Final answer:
To calculate the mass of dry ice needed to completely sublime in water, we can use the concept of stoichiometry and the ideal gas law equation. The moles of CO2 can be determined using the ideal gas law equation, and then converted to moles of water using the stoichiometry of the balanced equation. Finally, the mass of dry ice can be calculated using the moles of CO2 and the molar mass of CO2.
Explanation:
In order to calculate the mass of dry ice that should be added to the water, we need to use the concept of stoichiometry. The equation for the sublimation of carbon dioxide is CO2(s) -> CO2(g). We can use the molar ratio between CO2(s) and CO2(g) to convert the volume of water to moles of CO2. From there, we can use the molar mass of CO2 to calculate the mass of dry ice needed.
First, we need to calculate the moles of CO2 that can be produced using the ideal gas law equation:
n = PV / RT
where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature. Since temperature and pressure are not given, we cannot directly solve for moles of CO2. However, we can assume that the pressure and temperature remain constant throughout the process, so the ideal gas equation can still be used to find the moles of CO2.
The next step is to use stoichiometry to convert moles of CO2 to moles of water. The balanced equation is:
CO2(s) + H2O(l) -> H2CO3(aq)
From the balanced equation, we can see that the mole ratio between CO2 and water is 1:1. Therefore, if we know the moles of CO2, we also know the moles of water. The molar mass of CO2 is 44.01 g/mol, so we can calculate the mass of dry ice needed using the equation:
mass = moles * molar mass
Since we know the volume of water (15.0 L) and the final temperature (28°C), we can plug those values into the ideal gas equation and calculate the moles of CO2. Then, we can use the moles of CO2 to calculate the mass of dry ice needed. The final mass will depend on the pressure and temperature at which the dry ice is added, as well as the final temperature of the water.
Final answer:
The student's question involves calculating the mass of dry ice needed to produce fog in water until the temperature drops to 28°C. To perform this calculation, details such as the heat capacity of water, the heat of sublimation of CO2, and the complete temperature value are critical, but these are not provided in the question.
Explanation:
Dry ice, known chemically as solid carbon dioxide (CO2), sublimes at room temperature and pressure, meaning it converts directly from solid to gas. In the context of the question, when dry ice is added to warm water, the heat transfer from the water to the dry ice accelerates this sublimation process. The interaction cools the water down while providing a fog effect until either all the dry ice sublimes or the water temperature drops too much to sustain rapid sublimation.
To calculate the mass of dry ice required to create these special effects until the water cools to 28°C, one would need to determine the amount of heat the water can provide before reaching this temperature, and then use the heat of sublimation for CO2 to find the corresponding mass of dry ice. However, since the question doesn't provide specific heat capacity, heat of sublimation values, or the completion temperature value (presumably 28°C), an exact calculation cannot be performed without this information.
Handling dry ice requires caution due to its extremely cold temperature, and it is essential to be in a well-ventilated area since the resulting CO2 gas is heavier than air and can displace oxygen.
A 30.0 mL sample of hydrogen gas (H2) is collected over water at 20.00∘C and has a total pressure of 700.0 torr. The partial pressure of water vapor at 20.00∘C is 17.5 torr. Calculate the mole fraction of H2 gas in the sample.
Answer: The mole fraction of hydrogen gas at 20°C is 0.975
Explanation:
We are given:
Vapor pressure of water vapor at 20°C = 17.5 torr
Total pressure at 20°C = 700.0 torr
Vapor pressure of hydrogen gas at 20°C = (700.0 - 17.5) torr = 682.5 torr
To calculate the mole fraction of hydrogen gas at 20°C, we use the equation given by Raoult's law, which is:
[tex]p_{H_2}=p_T\times \chi_{H_2}[/tex]
where,
[tex]p_{H_2}[/tex] = partial pressure of hydrogen gas = 682.5 torr
[tex]p_T[/tex] = total pressure = 700.0 torr
[tex]\chi_{H_2}[/tex] = mole fraction of hydrogen gas = ?
Putting values in above equation, we get:
[tex]682.5torr=700.0torr\times \chi_{H_2}\\\\\chi_{H_2}=\frac{682.5}{700.0}=0.975[/tex]
Hence, the mole fraction of hydrogen gas at 20°C is 0.975
Answer: 0.975
Explanation:
First, calculate the partial pressure of hydrogen gas. The partial pressure of hydrogen gas, PH2, is the difference between the total pressure and the partial pressure of water vapor.
PH2=700.0 torr−17.5torr=682.5torr
Now, use the following equation to calculate the mole fraction of hydrogen gas.
PH2 = XH2 × Ptotal
XH2 = PH2 / Ptotal
= 682.5 torr / 700.0 torr = 0.975
The equilibrium constant for the gas phase reaction: N2O5(g) ---> 2 NO2(g) + ½ O2(g) is 95 at 25ºC. What is the value of the equilibrium constant for the following reaction at 25ºC?
O2(g) + 4 NO2(g) ---> 2 N2O5(g)
1/95
(95)^2
1/(95)^2
(95)^½
Answer: The value of equilibrium constant for reverse reaction is [tex](\frac{1}{95})^2[/tex]
Explanation:
The given chemical equation follows:
[tex]N_2O_5(g)\rightarrow 2NO_2(g)+\frac{1}{2}O_2(g)[/tex]
The equilibrium constant for the above equation is 95.
We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:
[tex]O_2(g)+4NO_2(g)\rightarrow 2N_2O_5(g)[/tex]
The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.
If the equation is multiplied by a factor of '2', the equilibrium constant of the reverse reaction will be the square of the equilibrium constant of initial reaction.
