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.
what is time actually
Answer:
Time is something I constantly run out of-
Answer:
The concept of time is self-evident. An hour consists of a certain number of minutes, a day of hours and a year of days. ... Time is represented through change, such as the circular motion of the moon around Earth. The passing of time is indeed closely connected to the concept of space
ight energy can be described as nhf , where n is a number of photons, h is Planck's constant, and f is the frequency of the light, also denoted by the symbol ???? . Calculate the minimum number of photons, n , needed to make the reduction of 2 moles of NADP+ favorable for light absorbed at 680.000 nm . Assume that the amount of energy needed for the reduction of one mole of NADP+ to be favorable must exceed 219 kJ/mol .
Answer:
1.496x 10^24photons
Explanation:
wavelength λ= 680 X10^-9 nm
h = planks constant - 6.636*10 ^-34js
c- speed of light - 3.0x 10^8 m/s
I mole of Energy of NADP+ = 219Kj/mol
2 moles of Energy of NADP+ = 2x 219= 438kj/mol = 438x10^3j
/mol
E= nhc/λ
438x 10^3j/mol -= n x (6.636*10 ^-34 x 3x10^8) / 680*10^-9
n=438x10^3j x 680x 10^-9/ (6.636*10 ^-34 x 3.0x10^8
1.496x 10^24photons
To make the reduction of 2 moles of NADP+ favorable for light absorbed at 680.000 nm, you would need a minimum of approximately [tex]6.35 x 10^{15[/tex] photons, which are particles of light.
The energy required for the reduction of 2 moles of NADP+ is given as 2 moles x 219 kJ/mol = 438 kJ. To calculate the number of photons needed, we can use the formula E = nhf, where E is the energy required, h is Planck's constant (6.626 x [tex]10^{-34[/tex] J·s), f is the frequency of light, and n is the number of photons. First, we need to convert the given wavelength to frequency using the speed of light (c =3 x[tex]10^8[/tex] m/s):
λ = 680.000 nm = 680.000 x [tex]10^{-9[/tex] m
f = c / λ = (3 x [tex]10^8[/tex] m/s) / (680.000 x [tex]10^{-9[/tex] m) ≈ 4.41 x [tex]10^{14[/tex] Hz
Now, we can calculate the number of photons (n) using the energy formula:
E = nhf
438,000 J = n(6.626 x [tex]10^{-34[/tex] J·s)(4.41 x [tex]10^{14[/tex] Hz)
Solving for n:
n ≈ 6.35 x [tex]10^{15[/tex] photons
So, approximately 6.35 x [tex]10^15[/tex]photons are needed to make the reduction of 2 moles of NADP+ favorable for light absorbed at 680.000 nm.
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The molar heat capacity of solid aluminium is 24.4\text{ J K}^{-1}\text{ mol}^{-1} \text{ at } 25^{\circ}\text{C}24.4 J K −1 mol −1 at 25 ∘ C. What is the change in internal energy when 1\text{ mol}1 mol of solid aluminium is heated from a temperature of 20^{\circ}\text{C} \text{ to }30^{\circ}\text{C}20 ∘ C to 30 ∘ C?
Answer:
[tex]\Delta U = 244\,J[/tex]
Explanation:
The change in internal energy is given by the following expression:
[tex]\Delta U = n \cdot \bar c \cdot \Delta T[/tex]
[tex]\Delta U = (1\,mole)\cdot \left(24.4\,\frac{J}{mole\cdot K} \right)\cdot (10\,K)[/tex]
[tex]\Delta U = 244\,J[/tex]
what is the atomic number of an oxygen atom with 8 protons and 10 neutrons in the nucleus.
A. 8
B.10
C.18
D. not enough information to calculate
Answer
A. 8
Explanation:
The Atomic number is equal to the number of protons.
I took the test and got it right so this is 100% correct. It is NOT 18 like some people say
The atomic number of an oxygen atom with 8 protons is 8, regardless of the number of neutrons.
Explanation:The atomic number of an atom is determined by the number of protons in its nucleus. This number also sorts elements into their correct position on the Periodic Table. Therefore, an oxygen atom with 8 protons will have an atomic number of 8, irrespective of the number of neutrons it has.
This is because neutrons do not influence the atomic number, only the atomic mass. So the correct answer to your question is: A. 8.
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How many miles of gas are contained in 890.0 mL at 21.0 C and 0.987 atm
Answer:
0.036 moles of gas are contained in 890.0 mL at 21.0 C and 0.987 atm
Explanation:
Ideal gases are those gases whose molecules do not interact with each other and move randomly.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
where P represents the pressure of the gas, V its volume, n the number of moles of gas (which must remain constant), R the constant of the gases and T the temperature of the gas.
