + H₃O⁺ Ka = ?
i) 0.045 0 0
c) -y +y +y
e) 0.045 - y y y
Writting the Ka expression we have:
Ka = [C₃H₅O₃⁻] [H₃O⁺] / [HC₃H₅O₃]
Now, to calculate Ka we need the values of [C₃H₅O₃⁻] and [H₃O⁺] in equilibrium. Fortunately, we have the value of the pH, which is 2.63 and with this we can get the value of [H₃O⁺] and then, the value of y. With that value, we replace it in the Ka expression to calculate Ka:
[H₃O⁺] = 10^(-pH)
[H₃O⁺] = 10^(-2.63)
[H₃O⁺] = [C₃H₅O₃⁻] = x = 2.34x10⁻³ M
Now, let's replace this value in the Ka expression:
Ka = (2.34x10⁻³)² / (0.045 - 2.34x10⁻³)
Ka = 1.28x10⁻⁴
b) Now, let's calculate the pH with the obtained value of Ka. We will use the same expression of Ka so:
1.28x10⁻⁴ = y² / (0.045-y)
1.28x10⁻⁴ (0.045 - y) = y²
5.76x10⁻⁶ - 1.28*10⁻⁴y = y²
y² + 1.28x10⁻⁴y - 5.76x10⁻⁶ = 0
From here, we'll use the quadratic equation general formula, for solving y:
y = -1.28x10⁻⁴ ±√(1.28x10⁻⁴)² + 4 * 1 * 5.76x10⁻⁶ / 2
y = -1.28x10⁻⁴ ±√2.31x10⁻⁵ / 2
y = -1.28x10⁻⁴ ± 4.8x10⁻³ / 2
y₁ = 2.34x10⁻³ M
y₂ = -2.464x10⁻³ M
Now, as y₁ is positive this is the value that we will take.
This value would be the [H₃O⁺] in equilibrium.
The value of pH would be:
pH = -log[H₃O⁺]
pH = -log(2.34x10⁻³)
pH = 2.631
According to this value of pH we can actually expect an inprovement in the accuracy, basically because we obtain a value with more significant figures, and this are relationed with accuracy.
Use the pump to add 250 molecules of the heavy species gas. How does the temperature change?
Type the correct answer in each box. Use numerals instead of words.
When you begin to add the molecules, the temperature inside the container is K. After all 250 molecules have been added, the temperature inside the container is K.
Answer:
When you begin to add the molecules, the temperature inside the container is 300 K. After all 250 molecules have been added, the temperature inside the container is 300 K.
A 50.0 mL sample containing Cd 2 + and Mn 2 + was treated with 50.0 mL of 0.0400 M EDTA . Titration of the excess unreacted EDTA required 19.5 mL of 0.0270 M Ca 2 + . The Cd 2 + was displaced from EDTA by the addition of an excess of CN − . Titration of the newly freed EDTA required 17.1 mL of 0.0270 M Ca 2 + . What are the concentrations of Cd 2 + and Mn 2 + in the original solution?
Answer:
Check the explanation
Explanation:
VOLUME OF newly freed EDTA
=(VOLUME OF Ca2* STRENGTH OF Ca2)/ STRENGTH OF EDTA
=(24.9*0.0220)/ 0.0700
=7.825mL
STRENGTH OF Cd2
=( VOLUME OF newly freed EDTA * STRENGTH OF EDTA) / VOLUME OF SAMPLE
=(7.825*0.0700)/50
=0.0109 M
VOLUME OF excess unreacted EDTA
=(VOLUME OF Ca2* STRENGTH OF Ca2)/ STRENGTH OF EDTA
=(19.5*0.0270)/ 0.0400
=13.16mL
VOLUME OF EDTA REQUIRED FOR SAMPLE CONTAINING Cd2 AND Mn2= (17.1-13.16) mL
=3.94 mL
VOLUME OF EDTA REQUIRED FOR Mn2
= (VOLUME OF EDTA REQUIRED FOR SAMPLE CONTAINING Cd2 AND Mn2 - VOLUME OF newly freed EDTA )
=40.02-7.825 mL
=32.20 mL
STRENGTH OF Mn2
=( VOLUME OF EDTA REQUIRED FOR Mn2* STRENGTH OF EDTA) / VOLUME OF SAMPLE
=(32.20*0.0700)/50
=0.045 M
When HClO2 is dissolved in water, it partially dissociates according to the equation
HClO2 → H+ + ClO2-. A solution is prepared that contains 5.850 g of HClO2 in 1.000 kg of water. Its freezing point is -0.3473 °C. Calculate the fraction of HClO2 that has dissociated.