The value of equilibrium constant for reverse reaction is:
[tex]K_{eq}'=(\frac{1}{95})^2[/tex]
Hence, the value of equilibrium constant for reverse reaction is [tex](\frac{1}{95})^2[/tex]
The equilibrium constant for the reaction O2(g) + 4 NO2(g) --> 2 N2O5(g) at 25ºC is (1/95)^2. This is due to Le Chatelier's principle stating that K for the reverse reaction is the reciprocal of the original one and it's squared when the reaction is doubled.
Explanation:The second reaction in your question is the reverse reaction of the first one, but also multiplied by two. According to Le Chatelier's principle, the equilibrium constant (K) of the reverse reaction is the reciprocal of the original one. So the K of the reverse of 1: N2O5(g) --> 2 NO2(g) + ½ O2(g) would be 1/95. However, the actual reaction to the question is twice that.
When you double a reaction, you square its equilibrium constant. Thus, the K for the reaction: O2(g) + 4 NO2(g) --> 2 N2O5(g) would be (1/95)^2.
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The complex [Zn(NH3)2Cl2]2+ does not exhibit cis-trans isomerism. The geometry of this complex must be ________.
a. tetrahedral
b. trigonal bipyramidal
c. square planar
d. octahedral
e. either tetrahedral or square planar
Answer:
A. tetrahedral
Explanation:
The only complex ion geometries that can exhibit cis-trans isomerim are the octahedral and square planar.With the above information in mind, the correct answer would then have to be either a) or b), however the complex [Zn(NH₃)₂Cl₂]²⁺ has 4 ligands. This means that the correct answer is a), because the trigonal bipyramidal geometry has 5 ligands.
Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How many milliliters of 5.50 M HCl (aq) are required to react with 6.25 g Zn (s) ?
Answer:
34.7mL
Explanation:
First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.
So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn
All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.
So now we need to go to HCl!
To do that we multiply by the molar coefficients in the chemical equation:
[tex]\frac{0.0956g Zn}{1 } (\frac{2 mol HCl}{1molZn})[/tex]
This leaves us with 2(0.0956) = 0.1912 mol HCl
Now we use the relationship M= moles / volume , to calculate our volume
Rearranging we get that V = moles / M
Now we plug in: V = 0.1912 mol HCl / 5.50 M HCl
V= 0.0347 L
To change this to milliliters we multiply by 1000 so:
34.7 mL
A chemist prepares a solution of mercury(II) iodide HgI2 by measuring out 0.0067μmol of mercury(II) iodide into a 350.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's mercury(II) iodide solution. Round your answer to 2 significant digits.
Final answer:
The concentration of the chemist's mercury(II) iodide solution is 0.019 μmol/L after rounding to two significant digits.
Explanation:
The concentration of the chemist's mercury(II) iodide solution can be calculated using the formula for molarity, which is the number of moles of solute divided by the volume of the solution in liters.
In this case, we are given 0.0067μmol of HgI2 and a volume of 0.350 L:
Molarity (μM) = amount of solute (in micromoles) / volume of solution (in liters)Molarity (μM) = 0.0067μmol / 0.350 L = 0.01914μmol/L
After rounding to two significant digits, the concentration of the solution is 0.019 μM.
The decomposition reaction of N2O5 in carbon tetrachloride is 2N2O5−→−4NO2+O2. The rate law is first order in N2O5. At 64 °C the rate constant is 4.82×10−3s−1. (a) Write the rate law for the reaction. (b) What is the rate of reaction when [N2O5]=0.0240M? (c) What happens to the rate when the concentration of N2O5 is doubled to 0.0480 M? (d) What happens to the rate when the concentration of N2O5 is halved to 0.0120 M?
Answer:
(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]
(b) rate = 1.16 x 10⁻⁴ M/s
(c) rate = 2.32 x 10⁻⁴ M/s
(d) rate = 5.80 x 10⁻⁵ M/s
Explanation:
We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so
(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]
(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s
(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s
(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to
1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s
Answer:
a) r = 4.82x10⁻³*[N2O5]
b) 1.16x10⁻⁴ M/s
c) The rate is doubled too (2.32x10⁻⁴ M/s)
d) The rate is halved too (5.78x10⁻⁴ M/s)
Explanation:
a) The rate law of a generic reaction (A + B → C + D) can be expressed by:
r = k*[A]ᵃ*[B]ᵇ
Where k is the rate constant, [X] is the concentration of the compound X, and a and b are the coefficients of the reaction (which can be different from the ones of the chemical equation).
In this case, there is only one reactant, and the reaction is first order, which means that a = 1. So, the rate law is:
r = k*[N2O5]
r = 4.82x10⁻³*[N2O5]
b) Substituing the value of the concentration in the rate law:
r = 4.82x10⁻³*0.0240
r = 1.16x10⁻⁴ M/s
c) When [N2O5] = 0.0480 M,
r = 4.82x10⁻³*0.0480
r = 2.32x10⁻⁴ M/s
So, the rate is doubled too.
d) When [N2O5] = 0.0120 M,
r = 4.82x10⁻³*0.0120
r = 5.78x10⁻⁴ M/s
So, the rate is halved too.
Gaseous cyclobutene undergoes a first-order reaction to form gaseous butadiene. At a particular temperature, the partial pressure of cyclobutene in the reaction vessel drops to one-eighth its original value in 124 seconds. What is the half-life for this reaction at this temperature?
Answer:
41.3 minutes
Explanation:
Since the reaction is a first order reaction, therefore, half life is independent of the initial concentration, or in this case, pressure.