In this case:
P= 0.987 atmV= 890 mL= 0.890 L (being 1 L= 1,000 mL)n= ?R= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]T= 21 °C= 294 °KReplacing:
0.987 atm* 0.890 L= n* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] * 294 K
Solving:
[tex]n=\frac{0.987 atm*0.890 L}{0.082\frac{atm*L}{mol*K}*294K }[/tex]
n= 0.036 moles
0.036 moles of gas are contained in 890.0 mL at 21.0 C and 0.987 atm
In what ways do coastal mountains affect a region’s climate? Select 3 correct choices.
1.They determine the way that large masses of cool, dry air move over land.
2.They control the amount of water vapor present in cool, dry air masses.
3.They increase precipitation on the windward sides of mountain ranges.
4.They force cool, moist air from oceans to rise as they move toward land.
5.They decrease precipitation totals on the leeward sides of mountain ranges.
Answer:
3)They increase precipitation on the windward sides of mountain ranges
4)They force cool, moist air from oceans to rise as they move toward land
5)They decrease precipitation totals on the leeward sides of mountain ranges.
Explanation:
During the day, as the ocean becomes heated up, there is always an increase in humidity of the air that is above the oceans in coastal areas. But Because of the higher heat capacity of water compare to the land,the ocean always remain cooler compare to the land. But because of lower density,the air on the land is replaced by the ocean's cool air as the land's air increases.
Whenever a mountain inland come in contact with cool
air with humidity,the sir get more cool. However,. there result a condensation and precipitation of the water that is present in the air at the part of the mountain.
Answer:
3.They increase precipitation on the windward sides of mountain ranges.
4.They force cool, moist air from oceans to rise as they move toward land.
5.They decrease precipitation totals on the leeward sides of mountain ranges.
Explanation:
The temperature on mountains usually become colder with a corresponding higher altitude. The Mountains also tend to have more precipitation than areas without them.
This is because in the day the ocean becomes heated up. This leads to an increase in humidity of the air that is above the oceans in coastal areas. Due to the higher heat capacity of water compared to land ,the ocean always remains cooler when compared to the land. The lower density also allows the air on the land to be replaced by the ocean's cool air as the land's air increases.
They force cool, moist air from oceans to rise as they move toward land.They increase precipitation on the windward sides of mountain ranges.They decrease precipitation totals on the leeward sides of mountain ranges.
if you had 0.867 miles of salt, NaCI , in a 0.69 L solution, what would be the molarity
Answer:
Approximately [tex]1.3\; \rm mol \cdot L^{-1}[/tex]. (Assuming that the question says [tex]0.867[/tex] moles of salt in this [tex]0.69\; \rm L[/tex] solution.)
Explanation:
The molarity of a solution gives the quantity of the solute in every unit volume of the solution. In this question:
Quantity of solute: [tex]n(\text{solute})= 0.867\; \rm mol[/tex] (with moles as the unit.)Volume of solution: [tex]V(\text{solution}) = 0.69\; \rm L[/tex] (with liters as the unit.)Note that in this question, liter is the unit for the volume of the solution. The molarity of the solution should thus give the amount of solute in every liter of the solution:
[tex]\begin{aligned} c & = \frac{n(\text{solute})}{V(\text{solution})} \\ &= \frac{0.867\; \rm mol}{0.69\; \rm mol} \approx 1.3\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
One possible mechanism for the gas phase reaction of hydrogen with nitrogen monoxide is: step 1: H2(g) + 2 NO (g) → N2O (g) + H2O (g) step 2: N2O (g) + H2 (g) → N2 (g) + H2O (g) Identify the molecularity of each step in the mechanism. (1, 2, or 3) Step 1 _______ Step 2 ________
Answer:
step 1 is trimolecular while step 2 is bimolecular.
Explanation:
Molecularity of an elementary reaction is defined the number of molecules that come together to react in a given elementary (single-step) reaction. It is equal to the sum of stoichiometric coefficients of reactants in that elementary reaction. Depending on the number of molecules that come together in an elementary step, a reaction can be designated as; unimolecular, bimolecular or trimolecular.
The kinetic order of any elementary reaction or reaction step is equal to its molecularity, and the rate equation of an elementary reaction can easily be determined by inspection, from the molecularity.
For a complex (multistep) reaction, the kinetic order of reaction is not determined from the molecularity since molecularity only describes elementary reactions or steps.
From our discussion above we can see that, step 1 is trimolecular while step 2 is bimolecular.
Predict whether the pHpH at the equivalence point for each titration will be acidic, basic, or neutral. Predict whether the at the equivalence point for each titration will be acidic, basic, or neutral. neutral for HFHF, and basic for HClHCl neutral for HClHCl, and basic for HFHF neutral for HFHF, and acidic for HClHCl neutral for HClHCl, and acidic for HFHF neutral for both
Answer:
pH at the equivalence point for titration of HF and HCl will be basic and neutral respectively.
Explanation:
pH at equivalence point depends on hydrolysis equilibrium of conjugated base present in mixture.