Answer:
x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates
Explanation:
Recall that , depression present in freezing point is calculated with the formulae = solute particles Molarity x KF
0.3473 = m * 1.86
Solving, m = 0.187 m
Moles of HClO2 = mass / molar mass = 5.85 / 68.5 = 0.0854 mol
Molality = moles / mass of water in kg = 0.0854 / 1 = 0.0854 m
Initial molality
Assuming that a % x of the solute dissociates, we have the ICE table:
HClO2 H+ + ClO2-
initial concentration: 0.0854 0 0
final concentration: 0.0854(1-x/100) 0.0854x/100 0.0854x / 100
We see that sum of molality of equilibrium mixture = freezing point molality
0.0854( 1 - x/100 + x/100 + x/100) = 0.187
2.1897 = 1 + x / 100
x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates
substances change temperature at different rates because of their
density
specific heat
melting point
boiling point
Answer:
ok
Explanation:
I have the same question
The pH of a solution is 6.5.
What is the pOH?
Answer:
7.5
Explanation:
pH + pOH = 14
pOH = 14 - 6.5
pOH = 7.5
The pOH of a solution, given a pH of 6.5 at temperature 25°C, is 7.5. This is calculated by subtracting the pH from 14.
Explanation:The pH and pOH of a solution are related by the equation pH + pOH = 14 at 25°C. Given that the pH of the solution is 6.5, to find the pOH you simply subtract the pH from 14. So, pOH = 14 - pH. In this case, pOH = 14 - 6.5 = 7.5. Therefore, the pOH of a solution with a pH of 6.5 is 7.5.
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21. A good transition state analog
A) binds covalently to the enzyme
B) binds to the enzyme more tightly than the substrate,
C) binds very weakly to the enzyme
D) is toonstable to solate
E) must be almost identical to the substrate
Answer:
hhgjghjfgjfghj
Explanation:
Which option is true about the effect of polarity on solubility?
Select all that apply.
1-Polar solutes dissolve better in oil than water.
2-Polar solutes have the greatest solubility in nonpolar solvents.
3-Polar solutes dissolve better in water than in oil.
4-Polar solutes have the greatest solkbility in polar solvents.
Explanation:
It is known that like dissolves like. Therefore, polar solutes are soluble in polar solvents.
For example, salt is a polar compound and water is a polar solvent. Therefore, salt will dissolve in water.
On the other hand, oil is an inorganic solvent and hence, it is unable to dissolve any polar solute.
Therefore, we can conclude that following is true about the effect of polarity on solubility.
Polar solutes dissolve better in water than in oil.Polar solutes have the greatest solubility in polar solvents.Hope this helps!
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The complete combustion of methane is: CH4 + 2O2 ! 2H2O + CO2 a. Calculate the standard Gibbs free energy change for the reaction at 298 K (i.e. ). b. Calculate the energetic (ΔH) and entropic contributions (TΔS) to the favorable standard Gibbs free energy change at 298 K and determine which is the dominant contribution,? c. Estimate the equilibrium constant at 298 K.
Answer:
a
The standard Gibbs free energy change for the reaction at 298 K is
[tex]\Delta G^o_{re} = -800.99 kJ/moles[/tex]
b
The energetic (ΔH) is [tex]\Delta H^o _{re} = -802.112 \ kJ/mole[/tex]
The entropic contributions is [tex]T \Delta S = -1.126 \ kJ/mole[/tex]
Energetic is the dominant contribution
c
The equilibrium constant at 298 K is [tex]K = 2.53[/tex]
Explanation:
From the question we are told that
The chemical reaction is
[tex]CH_4 + 2 O_2 ----> 2 H_2 O + CO_2[/tex]
Generally ,
The free energy of formation of [tex]CH_4[/tex] is a constant with a value
[tex]\Delta G^o_f __{CH_4}} = -50.794 \ kJ / moles[/tex]
The free energy of formation of [tex]O_2[/tex] is a constant with a value
[tex]\Delta G^o_f __{O_2}} = 0 \ kJ / moles[/tex]
The free energy of formation of [tex]H_2O[/tex] is a constant with a value
[tex]\Delta G^o_f __{H_2O}} = -228.59 \ kJ / moles[/tex]
The free energy of formation of [tex]CO_2[/tex] is a constant with a value