[tex]t_{1/2}= \frac{0.693}{K}[/tex]
So, fraction of original pressure = [tex]\frac{1}{2}^2[/tex]
n here is number of half life
therefore, [tex]\frac{1}{8}= \frac{1}{2}^3[/tex]
⇒ n= 3
it took 124 minutes to drop pressure to 1/8 of original value, half life = 124/3= 41.3 minutes.
Answer : The half-life of this reaction at this temperature is, 41.5 seconds.
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = ?
t = time passed by the sample = 124 s
a = let initial amount of the reactant = X
a - x = amount left after decay process = [tex]\frac{1}{8}\times (X)=\frac{X}{8}[/tex]
Now put all the given values in above equation, we get
[tex]k=\frac{2.303}{124s}\log\frac{X}{(\frac{X}{8})}}[/tex]
[tex]k=0.0167s^{-1}[/tex]
Now we have to calculate the half-life.
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]t_{1/2}=\frac{0.693}{0.0167s^{-1}}[/tex]
[tex]t_{1/2}=41.5s[/tex]
Therefore, the half-life of this reaction at this temperature is, 41.5 seconds.
Consider an AB3 molecule in which A and B differ in elec-tronegativity. You are told that the molecule has an over-all dipole moment of zero. Which of the following could be the molecular geometry of the molecule? (a) Trigonal pyramidal (b) Trigonal planar (c) T-shaped (d) Tetrahedral (e) More than one of the above
Answer:
(b) Trigonal planar
Explanation:
The molecular geometry is the one that stabilizes better the bonds and the free electron pairs. If the molecule is nonpolar (overall dipole moment zero), so, there's no free electron pairs at the central atom. So, the molecule has the central atom A surrounded by three atoms of B, which is the trigonal planar geometry.
In an AB3 molecule, where A and B differ in electronegativity, for the molecule to have an overall dipole moment of zero, the molecular shape must be calculated in a way that the polar bonds sum up to zero. The correct molecular shape in this case is Trigonal Planar. In this geometry, all three B atoms are symmetrically distributed in a plane around the atom A (120° apart), which leads to cancellation of dipole moments.
Explanation:In order for a molecule with disparity in electronegativity (polar bonds) to have an overall dipole moment of zero, the molecular structure needs to be in such a way that the dipole vectors cancel each other out. This typically happens when the molecule is symmetric. In the case of an AB3 molecule where A and B differ in electronegativity, the geometrical structures under consideration that could meet this condition are: Trigonal pyramidal, Trigonal planar, T-shaped, and Tetrahedral.
Upon reviewing these, the correct answer is (b) Trigonal planar. This is because this geometry allows for all three B atoms to be distributed in a plane around the A atom symmetrically (120° apart), allowing the dipole moments from the polar bonds to cancel each other out, which results in an overall zero dipole moment. The other geometries do not allow such cancellation, hence they can't be the correct answer.
To understand better, trying to draw the shapes out or use molecular 3D models can help visualise the planar shape and how the dipole vectors cancel each other out.
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Select the correct value for the indicated bond angle in each of the following compounds: O-S-O angle of SO2 F-B-F angle of BF3 Cl-S-Cl angle of SCI2 O-C-O angle of CO2 F-P-F angle of PF3 H-C-H angle of CH4
The bond angles are determined by their molecular structures - SO2 is 120°, BF3 is 180°, SCl2 is 120°, CO2 is 180°, PF3 is slightly less than 109.5°, and CH4 is 109.5°.
Explanation:The bond angles in various compounds are determined by the molecule's electron-pair geometry and molecular structure. For the SO2 compound, O-S-O angle corresponds to a bent molecular structure with a bond angle of 120° (electron-pair geometry: trigonal planar). The F-B-F angle in BF3 has a linear molecular structure that leads to a bond angle of 180° (electron-pair geometry: linear). In SCl2, the Cl-S-Cl angle is 120° due to its bent structure (electron-pair geometry: trigonal planar). For CO2, the O-C-O angle is 180° because of its linear structure (electron-pair geometry: linear). In PF3, the F-P-F angle is slightly less than 109.5° because of its trigonal pyramidal structure (electron-pair geometry: tetrahedral). Finally, in CH4, the H-C-H is 109.5° as it has a tetrahedral structure (electron-pair geometry: tetrahedral).
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A mixture of carbon dioxide and hydrogen gases contains carbon dioxide at a partial pressure of 401 mm Hg and hydrogen at a partial pressure of 224 mm Hg. What is the mole fraction of each gas in the mixture?
Answer:
[tex]X_{H_2}=\0.3584[/tex]
[tex]X_{CO_2}=\0.6416[/tex]
Explanation:
According to the Dalton's law of partial pressure, the total pressure of the gaseous mixture is equal to the sum of the pressure of the individual gases.
Partial pressure of carbon dioxide = 401 mmHg
Partial pressure of hydrogen gas = 224 mmHg
Total pressure,P = sum of the partial pressure of the gases = 401 + 224 mmHg = 625 mmHg
Also, the partial pressure of the gas is equal to the product of the mole fraction and total pressure.
So,
[tex]P_{CO_2}=X_{CO_2}\times P[/tex]
[tex]X_{CO_2}=\frac{P_{CO_2}}{P}[/tex]
[tex]X_{CO_2}=\frac{401\ mmHg}{625\ mmHg}[/tex]
[tex]X_{CO_2}=\0.6416[/tex]
Similarly,
[tex]X_{H_2}=\frac{P_{H_2}}{P}[/tex]
[tex]X_{H_2}=\frac{224\ mmHg}{625\ mmHg}[/tex]
[tex]X_{H_2}=\0.3584[/tex]
Estrone, which contains only carbon, hydrogen, and oxygen, is a female sexual hormone that occurs in the urine of pregnant women. Combustion analysis of a 1.893-g sample of estrone produced 5.543g of CO2 and 1.388g H2O. The molar mass of estrone is 270.36g/mol .Find the molecular formula for estrone.Express your answer as a chemical formula.