[tex]\Rightarrow[/tex] Neutralization reaction: [tex]HF+OH^{-}\rightleftharpoons F^{-}+H_{2}O[/tex]
Hence, at equilibrium, [tex]F^{-}[/tex] is present in mixture.
Hydrolysis reaction: [tex]F^{-}+H_{2}O\rightleftharpoons HF+OH^{-}[/tex]
Here HF is an weak acid and [tex]OH^{-}[/tex] is a strong base, So, resultant pH of the solution will be basic.
[tex]\Rightarrow[/tex] Neutralization reaction: [tex]HCl+OH^{-}\rightleftharpoons Cl^{-}+H_{2}O[/tex]
Hence, at equilibrium, [tex]Cl^{-}[/tex] is present in mixture.
Hydrolysis reaction: [tex]Cl^{-}+H_{2}O\rightleftharpoons HCl+OH^{-}[/tex]
Here HCl is a strong acid and [tex]OH^{-}[/tex] is a strong base, So, resultant pH of the solution will be neutral.
Final answer:
At the equivalence point, titration of HCl with NaOH results in a neutral pH, while titration of HF with NaOH results in a basic pH.
Explanation:
To predict whether the pH at the equivalence point for each titration will be acidic, basic, or neutral, we need to consider the nature of the acid and base involved in the reaction. For the titration of a strong acid like hydrochloric acid (HCl) with a strong base like sodium hydroxide (NaOH), the equivalence point occurs at a pH of 7.00, resulting in a neutral solution. This is due to the formation of water from the neutralization reaction between HCl and NaOH.
However, for the titration of a weak acid like hydrofluoric acid (HF) with a strong base like NaOH, the equivalence point will be at a pH greater than 7, resulting in a basic solution. This occurs because the salt formed from the reaction of HF with NaOH gives a solution of sodium fluoride, NaF, which is basic due to the hydrolysis of F- anions in water.
Therefore, the pH of the equivalent point will be neutral for the titration of HCl with NaOH, and basic for the titration of HF with NaOH. The correct prediction for the titrations in question is that the equivalence point will be neutral for HCl and basic for HF.
The diagram shows the scales used for recording temperatures. The labels for the scales are missing. 3 thermometers are oriented vertically labeled W, X, Y from left to right. The label Water Boils is connected by a dotted line to 212 degrees on W, 100 degrees on X, 373 degrees on Y. The label Water freezes is connected by a dotted line to 32 degrees on W, 0 degrees on W, 273 degrees on Y. The label Absolute Zero is connected by a dotted line to negative 460 degrees on W, negative 273 degrees on X, and 0 degrees on Y. Which labels complete the diagram? W: Fahrenheit X: Celsius Y: Kelvin W: Kelvin X: Celsius Y: Fahrenheit W: Celsius X:Fahrenheit Y: Kelvin W:Celsius X: Kelvin Y: Fahrenheit
Answer:
The Answer Is A. W: Fahrenheit Y: Kelvin X: Celsiuis
Answer:
a
Explanation:
How many milligrams of a 20mg sample of cesium-137 remain after 60 years
Approximately 0.237 milligrams of cesium-137 would remain after 60 years.
Explanation:The amount of a radioactive substance that remains after a certain amount of time can be calculated using the decay constant. For cesium-137, the decay constant is 0.0871 per year. To determine the amount remaining after 60 years, we can use the formula:
Amount remaining = initial amount * e^(-decay constant * time)
Substituting the values, we get:
Amount remaining = 20mg * e^(-0.0871 * 60) = 20mg * e^(-5.226) ≈ 0.237mg. Therefore, approximately 0.237 milligrams of cesium-137 would remain after 60 years.
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Use the drop-down menus to complete the statements. investigations allow for the control of variables and can be repeated. investigations are usually less time-consuming and less expensive. investigations make it possible to study a wide range of variables.
Answer:
Exper
des
com
Explanation:
Final answer:
Experimental investigations focus on manipulating one variable and controlling others to determine effects, while descriptive investigations observe natural occurrences without manipulation. Field experiments modify a variable in a natural environment with some control over extraneous factors.
Explanation:
In scientific investigations, it is essential to understand the roles of different types of variables and controls. Experimental investigations allow for the control of variables to ensure that only one variable is manipulated, which is the independent variable. This isolation helps in distinguishing the direct effects of the manipulation on the dependent variable, which is being measured and recorded.
Field investigations provide an opportunity to study phenomena in a natural setting, where controlling all extraneous variables is not always feasible. However, field experiments can also be conducted where one independent variable is intentionally altered while attempting to control extraneous factors, thus achieving a balance between external and internal validity.
Lastly, there are observational studies or descriptive investigations which do not manipulate variables, but rather observe and record variables as they naturally occur. These are typically less expensive, less time-consuming, and can encompass a wide range of variables, although they often lack the control of experimental studies.
