Answer:
%error = 0.32%
Explanation:
Let's answer both questions, by parts.
1. Percentage error:
In this case, I do not have the video, but I do have the reported melting point of naphtalene which is 80.26 °C.
The expression to calculate the percentage error is the following:
%Error = absolute error / actual percentage. (1)
And the absolute error is:
Abs error = actual value - experimental value (2)
But the experimental value is a range, so we just have to get a average of that:
Exp value = 77 + 83 / 2 = 80 °C
Now the absolute error:
Abs error = 80.26 - 80 = 0.26 °C
Finally the %error:
%error = (0.26 / 80.26) * 100
%error = 0.32%2. Meaning of melting point range and %error
The melting point range just means that the sample of naphtalene has impurities, and when a sample of any compound has impurities, melting point tends to be low. However, this decrease of temperature is a wider range. But usually a range of just 5° C means that compound has little traces of impurities but it can still be used for reactions.
The %error means that the impurities of the sample are really low, so the sample is practically pure with little traces of impurities.
Final answer:
The percent error for the melting point of naphthalene is calculated using the formula and considering the literature value of 80.2°C, yielding an approximate percent error of 0.25%. The broad melting range and low percent error suggest the sample is mostly pure with possible minor impurities.
Explanation:
The student's question is regarding the calculation of the percent error for the melting point of a sample of naphthalene and what the results suggest about the sample's purity.
Percent Error Calculation
To find the percent error, we use the formula:
Percent Error = (|Experimental Value - Literature Value| / Literature Value) x 100%
Assuming the literature value of naphthalene's melting point is approximately 80.2°C (the value will need to be verified as it can vary slightly in literature), we calculate percent error using the experimental value's mean (80°C, the midpoint of 77-83°C) as follows:
Percent Error = (|80°C - 80.2°C| / 80.2°C) x 100% = (0.2°C / 80.2°C) x 100% ≈ 0.25%
Suggestion About the Sample
The observed melting range of 77-83°C and the low percent error suggest that the sample is relatively pure but may contain minor impurities since a pure sample would have a narrower melting point range close to the literature value.
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
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.
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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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:
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
How much 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl?
Answer:
54 L
Explanation:
Given data
Initial concentration (C₁): 9.0 MInitial volume (V₁): 450 mLFinal concentration (C₂): 0.075 MFinal volume (V₂): to be calculatedWe have a concentrated NaCl solution and we will add water to it to obtain a diluted NaCl solution. We can find the volume of the final solution using the dilution rule.
[tex]C_1 \times V_1 = C_2 \times V_2\\V_2 = \frac{C_1 \times V_1}{C_2} = \frac{9.0M \times 450mL}{0.075M} = 5.4 \times 10^{4} mL = 54 L[/tex]
The volume of the final solution should be 54L.
Calculation of the volume:Since
Initial concentration (C₁): 9.0 M
Initial volume (V₁): 450 mL
Final concentration (C₂): 0.075 M
We know that
C1 * V1 = C2 * V2
V2 = C1 * V1/ C2
= 9.0M * 450 mL/ 0.075 M
= 54 L
hence, The volume of the final solution should be 54L.
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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:
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.
Which anion would bond with K+ in a 1:1 ratio to form a neutral ionic compound?
Select the correct answer below:
A. O2-
B. F-
C. N3-
D. S2-
The anion that would bond with K+ in a 1:1 ratio to form a neutral ionic compound is F- (fluoride ion)
Explanation:The anion that would bond with K+ in a 1:1 ratio to form a neutral ionic compound is B. F- (fluoride ion).
Potassium (K+) is a cation with a +1 charge, and to form a neutral ionic compound, it needs to bond with an anion with a -1 charge. The fluoride ion (F-) has a -1 charge, and therefore, it can form a 1:1 ratio with K+ to create a neutral compound.
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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.
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
Draw the structure(s) of the product(s) of the Claisen condensation reaction between ethyl propanoate and ethyl formate. Draw only the condensation product, including the self-condensation product if applicable, do not draw the structure of the leaving group. Do not consider stereochemistry.
Answer:
See explanation below
Explanation:
A claisen reaction, is often used between two esters (One of them, usually having alpha hydrogens atoms) to form Beta Keto esters. Depending of the reagents, this reaction is taking place with base where one ester acts as nucleophile and the other as electrophile.