Answer:
The formula of Estrone = [tex]C_{18}H_{22}O_2[/tex]
Explanation:
Mass of water obtained = 1.388 g
Molar mass of water = 18 g/mol
Moles of [tex]H_2O[/tex] = 1.388 g /18 g/mol = 0.07711 moles
2 moles of hydrogen atoms are present in 1 mole of water. So,
Moles of H = 2 x 0.07711 = 0.1542 moles
Molar mass of H atom = 1.008 g/mol
Mass of H in molecule = 0.1542 x 1.008 = 0.1555 g
Mass of carbon dioxide obtained = 5.543 g
Molar mass of carbon dioxide = 44.01 g/mol
Moles of [tex]CO_2[/tex] = 5.543 g /44.01 g/mol = 0.126 moles
1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,
Moles of C = 0.126 moles
Molar mass of C atom = 12.0107 g/mol
Mass of C in molecule = 0.126 x 12.0107 = 1.5133 g
Given that the Estrone only contains hydrogen, oxygen and carbon. So,
Mass of O in the sample = Total mass - Mass of C - Mass of H
Mass of the sample = 1.893 g
Mass of O in sample = 1.893 - 1.5133 - 0.1555 = 0.2242 g
Molar mass of O = 15.999 g/mol
Moles of O = 0.2242 / 15.999 = 0.01401 moles
Taking the simplest ratio for H, O and C as:
0.1542 : 0.01401 : 0.126
= 11 :1 : 9
The empirical formula is = [tex]C_9H_{11}O[/tex]
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 9×12 + 11×1 + 16= 135 g/mol
Given, Molar mass = 270.36 g/mol
So,
Molecular mass = n × Empirical mass
270.36 = n × 135
⇒ n = 2
The formula of Estrone = [tex]C_{18}H_{22}O_2[/tex]
Final answer:
To find the molecular formula of estrone, calculate the empirical formula first by finding the moles of carbon and hydrogen. Then, determine the mole ratio between carbon and hydrogen and round to the nearest whole number to get the empirical formula. Finally, divide the molar mass of estrone by the molar mass of the empirical formula to find the ratio, and round to the nearest whole number to get the molecular formula.
Explanation:
To find the molecular formula for estrone, we need to determine the empirical formula first. The empirical formula represents the simplest whole-number ratio of the elements in a compound. We can use the information from the combustion analysis to calculate the empirical formula.
First, we need to determine the moles of carbon and hydrogen in the estrone sample.
Moles of carbon = mass of CO₂ / molar mass of CO₂
Moles of carbon = 5.543 g / 44.01 g/mol = 0.1259 mol
Moles of hydrogen = mass of H₂O / molar mass of H₂O
Moles of hydrogen = 1.388 g / 18.015 g/mol = 0.0770 mol
Next, we need to find the mole ratio between carbon and hydrogen by dividing the moles of each element by the smallest mole value.
Dividing both moles of carbon and hydrogen by 0.0770 mol, we get the ratio 1.6367 : 1.
Finally, we round the ratio to the nearest whole number to obtain the empirical formula. In this case, the empirical formula is C2H2O.
To find the molecular formula, we need to know the molar mass of estrone. The molar mass of estrone is given as 270.36 g/mol. Dividing the molar mass of estrone by the molar mass of the empirical formula (42.08 g/mol), we get a ratio of approximately 6.42.
Rounding the ratio to the nearest whole number, we find that the molecular formula for estrone is C6H6O3.
Write complete ionic and net ionic equations for the reaction between sulfuric acid (H2SO4) and calcium carbonate (CaCO3).
H2SO4(aq) + CaCO3(s) —> H2O(I) + CO2(g) + CaSO4(aq)
Answer:
Complete ionic equation:
2H²⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
Net ionic equation:
2H²⁺(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq)
Explanation:
Chemical equation:
H₂SO₄(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + CaSO₄(aq)
Balanced chemical equation:
H₂SO₄(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + CaSO₄(aq)
Complete ionic equation:
2H²⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
Net ionic equation:
2H²⁺(aq) + CaCO₃(s) → H₂O(l) + CO₂(g) + Ca²⁺(aq)
Answer:
Complete ionic equation: 2 H⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
Net ionic equation: 2 H⁺(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq)
Explanation:
The molecular equation includes all the species in the molecular form.
H₂SO₄(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + CaSO₄(aq)
The complete ionic equation includes all the ions and the molecular species.
2 H⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.
2 H⁺(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq)
Nitrogen became the most abundant gas in today's atmosphere because Choose one: nitrogen is highly reactive. nitrogen is relatively inert. nitrogen was the most abundant gas left over from the solar nebula. nitrogen is the only gas given off by volcanic outgassing.
Answer:
The correct answer is: 'Nitrogen is relatively inert'.
Explanation:
As, atomic number of nitrogen is 7 and its electronic distribution is 2, 5. Hence, there are 5 valence electrons present in a nitrogen atom. And, to attain stability it will gain three electrons from a donor atom. Hence, it will make a triple bond.
[tex][N]=1s^22s^22p^3[/tex]
Therefore, nitrogen has 5 valence electrons and makes 3 bonds in neutral compounds.Thus nitrogen will combine with another nitrogen atom to completes its octet to form [tex]N_2[/tex] gas molecule. Due to the presence of this triple bond, the gas molecule is almost inert as its bond dissociation energy is very high.