Logical steps to do the investigation involve identifying the independent, dependent, and controlled variables, establishing controls and a control group if applicable, and following an experimental procedure that ensures repeatability and reliability of the results.
Calculate the pressure exerted by 66.0 g of CO2 gas at -14.5°C that occupies a volume of 50.0 L
The pressure exerted by 66.0 g of CO₂ gas at -14.5°C that occupies a volume of 50.0 L is 0.636 atm.
How do we calculate pressure?Pressure of any gas will be calculated by using the ideal gas equation as:
PV = nRT, where
P = pressure of gas = ?
V = volume of gas = 50L
R = universal gas constant = 0.082 L.atm/K.mol
T = temperature of gas = -14.5°C = 258.65 K
n is moles of gas and it will be calculated as:
n = W/M, where
W = given mass of CO₂ = 66g
M = molar mass of CO₂ = 44 g/mol
n = 66/44 = 1.5 moles
On putting values we get
P = (1.5)(0.082)(258.65) / (50)
P = 0.636 atm
Hence required pressure is 0.636 atm.
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To calculate the pressure exerted by a gas, we can use the Ideal Gas Law equation and convert the given values to the appropriate units. Then, plug these values into the equation to calculate the pressure.
Explanation:To calculate the pressure exerted by a gas, we can use the Ideal Gas Law equation:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the equation to solve for pressure, we have:
P = (nRT) / V
In this case, we are given the mass of CO2 gas (66.0 g), the volume (50.0 L), and the temperature in Celsius (-14.5°C). First, we need to convert the mass to moles using the molar mass of CO2.
Next, we need to convert the temperature to Kelvin by adding 273.15. Once we have the number of moles and the temperature in Kelvin, we can plug these values into the equation to calculate the pressure.
Consider the reaction 2H2S(g)⇌2H2(g)+S2(g),Kp=2.4×10−4 (at 1073 K) A reaction mixture contains 0.111 atm of H2, 0.051 atm of S2, and 0.566 atm of H2S. Determine how these conditions compare to equilibrium conditions. Match the words in the left column to the appropriate blanks in the sentences on the right.
Answer:
Q> Kp
The reaction of the system, will be a shift to the left, the side of the reactants.
Explanation:
Step 1: Data given
Kp = 2.4 * 10^-4
Partial pressure H2 = 0.111 atm
Partial pressure S2 = 0.051 atm
Partial pressure H2S = 0.566 atm
Step 2: The balanced equation
2H2S(g) ⇌ 2H2(g) + S2(g)
Step 3: Calculate Q
Q = (pS2 * (pH2)²) / (pH2S)²
Q = (0.051 * 0.111²) / (0.566²)
Q = 0.00196 =1.96 *10^-3
Q> Kp
Since Q>K, we have more products than reactants (pressure). The reaction of the system, will be a shift to the left, the side of the reactants.
To compare the given conditions to the equilibrium conditions for the reaction, you calculate the reaction quotient (Qp) with the given pressures. Because Qp is less than Kp, the reaction isn't in equilibrium and will shift towards the products to get there.
Explanation:To determine how these conditions compare to equilibrium conditions for the reaction 2H2S(g)⇌2H2(g)+S2(g), you first need to calculate the reaction quotient Qp at the given conditions. You use the equation Qp = [H2]^2*[S2]/[H2S]^2. Insert the given pressures into the equation, so Qp becomes (0.111)^2*(0.051)/(0.566)^2 = 1.13x10^-4.
Since Kp = 2.4*10^-4, and Qp < Kp, the system is not at equilibrium and the reaction will shift towards the products [H2] and [S2] in order to reach equilibrium.
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What is the total amount of kinetic and potential energy of a substance?
Answer:
I THINK mechanical energy
A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next
few minutes?
O Molecules in both the metal and the surrounding air will start moving at lower speeds.
O Molecules in both the metal and the surrounding air will start moving at higher speeds.
O The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up.
O The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Answer:
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Explanation:
Because the heat of the plate will be releases warming up the air making it move faster
Convert 26.02 x 1023 molecules of C2H8 to grams. Round your answer to the hundredths place.
Answer:
x= 138.24 g
Explanation:
We use the avogradro's number
6.023 x 10^23 molecules -> 1 mol C2H8
26.02 x 10^23 molecules -> x
x= (26.02 x 10^23 molecules * 1 mol C2H8 )/6.023 x 10^23 molecules
x= 4.32 mol C2H8
1 mol C2H8 -> 32 g
4.32 mol C2H8 -> x
x= (4.32 mol C2H8 * 32 g)/ 1 mol C2H8
x= 138.24 g
The correct answer is 156.69 * 10^46 grams.
How to convert molecules to grams?