[tex]\Delta G^o_f __{H_2O}} = -394.6 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]CH_4[/tex] at standard condition i is a constant with a value
[tex]\Delta H^o_f __{CH_4}} = -74.848 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]CO_2[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{CO_2}} = -393.3 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]O_2[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{O_2}} = 0 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]H_2O[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{H_2O}} = -241.83 \ kJ / moles[/tex]
The standard Gibbs free energy change for the reaction at 298 K is mathematically represented as
[tex]\Delta G^o_{re} = (\Delta G^o_f __{H_2O}} + (2 * \Delta G^o_f __{H_2O}} )) - ((\Delta G^o_f __{CH_4}} + (2 * \Delta G^o_f __{O_2}}))[/tex]
Substituting values
[tex]\Delta G^o_{re} =\Delta G= ( (-394.6 ) + (2 * (-228.59)) ) - ((-50.794) +(2* 0))[/tex]
[tex]\Delta G^o_{re} = -800.99 kJ/moles[/tex]
The Enthalpy of formation of the reaction is
[tex]\Delta H^o _{re} =( \Delta H^o_f __{CH_4}} + (2 * (\Delta H^o_f __{H_2O}} ))) - ( \Delta H^o_f __{CH_4}} + (2 * \Delta H^o_f __{O_2}}))[/tex]
Substituting values
[tex]\Delta H^o _{re} = \Delta H = ((-393.3) + 2 * ( -241.83)) - ( -74.848 + (2 * 0))[/tex]
[tex]\Delta H^o _{re} = -802.112 \ kJ/mole[/tex]
The entropic contributions is mathematically represented as
[tex]T \Delta S = \Delta H -\Delta G[/tex]
Substituting values
[tex]T \Delta S =-802 .112-(-800.986)[/tex]
[tex]T \Delta S = -1.126 \ kJ/mole[/tex]
Comparing the values of [tex]T \Delta S \ and \ \Delta G[/tex] we see that energetic is the dominant contribution
The standard Gibbs free energy change for the reaction at 298 K can also be represented mathematically as
[tex]\Delta G = -RT lnK[/tex]
Where R is the gas constant with as value of [tex]R = 8.314 *10^{-3} kJ/mole[/tex]
K is the equilibrium constant
T is the temperature with a given value of [tex]T = 298K[/tex]
Making K the subject we have
[tex]K = e ^{- \frac{\Delta G }{RT} }[/tex]
Substituting values
[tex]K = e ^{- \frac{-800.99 }{(8.314 *10^{-3} ) * (298)} }[/tex]
[tex]K = 2.53[/tex]
Final answer:
To calculate the standard Gibbs free energy change for the combustion of methane, use the formula ΔG = ΔH - TΔS, where ΔG is the standard Gibbs free energy change, ΔH is the standard enthalpy change, T is the temperature in Kelvin, and ΔS is the standard entropy change.
Explanation:
The standard Gibbs free energy change for a reaction can be calculated using the formula:
ΔG = ΔH - TΔS
where ΔG is the standard Gibbs free energy change, ΔH is the standard enthalpy change, T is the temperature in Kelvin, and ΔS is the standard entropy change.
In this case, we are given the thermochemical equation for the combustion of methane as: CH4 + 2O2 → 2H2O + CO2
To calculate the standard Gibbs free energy change, we need the values of ΔH and ΔS for this reaction at 298 K. Once we have these values, we can plug them into the formula to get the answer.
The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant. Deuterium, D, is an isotope of hydrogen. 2 HD(g) ⇌ H2(g) + D2(g) Kc = 0.28 5 H2(g) + 5 D2(g) ⇌ 10 HD(g) Kc = ? The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant. Deuterium, D, is an isotope of hydrogen. 2 HD(g) ⇌ H2(g) + D2(g) Kc = 0.28 5 H2(g) + 5 D2(g) ⇌ 10 HD(g) Kc = ? 581 0.73 1.3 0.0017 2.06
Answer:
Answer choice 'D' => Kc(2) = 581
Explanation:
2HD ⇄ H₂ + D₂ => Kc(1) = [H₂][D₂]/[HD]² => 0.28
5H₂ + 5D₂ ⇄ 10HD => Kc(2) = [HD]¹⁰/[H₂]⁵[D₂]⁵ = 1/(Kc(1))⁵ = 1/(0.28)⁵ = 581
For these type problems, one should 1st write the empirical Kc expression for each equation given. Then compare the expressions and ask 'how can the 1st Kc expression be changed into the 2nd Kc expression?' Apply and substitute given Kc(1) into the Kc(2) expression and solve for numerical results.
The equilibrium constant for the second reaction 5 H2(g) + 5 D2(g) ⇌ 10 HD(g) is 0.0017 when the equilibrium constant for the first reaction 2 HD(g) ⇌ H2(g) + D2(g) is 0.28. This is calculated by taking the first reaction's Kc and raising it to the power of 5.