Hence, the correct option is:- nitrogen is relatively inert.
The statement that "the lowest energy configuration for an atom is the one having the maximum number of unpaired electrons allowed by the Pauli principle in a particular set of degenerate orbitals" is known as: A. the quantum model B. Hund's rule C. the aufbau principle D. Heisenberg uncertainty principle E. the Pauli exclusion principle
Answer:
B. Hund's rule
Explanation:
Hund's rule -
According to Hund's rule ,
As, the electrons are negatively charged and hence , like poles repel , so they repel each other in order to stabilizes themselves and hence , the electron firstly occupies a vacant orbital , before actually pairing up , so as to reduce repulsion .
The rule of maximum multiplicity states that ,
The term with the lowest energy is the one with the highest value of the spin multiplicity .
hence , the statement given in the question is about Hund's rule .
Answer:
its actually A
Explanation:
Calculate the number of free electrons per cubic meter for silver, assuming that there are 1.3 free electrons per silver atom. The electrical conductivity and density for Ag are 6.8 × 107(Ω.m)–1and 10.5 g/cm3, respectively.
Answer : The number of free electrons per cubic meter for silver are [tex]7.62\times 10^{28}m^{-3}[/tex]
Explanation :
To calculate the number of free electrons per cubic meter for silver by using the following equation:
[tex]n=1.3\times N_{Ag}[/tex]
[tex]n=1.3\times \frac{\rho_{Ag}\times N_A}{A_{Ag}}[/tex]
where,
[tex]\rho_{Ag}[/tex] = density of Ag = [tex]10.5g/cm^3[/tex]
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}atoms/mol[/tex]
[tex]A_{Ag}[/tex] = 107.87 g/mol
Now put all the given values in the above formula, we get:
[tex]n=1.3\times \frac{\rho_{Ag}\times N_A}{A_{Ag}}[/tex]
[tex]n=1.3\times \frac{(10.5g/cm^3)\times (6.022\times 10^{23}atoms/mol)}{107.87g/mol}[/tex]
[tex]n=7.62\times 10^{22}cm^{-3}[/tex]
[tex]n=7.62\times 10^{28}m^{-3}[/tex]
conversion used : [tex]1cm^{-3}=10^6m^{-3}[/tex]
Therefore, the number of free electrons per cubic meter for silver are [tex]7.62\times 10^{28}m^{-3}[/tex]
Final answer:
To calculate the number of free electrons per cubic meter for silver, we can use the density and molar mass of silver to determine the number of silver atoms per cubic meter. Since there is one free electron per silver atom, we can then calculate the number of free electrons per cubic meter.
Explanation:
To calculate the number of free electrons per cubic meter for silver, we first need to calculate the number of silver atoms per cubic meter. We can do this by using the density of silver, which is given as 10.5 g/cm³. Converting this to kg/m³ gives us a density of 10500 kg/m³. Next, we need to find the molar mass of silver, which is 107.87 g/mol. Using Avogadro's number (6.02 × 10^23 atoms/mol), we can calculate the number of silver atoms per cubic meter.
n = (10500 kg/m³) / (107.87 g/mol * (1 kg / 1000 g) * (6.02 × 10^23 atoms/mol))
n ≈ 9.48 x 10^28 atoms/m³
Since there is one free electron per silver atom, we can conclude that there are approximately 9.48 x 10^28 free electrons per cubic meter of silver.
What is the [H3O+] and the pH of a benzoic acid-benzoate buffer that consists of 0.17 M C6H5COOH and 0.42 M C6H5COONa? (Ka of benzoic acid = 6.3 × 10^−5). Be sure to report your answer to the correct number of significant figures.
Answer:
[tex][H_{3}O^{+}]=x M = 2.5\times 10^{-5}M[/tex] and pH = 4.6
Explanation:
Construct an ICE table to calculate changes in concentration at equilibrium.
[tex]C_{6}H_{5}COOH+H_{2}O\rightleftharpoons C_{6}H_{5}COO^{-}+H_{3}O^{+}[/tex]
I(M): 0.17 0.42 0
C(M): -x +x +x
E(M): 0.17-x 0.42+x x
So, [tex]\frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}=K_{a}(C_{6}H_{5}COOH)[/tex]
or, [tex]\frac{(0.42+x)x}{(0.17-x)}=6.3\times 10^{-5}[/tex]
or, [tex]x^{2}+0.4201x-(1.071\times 10^{-5})=0[/tex]
So, [tex]x=\frac{-0.4201+\sqrt{(0.4201)^{2}+(4\times 1\times 1.071\times 10^{-5})}}{(2\times 1)}M[/tex]
([tex]ax^{2}+bx+c=0\Rightarrow x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a};x< 0.17M[/tex])
So, [tex]x=2.5\times 10^{-5}[/tex]M
Hence [tex][H_{3}O^{+}]=x M = 2.5\times 10^{-5}M[/tex]
[tex]pH=-log[H_{3}O^{+}]=-logx=-log(2.5\times 10^{-5})=4.6[/tex]
The pH is 4.44, and the [H3O+] is 3.63 × 10^-5 M.
The question asks about the calculation of the hydronium ion concentration ([H3O+]) and the pH of a buffer solution consisting of benzoic acid (C6H5COOH) and sodium benzoate (C6H5COONa).
Write the equilibrium expression for benzoic acid dissociation:
C6H5COOH ↔ C6H5COO- + H3O+.