To convert from molecules to grams, it is necessary to first convert the number of molecules of a substance by dividing by Avogadro’s number to find the number of moles, and then multiply the number of moles by the molar mass of this substance.Avogadro’s number is given as 6.022 x 10^23
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Evaluate each scenario described to determine the direction of heat flow.
ice cube to tap water
tap water to ice cube
Answer: tap water to ice cube
Explanation:
Answer: tap water to ice cube
Explanation:
How much heat (in Joules) will be needed to vaporize 18.015 grams of liquid water at 100°C?
Answer:
40659.855 J
Explanation:
From the question given above, we obtained the following:
Mass (m) = 18.015g
Heat of vaporisation (ΔHv) = 2257 J/g
Heat (Q) =?
The heat required to vaporise the water can be calculated as follow:
Q = mΔHv
Q = 18.015 x 2257
Q = 40659.855 J
Therefore, the heat required to vaporise the water is 40659.855 J
A chemist titrates 80.0mL of a 0.3184M pyridine C5H5N solution with 0.5397M HBr solution at 25°C . Calculate the pH at equivalence. The pKb of pyridine is 8.77.
Answer:pH = 2.96
Explanation:
C5H5N + HBr --------------> C5H5N+ + Br-
millimoles of pyridine = 80 x 0.3184 =25.472mM
25.472 millimoles of HBr must be added to reach equivalence point.
25.472 = V x 0.5397
V =25.472/0.5397= 47.197 mL HBr
total volume = 80 + 47.197= 127.196 mL
Concentration of [C5H5N+] = no of moles / volume=
25.472/ 127.196= 0.20M
so,
pOH = 1/2 [pKw + pKa + log C]
pKb = 8.77
pOH = 1/2 [14 + 8.77 + log 0.20]
pOH = 11.0355
pH = 14 - 11.0355
pH = 2.96
Final answer:
To calculate the pH at equivalence, we need to determine the concentration of pyridine and its conjugate acid. The pH at equivalence can be calculated using the Henderson-Hasselbalch equation, which relates the pH to the pKa and the concentration of the conjugate acid and base. The pKa value for the pyridinium ion can be determined by subtracting the pKb of pyridine from the pKw.
Explanation:
To calculate the pH at equivalence, we need to determine the concentration of pyridine and its conjugate acid. From the given information, we know that the initial volume of pyridine solution is 80.0 mL and its concentration is 0.3184 M. We also have the concentration of HBr solution, which is 0.5397 M. The reaction between pyridine and HBr is:
C5H5N (aq) + HBr (aq) → C5H5NH+Br- (aq)
This reaction forms the pyridinium ion (C5H5NH+) which is the conjugate acid of pyridine. At equivalence, the moles of pyridine and pyridinium ion are equal. Using the stoichiometry of the reaction, we can calculate the number of moles of pyridine:
Moles of pyridine = Volume of pyridine solution * Concentration of pyridine = 80.0 mL * 0.3184 M = 25.472 moles
Since the reaction is 1:1, the moles of pyridine also correspond to the moles of pyridinium ion. Therefore, the concentration of pyridinium ion is:
Concentration of pyridinium ion = Moles of pyridinium ion / Volume of pyridinium ion solution = 25.472 moles / 80.0 mL = 0.3184 M
Now, we can use the Henderson-Hasselbalch equation to calculate the pH at equivalence:
pH = pKa + log10 ([A-] / [HA])
Given that the pKb of pyridine is 8.77, we can determine the pKa of pyridinium ion:
pKa = 14.00 - pKb = 14.00 - 8.77 = 5.23
Substituting the values into the Henderson-Hasselbalch equation:
pH = 5.23 + log10 (0.3184 / 0.3184) = 5.23 + 0 = 5.23
Therefore, the pH at equivalence is 5.23.
A compound contains only carbon, hydrogen, and oxygen. Combustion of 139.1 g of the compound yields 208.6 g of CO2 and 56.93 g of H2O. The molar mass of the compound is 176.1 g/mol. *Each part of this problem should be submitted separately to avoid losing your work* 1. Calculate the grams of carbon (C) in 139.1 g of the compound: grams 2. Calculate the grams of hydrogen (H) in 139.1 g of the compound. grams 3. Calculate the grams of oxygen (O) in 139.1 g of the compound. grams
Answer:
1. Mass of Carbon is 56.89g
2. Mass of Hydrogen is 6.33g
3. Mass of Oxygen is 75.88
Explanation:
The following were obtained from the question.