Explanation:The chemical reaction you have given is a perfect example of equilibrium in chemistry. In the first reaction, 2 HD(g) ⇌ H2(g) + D2(g), the equilibrium constant Kc is already given as 0.28. If we look at the second reaction, 5 H2(g) + 5 D2(g) ⇌ 10 HD(g), we can see that it's technically the first reaction multiplied by 5. Therefore, to find Kc for the second reaction, we simply raise the Kc of the first reaction to the power of 5. So, Kc for the second reaction would be (0.28)^5 = 0.0017.
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A quantity of 25.0 mL of a solution containing both Fe2+ and Fe3+ ions is titrated with 23.0 mL of 0.0200 M KMnO4 (in dilute sulfuric acid). As a result, all of the Fe2+ ions are oxidized to Fe3+ ions. Next, the solution is treated with Zn metal to convert all of the Fe3+ ions to Fe2+ ions. Finally, the solution containing only the Fe2+ ions requires 40.0 mL of the same KMnO4 solution for oxidation to Fe3+. Calculate the molar concentrations of Fe2+ and Fe3+in the original solution. The net ionic equation is Mno4- +5Fe2+ + 8H+ >>>> Mn2+ + 5Fe3+ +4H2o
The molar concentrations of Fe2+ and Fe3+ in the original solution are 18.4 mM.
Explanation:To calculate the molar concentrations of Fe2+ and Fe3+ in the original solution, we can use stoichiometry based on the net ionic equation provided:
MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O
From the equation, we can see that one mole of MnO4- reacts with 5 moles of Fe2+. In the first titration, 23.0 mL of 0.0200 M KMnO4 solution is required to react with all of the Fe2+ ions, which means there is a total of 23.0 mL x 0.0200 M = 0.460 mmol of Fe2+ in the original solution.
In the second titration, 40.0 mL of the same KMnO4 solution is required to react with all of the Fe2+ ions converted from Fe3+. This means there is a total of 40.0 mL x 0.0200 M = 0.800 mmol of Fe2+ in the original solution.
Since all of the Fe2+ ions are converted to Fe3+ in the first titration, the total amount of Fe3+ in the original solution is also 0.460 mmol.
Therefore, the molar concentrations of Fe2+ and Fe3+ in the original solution are:
[Fe2+] = 0.460 mmol / 25.0 mL = 18.4 mM
[Fe3+] = 0.460 mmol / 25.0 mL = 18.4 mM
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Atoms can be seen with the naked eye.
Atoms do not have mass.
Atoms make up every substance around you.
Answer:
Atoms make up every substance around you. True
Explanation:
Atoms can be seen with the naked eye. False because atoms are very very very smallllll....
Atoms do not have mass. False, atoms do have mass although it is very smallllllllll as welll
Atoms make up every substance around you. True, similar to how cells make up different things like tissues, and tissues make organs and so on, atoms make up everything around us.
Practice the Skill 21.15b When the following ketone is treated with aqueous sodium hydroxide, the aldol product is obtained in poor yields. In these cases, special distillation techniques are used to increase the yield of aldol product. Predict the aldol addition product that is obtained, and propose a mechanism for its formation. For the mechanism, draw the curved arrows as needed. Include lone pairs and charges in your answer. Do not draw out any hydrogen explicitly in your products. Do not use abbreviations such as Me or Ph.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below for the step by step explanation to the question above.
A 175 ml sample of neon had its pressure changed from 75 kPa to 150 kPa. What is its new volume?
Answer:
The new volume is 87.5 ml.
Explanation:
We have,
Volume of sample of neon is 175 ml
The pressure changed from 75 kPa to 150 kPa.
We need to find new volume.
It is based on the concept of Boyle's law. Let V₂ is new volume. So,
[tex]P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{175\times 75}{150}\\\\V_2=87.5\ ml[/tex]
So, new volume is 87.5 ml.
Using Boyle's Law, the new volume of the neon gas when the pressure is changed from 75 kPa to 150 kPa is calculated to be 87.5 mL.
To find the new volume of a gas when its pressure is changed, we use Boyle's Law.
Boyle's Law states that the pressure and volume of a gas are inversely proportional when temperature and the number of gas molecules remain constant.
The formula is expressed as:
P₁V₁ = P₂V₂
In this problem:
Initial Pressure, P₁ = 75 kPaInitial Volume, V₁ = 175 mLFinal Pressure, P₂ = 150 kPaWe need to find the final volume, V₂. Rearranging Boyle's Law, we get:
V₂ = (P₁V₁) / P₂
Plugging in the values:
V₂ = (75 kPa * 175 mL) / 150 kPa
V₂ = 13125 mL / 150 kPa
V₂ = 87.5 mL
Therefore, the new volume of the neon gas sample is 87.5 mL.