Use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA]),
where [A-] = concentration of benzoate ion and [HA] = concentration of benzoic acid.
Insert the values into the Henderson-Hasselbalch equation:
pH = 4.20 + log (0.42/0.17).
Calculate the pH.
Determine [H3O+] from pH:
[H3O+] = 10-pH.
Using the provided concentrations and the pKa of benzoic acid (4.20), the pH and [H3O+] can be calculated with high precision. For this specific buffer:
The pH is calculated to be 4.44.
The [H3O+] can then be found as 3.63 × 10-5 M.
In an aqueous mixture of aluminum, lead, and iron salts, which of these will be reduced first upon application of an electric current?
(A) Al3+ + 3e− → Al (s) Eo = −1.66V
(B) Pb2+ + 2e− → Pb (s) Eo = −0.13V
(C) Fe2+ + 2e− → Fe (s) Eo = −0.45V
(D) Fe3+ + 3e− → Fe (s) Eo = −0.036V
Answer:
D) Fe3+ + 3e− → Fe (s) Eo = −0.036V
Explanation:
In electrolysis we have to add electrical energy for redox reactions non-spontaneous because ΔºG is positive.
These 4 reduction reactions are all non-spontaneous because their reduction potentials are negative (ΔGº = - nFεº , ΔGº will be positive). So the ion most easily reduced is the least negative, in this case Fe3+ + 3e− → Fe (s) Eo = −0.036V.
the freezing point depression constants of the solvents cyclohexane and naphthalene are 20.1°C/m and 6.94°C/m respectively. Which would give a more accurate determination by freezing point depression of the molar mass of a substance that is soluble in either solvent? Why?
Explanation:
It'd be better to use cyclohexane. The possible explanation is that the freezing temperature will change by 20.1 degrees for each mole of substance added to 1 kg of cyclohexane, although the same amount added to naphthalene will change its freezing point just by 6.94 degrees.
It is so much easier to identify a larger change more adequately than a smaller one. You would actually not have a 1 molal solution in operation, so the variations in freezing points would be even smaller than the ones already described.
Final answer:
Cyclohexane, with its higher freezing point depression constant, would give a more accurate determination of molar mass than naphthalene because it results in more significant temperature changes, making precise measurements easier.
Explanation:
The determination of the molar mass of a substance using freezing point depression will be more accurate with a solvent that has a higher freezing point depression constant (Kƒ). In this case, cyclohexane, with a Kƒ of 20.1°C/m, would give a more accurate determination of molar mass compared with naphthalene, which has a Kƒ of 6.94°C/m, assuming that the substance is equally soluble in both. This is because a higher Kƒ value indicates a greater change in freezing point per mole of solute, making the temperature change easier to measure precisely.
A flask is charged with 3.00 atm of dinitrogen tetroxide gas and 2.00 atm of nitrogen dioxide gas at 25ºC and allowed to reach equilibrium.
N2O4(g)-->2 NO2(g) Kp = 0.316
Which of the following best describes the system within the flask once equilibrium has been established?
The rate of the decomposition of N2O4(g) is equal to the rate of formation of NO2(g).
The partial pressure of N2O4(g) is equal to the partial pressure of NO2(g).
The rate of the decomposition of N2O4(g) is greater than the rate of formation of NO2(g).
The rate of the decomposition of N2O4(g) is less than the rate of formation of NO2(g).
Answer:
The correct answer is The rate of decomposition of N2O4 is equal to the rate of formation of NO2.
Explanation:
According to theory of reaction kinetics we all know that at equilibrium the amount of reactant is equal to the amount of product.
In other word it can be stated that the rate of the decomposition of N2O4 is equal to the rate of the formation of NO2.
Final answer:
At equilibrium, the partial pressure of N2O4 is equal to the partial pressure of NO2 in the flask.
Explanation:
Once equilibrium has been established, the partial pressure of N2O4(g) is equal to the partial pressure of NO2(g) in the flask. This is because the equilibrium constant (Kp) expression for the decomposition of N2O4 is (PNO2)^2 / PN2O4, which means the partial pressures of the two gases are directly related. Therefore, answer choice B, 'The partial pressure of N2O4(g) is equal to the partial pressure of NO2(g),' best describes the system within the flask at equilibrium.
The following reaction is found to be at equilibrium at 25 celcius: 2SO3--->O2 + 2SO2 + 198kJ/mol.
What is the expression for the equilibrium constant, Kc?
A.) [SO3]^2/[O2][SO2]^2
B.) 2[SO3]/[O2]2[SO2]
C.) [O2][SO2]^2/[SO3]^2
D.) [O2]2[SO2]/2[SO3]
Final answer:
The expression for the equilibrium constant (Kc) for the reaction 2SO₃ --> O₂ + 2SO₂ + 198 kJ/mol is C.) [O₂][SO₂]^2/[SO₃]^2, according to the law of equilibrium.
Explanation:
The question involves understanding how to write the expression for the equilibrium constant (Kc) for a given chemical reaction. Given the reaction 2SO₃ --> O₂ + 2SO₂ + 198 kJ/mol, the correct expression for Kc reflects the concentration of products over reactants, raised to the power of their stoichiometric coefficients in the balanced equation.
The correct expression for Kc is therefore C. [O₂][SO₂]^2/[SO₃]^2. This follows from the equilibrium law, which states that Kc is calculated by taking the concentration of the products, [O₂] and [SO₂] squared (because the coefficient of SO₂ is 2), over the concentration of the reactants, [SO₃] squared (because the coefficient of SO₃ is also 2).
What is the percentage composition of each element in hydrogen peroxide, H2O2?