Mass of the compound = 139.1g
Mass of CO2 produced = 208.6g
Mass of H2O produced = 56.93
1. Determination of mass of Carbon (C). This is illustrated below:
Molar Mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 208.6
Mass of C = 56.89g
2. Determination of the mass of Hydrogen (H). This is illustrated below:
Molar Mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 56.93
Mass of H = 6.33g
3. Determination of the mass of oxygen (O).
This is illustrated below:
Mass of the compound = 139.1g
Mass of C = 56.89g
Mass of H = 6.33g
Mass of O = Mass of compound - (mass of C + Mass of H)
Mass of O = 139.1 - (56.89 + 6.33)
Mass of O = 139.1 - 63.22
Mass of O = 75.88
Consider your experimental results from Part A of this lab. Suppose your strongest reducing agent were added to your strongest oxidizing agent. (Use the lowest possible coefficients. Omit states-of-matter from your answers.) (a) Write the half-reaction for your strongest reducing agent. chemPadHelp (b) Write the half-reaction for your strongest oxidizing agent. chemPadHelp (c) Note the number of electrons in each half reaction. In order to balance the number of electrons lost and gained, the oxidation half-reaction must be multiplied by and the reduction half-reaction must be multiplied by (d) Write the net redox reaction. chemPadHelp
Answer:
See Explaination
Explanation:
We can define an oxidizing agent as a reactant that removes electrons from other reactants during a redox reaction. The oxidizing agent typically takes these electrons for itself, thus gaining electrons and being reduced. An oxidizing agent is thus an electron acceptor.
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In this example, the strongest reducing agent is Aluminum (Al) and the strongest oxidizing agent is the dichromate ion (Cr₂O₇²¯). They both involve 6 electrons in their respective half-reactions. Balancing and summing these half-reactions give the net redox reaction.
Explanation:To answer your experimental lab results query about redox reactions, we first identify our strongest oxidizing agent and reducing agent. An oxidizing agent is a substance that tends to oxidize other substances, meaning it is reduced. Conversely, a reducing agent reduces other substances while being oxidized itself.
Your strongest reducing agent might be represented by this half-reaction: 2Al(s) → 2Al³+ + 6e⁻, and your strongest oxidizing agent might be represented by: Cr₂O₇²¯ + 14H⁺ + 6e⁻ → 2Cr³+ + 7H₂0. Here, aluminum (Al) is losing electrons, so it's oxidized, and dichromate ion (Cr₂O₇²¯) is gaining electrons, so it's reduced.
So, identifying the number of electrons involved: Reducing half-reaction (Aluminum) = 6 electrons, Oxidizing half-reaction (Chromium) = 6 electrons. As the number of electrons in both half-reactions is equal, no multiplication is needed to balance them.
Now, combining these half-reactions to form the net redox reaction gives us: 2Al(s) + Cr₂O₇²¯ + 14H⁺ → 2Al³+ + 2Cr³+ + 7H₂0.
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Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the following experimental values of virial coefficients: B = −152.5 cm3·mol−1 C = −5800 cm6·mol−2 (b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)]. (c) The steam tables (App. E).
Answer:
Explanation:
Given that:
the temperature [tex]T_1[/tex] = 250 °C= ( 250+ 273.15 ) K = 523.15 K
Pressure = 1800 kPa
a)
The truncated viral equation is expressed as:
[tex]\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}[/tex]
where; B = - [tex]152.5 \ cm^3 /mol[/tex] C = -5800 [tex]cm^6/mol^2[/tex]
R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹
Plugging all our values; we have
[tex]\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}[/tex]
[tex]4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}[/tex]
Multiplying through with V² ; we have
[tex]4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0[/tex]
[tex]4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0[/tex]
V = 2250.06 cm³ mol⁻¹
Z = [tex]\frac{PV}{RT}[/tex]
Z = [tex]\frac{1800*2250.06}{8.314*10^3*523.15}[/tex]
Z = 0.931
b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].
The generalized Pitzer correlation is :
[tex]T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345[/tex]
[tex]T__{\gamma}} = \frac{T}{T_c}[/tex]
[tex]T__{\gamma}} = \frac{523.15}{647.1}[/tex]
[tex]T__{\gamma}} = 0.808[/tex]
[tex]P__{\gamma}} = \frac{P}{P_c}[/tex]
[tex]P__{\gamma}} = \frac{1800}{22055}[/tex]
[tex]P__{\gamma}} = 0.0816[/tex]
[tex]B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}[/tex]
[tex]B_o = 0.083 - \frac{0.422}{0.808^{1.6}}[/tex]
[tex]B_o = 0.51[/tex]
[tex]B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}[/tex]
[tex]B_1 = -0.282[/tex]
The compressibility is calculated as:
[tex]Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}[/tex]
[tex]Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}[/tex]
Z = 0.9386
[tex]V= \frac{ZRT}{P}[/tex]
[tex]V= \frac{0.9386*8.314*10^3*523.15}{1800}[/tex]
V = 2268.01 cm³ mol⁻¹
c) From the steam tables (App. E).