Now, we have here ethyl propanoate and ethyl formate. The ethyl propanoate has two alpha hydrogens, while ethyl formate do not have that. So, the base will react with the ethyl propanoate first, and then, it will react as nucleophile with the formate.
Now, the base to be used in this case will have to be a relatively strong base such sodium ethanoate, because if we use a strong base such NaOH, this will cause the saponification of the ester and not the condensation. So, in order to promove the claisen condensation, we need to use a base not too strong.
In the picture attached you have the mechanism and structure of the final product.
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.
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]
Despite the destruction from Hurricane Katrina in September 2005, the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered an atmospheric pressure of 88.2 kPa on October 19, 2005, that was 2.0 kPa lower than Hurricane Katrina. What was the difference in pressure between the two hurricanes?
Complete Question
Despite the destruction from Hurricane Katrina in Sept 2005,the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered at atmospheric pressure of 88.2 kPa on Oct 19, 2005,about 2 kPa lower than Hurricane Katrinia. What was the difference between the two hurricanes in:
millimeters of Hg? ___ mm Hg
atmospheres? ___ atm
millibars? ___ mb
Answer:
The difference in pressure in mm of Hg is [tex]Pg = 15.05 \ mm\ Hg[/tex]
The difference in pressure in atm is [tex]P_{atm} = 0.0198 \ atm[/tex]
The difference in pressure in millibar is [tex]P_{bar} = 20 \ mbar[/tex]
Explanation:
From the question we are told that
The pressure of Hurricane Wilma is [tex]P = 88 kPa[/tex]
The difference is pressure is [tex]\Delta P =2\ k Pa[/tex]
Generally
[tex]1\ atm = 1.01 *10^5 \ Pa[/tex]
[tex]1 \ atm = 760\ mmHg[/tex]
The pressure difference in mm Hg is mathematically evaluated as
[tex]Pg = \Delta P * \frac{1000 Pa}{1kPa} * \frac{1\ atm}{1.01 *10^5} * \frac{760 mm Hg}{1 \ atm}[/tex]
Substituting the value
[tex]Pg = 2 kPa * \frac{1000 Pa}{1kPa} * \frac{1\ atm}{1.01 *10^5} * \frac{760 mm Hg}{1 \ atm}[/tex]
[tex]Pg = 15.05 \ mm\ Hg[/tex]
The pressure difference in atm is
[tex]P_{atm } = 2kPa * \frac{1000}{1\ kPa} *\frac{1 \ atm}{1.01 *10^5 Pa}[/tex]
[tex]= 0.0198 \ atm[/tex]
The pressure in millibars is
[tex]= 2kPa * \frac{1000Pa}{1 kPa} * \frac{1\ bar }{1*10^5 Pa} * \frac{1000 \ milli bar}{1 bar}[/tex]
[tex]= 20 \ m bar[/tex]
Final answer:
The pressure difference between Hurricane Wilma and Hurricane Katrina was 2.0 kPa, with Wilma's pressure being 88.2 kPa and Katrina's being 90.2 kPa. The pressure drop during a hurricane on the northeast U.S. coast in torr is 50.8 torr. A nonvolatile liquid is used in barometers and manometers because volatility would affect the accuracy of pressure measurements.
Explanation:
The difference in atmospheric pressure between Hurricane Wilma and Hurricane Katrina can be calculated using the information provided. If Wilma's pressure was 88.2 kPa and that was 2.0 kPa lower than Katrina's, then Katrina's pressure was 88.2 kPa + 2.0 kPa = 90.2 kPa. Therefore, the difference in pressure between the two hurricanes was 2.0 kPa.
To calculate the drop in pressure in torr during a hurricane on the northeastern United States coast: the pressure usually at 30.0 in. Hg can drop to 28.0 in. Hg. Since 1 in. Hg is equivalent to 25.4 mm Hg, and 1 mm Hg is the same as 1 torr, the drop in pressure is (30.0 in. Hg - 28.0 in. Hg) x 25.4 mm Hg/in. Hg = 50.8 torr.
It is necessary to use a nonvolatile liquid in a barometer or manometer because a volatile liquid would evaporate and form a vapor that would affect the pressure measurement, leading to inaccurate readings.