A)7.01% H and 92.99% O
B)7.22% H and 92.78% O
C)6.32% H and 93.68% O
D)5.88% H and 94.12% O
Answer:
The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).
Explanation:
Step 1: Data given
Molar mass of H = 1.0 g/mol
Molar mass of O = 16 g/mol
Molar mass of H2O2 = 2*1.0 + 2*16 = 34.0 g/mol
Step 2: Calculate % hydrogen
% Hydrogen = ((2*1.0) / 34.0) * 100 %
% hydrogen = 5.88 %
Step 3: Calculate % oxygen
% Oxygen = ((2*16)/34)
% oxygen = 94.12 %
We can control this by the following equation
100 % - 5.88 % = 94.12 %
The percentage composition of each element in H2O2 is 5.88% H and 94.12% O (Option D).
The percentage composition of hydrogen peroxide (H₂O₂) is D) 5.88% H and 94.12% O.
The calculation can be represented as follows:
1) Molar masses:
H = 1.008 g/mol O = 15.999 g/mol2) Molar mass of [tex]H_2O_2:[/tex]
[tex]$\ce{H_2O_2} = (2 \times 1.008) + (2 \times 15.999) = 34.014$ g/mol[/tex]3) Percentage of hydrogen:
[tex]\[ \% \ce{H} = \frac{\text{mass of H}}{\text{total mass}} \times 100\% = \frac{2 \times 1.008}{34.014} \times 100\% \approx 5.88\% \][/tex]4) Percentage of oxygen:
[tex]\[ \% \ce{O} = 100\% - \% \ce{H} = 100\% - 5.88\% = 94.12\% \][/tex]Therefore, the percentage composition of [tex]$\ce{H_2O_2}$[/tex] is approximately 5.88% H and 94.12% O.
Comparing with the given options, we can see that the closest match is D) 5.88% H and 94.12% O
Which of the following conclusions can be drawn from JJ Thomson's cathode ray experiments?A.) Atoms contain electronsB.) Practically all the mass of an atom is contained in its nucleusC.) Atoms contain protons, neutrons, and electronsD.) Atoms have positively charged nuclues surrounded by electron cloud
Answer:
A.) Atoms contain electrons
Explanation:
J.J. Thomson conducted an experiment in which he took a gas at low pressure in a discharge tube and applied high voltage current. When he did so, he noticed that there are some particles emitting from cathode going towards the anode. Then he concluded that they are negatively charged particles and coined them electrons.
Hence, out of the options:- A.) Atoms contain electrons is correct.
All of the following statements are consistent with the kinetic molecular theory of gases EXCEPT:
A. The gas molecules collide with each other and with the surfaces around them.
B. Strong attractive forces hold the gas molecules together.
C. The average kinetic energy of the molecules of a gas is proportional to the temperature of the gas in kelvins.
D. The size of the gas molecules is negligible compared to the total volume of the gas.
E. none of the above.
Answer:
B. Strong attractive forces hold the gas molecules together.
Explanation:
Kinetic molecular theory postulates:-
The gas is composed of small molecules are they are in continuous random motion and having elastic collisions with one another and also with the walls of the container. The molecules of the gas does not exert any kind of repulsive or attractive forces on each other and they their size is negligible as compared to the difference between them. Pressure exerted by the molecules of the gas results from the collisions which is happening between the molecules of the gas and the walls of the container. Average kinetic energy of molecules of the gas is directly proportional to absolute temperature.Hence, The correct option which is not a postulate of kinetic molecular theory is:- B. Strong attractive forces hold the gas molecules together.
Consider the reaction 2Na(s) + 2H2O(l)2NaOH(aq) + H2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.74 moles of Na(s) react at standard conditions. S°surroundings = J/K g
The entropy change for the surroundings in the given reaction can be calculated using the formula ΔS° surroundings = -ΔH° reaction / T.
Explanation:The entropy change for the surroundings in the given reaction can be calculated using the formula: ΔS° surroundings = -ΔH° reaction / T. In this case, the enthalpy change of the reaction is -802 kJ mol⁻¹ and the temperature is 298 K. Plugging in these values, we can calculate the entropy change for the surroundings.
ΔS° surroundings = -(-802 kJ mol⁻¹) / 298 K
ΔS° surroundings = 2.69 kJ K⁻¹ mol⁻¹
Therefore, the entropy change for the surroundings is 2.69 kJ K⁻¹ mol⁻¹.
Use the reaction:
2AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2H2O(l)
What volume of 0.123 M AgNO3(aq) is needed to form 0.657 g of Ag2SO4(s)?
(A) 53.4 mL
(B) 10.7 mL
(C) 17.1 mL
(D) 34.2 L
(E) 34.2 mL
Answer:
It’s either 34.2 OR 34.3 either would be correct!
Explanation:
The correct answer is (E) [tex]34.17\text{ mL}[/tex].