At [tex]T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa[/tex]
V = 0.1249 m³/ kg
M (molecular weight) = 18.015 gm/mol
V = 0.1249 × 10³ × 18.015
V = 2250.07 cm³/mol⁻¹
R = 729.77 J/kg.K
Z = [tex]\frac{PV}{RT}[/tex]
Z = [tex]\frac{1800*10^3 *0.1249}{729.77*523.15}[/tex]
Z = 0.588
Final answer:
To determine Z and V for steam at 250°C and 1800 kPa, we can use the truncated virial equation with the given experimental values of the virial coefficients, or we can use the generalized Pitzer correlation to obtain the B value and then use the truncated virial equation. Alternatively, we can look up the values in the steam tables.
Explanation:
The question asks us to determine Z and V for steam at 250°C and 1800 kPa using three different methods: (a) The truncated virial equation with given experimental values of the virial coefficients, (b) The truncated virial equation with B value obtained using the generalized Pitzer correlation, and (c) The steam tables.
(a) To determine Z and V using the truncated virial equation with B and C values, we substitute the given temperature and pressure into the equation and solve for Z and V.
(b) To determine Z and V using the truncated virial equation with the B value obtained from the generalized Pitzer correlation, we substitute the given temperature and pressure into the equation and solve for Z and V.
(c) To determine Z and V using the steam tables, we look up the values for Z and V at the given temperature and pressure.
What additional information is needed to solve this problem: If a sample of a gas 12.0 o C and 1.06 atm pressure is moved to a 2.30 L container at 24.9 0 C, what is the final pressure of the gas?
Answer:
The additional information required to solve this problem is the initial volume.
the final pressure P₂ of the gas is 1.108 atm
Explanation:
Given that :
A sample of gas at initial temperature [tex]T_1 = 12.0^0 \ C[/tex] = (12+273)K = 285 K
Pressure (P₁) = 1.06 atm
Initial Volume (V₁) = unknown ???
Final Volume (V₂) = 2.30 L
final temperature [tex]T_2 = 24.9^0 \ C[/tex] = (24.9 +273)K = 297.9 K
Find the final Pressure (P₂)
The relation between: Pressure, Volume and Temperature can be gotten from the ideal gas equation :
PV = nRT
The Ideal Gas Equation is also reduced to the General Gas Law or the combined Gas Law by assuming that n= 1 .
From ; PV = nRT
[tex]\frac{PV}{T} = R \ \ ( constant) \ if \ n=1[/tex]
∴ [tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = \frac{P_3V_3}{T_3}...= \frac{P_nV_n}{T_n} \ \ \ ( n \ constant)[/tex]
The additional information required to solve this problem is the initial volume.
This expression is a combination of Boyle's Law and Charles Law. From the combined Gas Law , it can be deduced that at constant volume, the pressure of a given mass(mole) of gas varies directly with absolute temperature.
∴ [tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex] if n & Volume (V) are constant .
[tex]P_2 = \frac{1.06*297.9}{285}[/tex]
P₂ = 1.108 atm
Thus, the final pressure P₂ of the gas is 1.108 atm
A basketball has a volume of 7.1 L at a temperature of 300 k what is the volume of the basketball at 273 k
Answer:
The volume of the basketball at 273 K is 6.461 L
Explanation:
When the gas temperature increases, the molecules move faster and take less time to reach the walls of the container. This means that the number of crashes per unit of time will be greater. That is, there will be an increase in the pressure inside the container and the volume will increase.
Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. So, the ratio between volume and temperature will always have the same value:
[tex]\frac{V}{T} =k[/tex]
In this case, you know:
V1= 7.1 LT1= 300 KV2= ?T2= 273 KReplacing:
[tex]\frac{7.1 L}{300K} =\frac{V2}{273 K}[/tex]
Solving:
[tex]V2=\frac{7.1 L}{300K} *273 K[/tex]
V2= 6.461 L
The volume of the basketball at 273 K is 6.461 L
According to Charles's law, the volume of a gas is directly proportional to its temperature at constant pressure. Using the formula V1/T1 = V2/T2, we can calculate the volume of the basketball at 273 K.
Explanation:According to Charles's law, the volume of a gas is directly proportional to its temperature at constant pressure. We can use the formula V1/T1 = V2/T2 to solve this problem.
Given:
V1 = 7.1 LT1 = 300 KT2 = 273 KV2 = ?Using the formula, we can substitute the values and solve for V2:
V1/T1 = V2/T2
7.1 L / 300 K = V2 / 273 K
Cross multiplying, we get:
V2 = (7.1 L * 273 K) / 300 K
V2 = 6.46 L (rounded to two decimal places)
Which of the following has nonvolatile bonds
Answer:
I can provide a proper answer since there are no bonds specified.
Explanation:
Need help with this chemistry problem
Answer:
40g/mol
Explanation:
Step 1:
Data obtained from the question. This includes the following:
Volume (V) = 500mL = 500/1000 = 0.5L
Mass of gass = 1g
Temperature (T) = – 23°C = – 23°C + 273 = 250K
Pressure (P) = 105 KPa = 105/101.325 = 1.04 atm
Number of mole (n) =?