The rate constant of the elementary reaction C2H5CN(g) → CH2CHCN(g) + H2(g) is k = 7.21×10-3 s-1 at 634 °C, and the reaction has an activation energy of 247 kJ/mol. (a) Compute the rate constant of the reaction at a temperature of 741 °C. _________ s-1 (b) At a temperature of 634 °C, 96.1 s is required for half of the C2H5CN originally present to be consume. How long will it take to consume half of the reactant if an identical experiment is performed at 741 °C? (Enter numbers as numbers, no units. For example, 300 minutes would be 300. For letters, enter A, B, or C. Enter numbers in scientific notation using e# format. For example 1.43×10-4 would be 1.43e-4.)
Answer:
the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]
it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C
Explanation:
Given that :
[tex]k_1 = 7.21*10^{-3} s^{-1} \\ \\ E_a = 247 kJ/mol \ \ \ = 247*10^3 \ J/mol \\ \\ T_1 = 634 ^ {^0} C= (273 + 634) K = 907 \ K \\ \\ T_2 = 741^{^0 } C = (273+ 741) K = 1014 \ K \\ \\ R =8.314 \ \ J/mol/K[/tex]
a)
According to Arrhenius Equation ;
[tex]In\frac{k_2}{k_1} = -\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]
[tex]In\frac{k_2}{7.21*10^{-3}} = 3.456412[/tex]
[tex]\frac{k_2}{7.21*10^{-3}} =e^{ 3.456412[/tex]
[tex]k_2 = 31.70302 * 7.21*10^{-3}[/tex]
[tex]k_2 = 0.22858 \ s^{-1}[/tex]
Therefore , the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]
b) Given that :
[tex]t_{(1/2)} = 96.1 \ s[/tex]
[tex]k_1 = \frac{0.693}{ t_{(1/2)}}[/tex]
[tex]k_1 = \frac{0.693}{ 96.1 \ s}[/tex]
[tex]k_1 = 7.211*10^{-3} \ s^{-1}[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]
[tex]In\frac{k_2}{7.211*10^{-3}} = 3.456412[/tex]
[tex]\frac{k_2}{7.211*10^{-3}} =e^{ 3.456412[/tex]
[tex]k_2 =e^{ 3.456412 * 7.21*10^{-3}[/tex]
[tex]k_2 = 0.22862 \ s^{-1}[/tex]
[tex]t_{(1/2)_2}} = \frac{0.693}{k_2}[/tex]
[tex]t_{(1/2)_2}} = \frac{0.693}{0.22862}[/tex]
[tex]t_{(1/2)_2}} =3.0313 \ s[/tex]
it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C
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.
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.
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.
In your lab you are studying the kinetics of the degradation of a pain killer in the human liver. You are monitoring the concentration of the pain killer over a period of time. The initial concentration of the pain killer in your experiment was 1.59 M. After 0.80881 hours the concentration was found to be 0.795 M. In another 0.80881 hours the concentration was found to be 0.3975 M (t=27.868 hours overall).
If another experiment were set up where the initial concentration of the pain killer was 0.479 M, how long would it take for the pain killer concentration to reach 0.0161 M?
Answer:
Explanation:
From the data it is clear that after each .80881 hours , the concentration becomes half . so its decomposition appears to be first order .
k = 1 / t ln A₀ / A
= (1/.80881) ln 2
= .857
for the given case
A = .0161 , A₀ = .479
t = ?
t = 1/k ln( .479 / .0161)
= (1 / .857 )ln (.479 / .0161 )
= (1 / .857 ) x ln 29.75
= 3.959 hours
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.
substances change temperature at different rates because of their
density
specific heat
melting point
boiling point
Answer:
ok
Explanation:
I have the same question
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.
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
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.
How much energy is released when 50.00 mol of glycerin melts? (Heat of fusion glycerin-91.7 KJ/mol)
Answer:
4585 KJ
Explanation:
Data obtained from the question. This includes following:
Number of mole (n) of glycerin = 50 moles
Heat of fusion glycerin (Hf) = 91.7 KJ/mol
Heat (Q) released =?
The heat released can be obtained as follow:
Q = n•Hf
Q = 50 x 91.7
Q = 4585 KJ
Therefore, 4585 KJ of heat is released.
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
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.