A concentration cell is constructed with copper electrodes and Cu2+ in each compartment. In one compartment, the [Cu2+] = 1.0 × 10–3 M and in the other compartment, the [Cu2+] = 2.0 M. Calculate the potential for this cell at 25°C. The standard reduction potential for Cu2+ is +0.34 V.
a. 0.78 V
b. –0.098 V
c. –0.44 V
d. 0.44 V
e. 0.098 V
The potential of this concentration cell can be calculated using the Nernst Equation, which provides an approximation of about 0.098 V for the potential of the cell. This corresponds to choice (e).
Explanation:The potential for this concentration cell can be calculated using the Nernst equation, which incorporates the standard reduction potential and the concentrations of the reaction species in its equilibrium state. The Nernst Equation can be written as:
E = E0 - (0.05916/n) * logQ
Where E is the cell potential, E0 is the standard cell potential (+0.34 V for Cu2+/Cu), n is the number of moles of electrons exchanged in the reaction (which is 2 for this reaction), and Q is the reaction quotient, which is the ratio of concentration of products to reactants.
Since this cell involves Cu2+ being reduced to Cu, the concentration in the cell where [Cu2+]=2.0M would be considered the product, and the one where [Cu2+]=1.0×10–3M would be considered the reactant. Therefore, Q=[product]/[reactant]=2.0/ (1.0×10–3) = 2000.
Plugging these values into the Nernst Equation, we get:
E = 0.34V - (0.05916/2) * log2000
After calculation, gives us an E value of approximately V, which correlates to option (e).
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5. When doing an experiment, aside from doing calculations, what can you do to
determine which reactant is the limiting reactant and which is the excess
reactant? *
Answer:
check which reactant is totally consumed and which one remains in the mixture
Explanation:
Apart from doing calculations during an experiment, one can determine which reactant is limiting and which one is in excess by checking the resulting mixture for the presence of reactants.
A limiting reactant is one that determines the amount of product formed during a reaction. It is usually a reactant that is lower than stoichiometry amount.
On the other hand, an excess reactant is one that is present in more than the stoichiometrically required amount during a reaction.
Limiting reactants will be totally consumed in a reaction while excess reactant would still be seen present in mixture after the reaction has stopped.
Hence, apart from using stoichiometric calculation to determine which reactant is limiting or in excess during an experiment, one can just check the final mixture of the reaction for the presence of any of the reactants. The reactant that is detected is the excess reactant while the one without traces in the final mixture is the limiting reactant.
Which scenario best describes an increase in the entropy of a system?
Answer:
A) A solid salt dissolves in water.
Explanation:
A solid, like a salt, dissociates into ions as it dissolves in liquid. The particles (ions) become more spaced apart and with greater randomness. This is increasing entropy.
Carbon 14 (14C) dating assumes that the carbon dioxide on Earth today has the same radioactive content as it did centuries ago. If this is true, the amount of 14C absorbed by a tree that grew several centuries ago should be the same as the amount of 14C absorbed by a tree growing today. A piece of ancient charcoal contains only 19% as much radioactive carbon as a piece of modern charcoal. How long ago was the tree burned to make the ancient charcoal given that the half-life of 14C is 5700 years?
Answer:
23388 years
Explanation:
Now we have to use the formula;
0.693/t1/2 = 2.303/t log No/N
Where
t1/2 = half life of the C-14 = 5700
t= age of the sample
No= radioactive material in a modern sample = No
N= radioactive material in a sample being studied. = 0.81No( since the amount N= No-0.19No)
Hence;
0.693/5700 = 2.303/t log No/0.81No
0.693/5700 = 2.303/t log 1/0.81
0.693/5700 = 2.303/t × 1.235
0.693/5700= 2.844/t
1.216×10^-4= 2.844/t
t= 2.844/1.216×10^-4
t= 23388 years
Which activity is a way the circulatory system works with another body system
to maintain homeostasis?
A. breaking down food into nutrients
B. transporting hormones through the body
C. depositing fat on the inside walls of arteries
Answer:
A. breaking down food into nutrients
Explanation:
Homeostasis is the ability of the human body to regulate itself or achieve internal stability. If a body's conditions can not be regulated, it may result to the break down of its immune system leading to infections, diseases or disorder.
To maintain homeostasis, the circulatory system transports oxygen-rich blood to other parts of the body. So that the immediate condition can be regulated. It works with the respiratory system, delivering nutrients to other parts of the body. It also remove wastes from our body.
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