Gas constant (R) = 0.082atm.L/Kmol
Step 2:
Determination of the number of mole of the gas.
With the ideal gas equation, the number of mole of the gas can be obtained as follow:
PV = nRT
Divide both side by RT
n = PV / RT
n = (1.04 x 0.5)/(0.082 x 250)
n = 0.025mole
Step 3:
Determination of the molar mass of the gas:
This is illustrated below:
Mass of the gas = 1g
Number of mole of the gas = 0.025mole
Molar Mass of the gas =..?
Number of mole = Mass /Molar Mass
Molar Mass = Mass /number of mole
Molar Mass of the gas = 1/0.025
Molar Mass of the gas = 40g/mol
Therefore, the molar mass of the gas is 40g/mol
Calculate the cell potential for the reaction as written at 25.00 C given that [Cr2+ ]=0.862 M and [Fe2+ ]=0.0140M Use the standard reduction potentials in this table.
Answer:
0.497 V
Explanation:
We need to apply the Nernst equation here. According to the Nernst equation;
Ecell= E°cell - 0.0592/n log Q
Where;
Ecell= emf of the cell under the given conditions
E°cell= standard emf of the cell
n= number of electrons transferred
Q= reaction quotient= [products]/[Reactants]= [Cr^2+]/[Fe^2+]
Balanced redox reaction equation; Cr(s)+Fe2+(aq)---------->Cr2+(aq)+Fe(s)
Values of standard electrode potential
Fe II: -0.44 V
Cr II: -0.91 V
E°cell= (-0.44) - (-0.99)
E°cell= 0.55V
[Fe2+ ]=0.0140M
[Cr2+ ]=0.862 M
Number of electrons transferred (n)= 2
Substituting into the Nernst's equation;
Ecell= 0.55- 0.0592/2 log [0.862]/[0.0140]
Ecell= 0.55 - 0.053
Ecell= 0.497 V
Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO(s) + H2O(l) → Ca(OH)2(s) A 5.00-g sample of CaO is reacted with 4.83 g of H2O. How many grams of water remain after the reaction is complete? Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO(s) + H2O(l) → Ca(OH)2(s) A 5.00-g sample of CaO is reacted with 4.83 g of H2O. How many grams of water remain after the reaction is complete? 3.22 0.00 0.179 0.00991 1.04
We calculated that after the 5.00 g of calcium oxide reacts with water to produce calcium hydroxide, the rest of the water left is 3.23g.
Explanation:To solve this question, we first need to understand the stoichiometry of the reaction. The balanced equation clearly illustrates that one mole of calcium oxide (CaO) reacts with one mole of water (H2O) to form calcium hydroxide (Ca(OH)2). Therefore, this reaction is a 1:1 ratio.
Next, we need to convert the grams to moles. Since the molar mass of CaO is approximately 56.08 g/mol and H2O is 18.015 g/mol, we can calculate that 5.00 g of CaO is around 0.089 mol and 4.83 g of H2O is approximately 0.268 mol.
Considering the stoichiometry of the reaction, it is clear that not all of the water will react because it is present in excess. Only an amount equivalent to the moles of CaO will, so 0.089 mol of H2O will react. To convert this back to grams, simply multiply the moles of water reacted by the molar mass of water. This is approximately 1.60g.
Then, subtract the amount of water used in the reaction from the original amount to get the amount of water left. Hence, 4.83 g - 1.60 g = 3.23 g of water remain after the reaction is complete.
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Which of the following statements about bonding and hybridization is INCORRECT? (Select ALL incorrect statements) Group of answer choices Hybridization does not account for observed bond angles in molecules Single bonds are always pi bonds The length of a bond is determined by where the energy of the system is at its lowest point Multiple bonds always have a combination of sigma and pi bonds Pi bonds are always between unhybridized p orbitals
Answer:
-Hybridization does not account for observed bond angles in molecules.
-Single bonds are always pi bonds.
-The length of a bond is determined by where the energy of the system is at its lowest point
Explanation:
-The very first statement is incorrect because it does account for different bong angles as the hybrid orbitals are responsible for contributing for bond angles in a way that more the hybrid orbitals present the lesser the angles it forms.
-The second statement is incorrect because single bonds are considered as sigma bonds and not a pi bond.
-The third statement is incorrect because hybridization is responsible for deciding the bond length.
The incorrect statements about bonding and hybridization are that single bonds are always pi bonds and pi bonds are always between unhybridized p orbitals.
Explanation:The incorrect statements about bonding and hybridization are:
Single bonds are always pi bonds.Pi bonds are always between unhybridized p orbitals.Hybridization does not account for observed bond angles in molecules, so this statement is correct. The length of a bond is determined by where the energy of the system is at its lowest point, so this statement is also correct. Multiple bonds can have a combination of sigma and pi bonds, so this statement is correct as well.
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