To solve this problem, we will use stoichiometry to relate the moles of [tex]AgNO_3[/tex] needed to the moles of Ag2SO4 formed. Here is the step-by-step solution:
1. First, we need to write down the balanced chemical equation:
[tex]\[ 2\text{AgNO}_3(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Ag}_2\text{SO}_4(s) + 2\text{H}_2\text{O}(l) \][/tex]
2. From the equation, we can see that [tex]2[/tex] moles of [tex]AgNO_3[/tex] produce 1 mole of [tex]Ag_2SO_4[/tex] This gives us the mole ratio:
[tex]\[ \frac{2\text{ moles AgNO}_3}{1\text{ mole Ag}_2\text{SO}_4} \][/tex]
3. Calculate the molar mass of [tex]Ag_2SO_4[/tex] using the atomic masses of Ag [tex](107.87 g/mol)[/tex] and [tex]S (32.07 g/mol)[/tex]
[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 2 \times 107.87\text{ g/mol (Ag)} + 32.07\text{ g/mol (S)} + 4 \times 16.00\text{ g/mol (O)} \][/tex]
[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 2 \times 107.87 + 32.07 + 4 \times 16.00 \][/tex]
[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 215.74 + 32.07 + 64.00 \][/tex]
[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 311.81\text{ g/mol} \][/tex]
4. Use the given mass of [tex]Ag_2SO_4[/tex] to find the moles of [tex]Ag_2SO_4[/tex] produced:
[tex]\[ \text{Moles of Ag}_2\text{SO}_4 = \frac{\text{mass of Ag}_2\text{SO}_4}{\text{molar mass of Ag}_2\text{SO}_4} \][/tex]
[tex]\[ \text{Moles of Ag}_2\text{SO}_4 = \frac{0.657\text{ g}}{311.81\text{ g/mol}} \][/tex]
[tex]\[ \text{Moles of Ag}_2\text{SO}_4 \approx 0.002097\text{ moles} \][/tex]
5. Using the mole ratio from step 2, calculate the moles of AgNO3 needed:
[tex]\[ \text{Moles of AgNO}_3 = 2 \times \text{Moles of Ag}_2\text{SO}_4 \][/tex]
[tex]\[ \text{Moles of AgNO}_3 = 2 \times 0.002097 \][/tex]
[tex]\[ \text{Moles of AgNO}_3 = 0.004194\text{ moles} \][/tex]
6. Now, we can use the molarity of AgNO3 to find the volume needed:
[tex]\[ \text{Molarity (M)} = \frac{\text{moles of solute (mol)}}{\text{volume of solution (L)}} \][/tex]
[tex]\[ \text{Volume of AgNO}_3 = \frac{\text{moles of AgNO}_3}{\text{Molarity of AgNO}_3} \][/tex]
[tex]\[ \text{Volume of AgNO}_3 = \frac{0.004194\text{ moles}}{0.123\text{ M}} \][/tex]
[tex]\[ \text{Volume of AgNO}_3 \approx 0.03417\text{ L} \][/tex]
7. Convert the volume from liters to milliliters (since 1 L = 1000 mL):
[tex]\[ \text{Volume of AgNO}_3 = 0.03417\text{ L} \times 1000\text{ mL/L} \][/tex]
[tex]\[ \text{Volume of AgNO}_3 = 34.17\text{ mL} \][/tex]
WILL MARK BRAINLIEST
What pressure (in atm and in bars) is exerted by a column of methanol (CH3OH) 305 m high? The density of methanol is 0.787 g/cm3.
Answer
23.52 atm pressure is exerted by a column of methanol
Explanation:
formula for pressure
Pressure = force / area
And we know that Force F= mass x acceleration (ma)
And mass = volume x density
And volume = cross sectional area x height?
so that...
P = F/A
Putting value of F
P = (mg)/A
Putting value for mass
P = (ρ V) g / A
Putting value for volume
P = ρ (A h) g / A
Canceling A
P = ρgh
Now add the value
now watch the units...
P = (0.787g/cm³) x 9.80 m/s² x 305m... units= g x m² / (cm³ s²)
and if you recall for pressure unit Pressure ...
1 Pa = 1 N / m² = (1 kg m / s²) / m²
so know
[g x m² / (cm³ s²)] x (1 kg / 1000g) x (100 cm / m)³
converts the units to (1 kg m² / s²) / m
= (1 kg m / s²) / m²
Addition of all values
P = (0.787g/cm³) x 9.80 m/s² x 305m x (1 kg / 1000g) x (100 cm / m)³ x (1 atm / 101325 Pa)
Pressure= 23.52 atm
A chemist prepares a solution of potassium dichromate K2Cr2O7 by measuring out 3.46μmol of potassium dichromate into a 50.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /mmolL of the chemist's potassium dichromate solution. Round your answer to 2 significant digits.
Answer: 0.069 mmol/L
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute = [tex]3.46\mu mol[/tex] = [tex]0.001\times 3.46=3.46\times 10^{-3}mmol[/tex]
1micro (µ) = 0.001 milli (m)
[tex]V_s[/tex] = volume of solution in ml= 50 ml
[tex]Molarity=\frac{3.46\times 10^{-3}\times 1000}{50ml}=0.069mmol/L[/tex]
Thus concentration in mmol/L of the chemist's potassium dichromate solution is 0.069.
Final answer:
To calculate the concentration of the potassium dichromate solution in mmol/L, convert 3.46μmol to mmol, then divide by the volume in liters, resulting in a concentration of 0.069 mmol/L.
Explanation:
The question asks to calculate the concentration of a potassium dichromate solution in millimoles per liter (mmol/L) given that 3.46μmol of potassium dichromate is dissolved in a 50 mL volumetric flask.
First, convert micromoles (μmol) to millimoles (mmol). Recall that 1μmol = 0.001 mmol:
3.46μmol = 3.46 x 0.001 = 0.00346 mmol
Next, we need to find the concentration in mmol/L. To do this, we use the formula for concentration:
Concentration (mmol/L) = (amount of solute in mmol) / (volume of solution in L)
Since the volume of the solution is 50 mL, we convert this to liters (L) because concentration is expressed in mmol/L. Recall that 1000 mL = 1 L.
50 mL = 0.05 L
Now, we can calculate the concentration:
Concentration = 0.00346 mmol / 0.05 LConcentration = 0.0692 mmol/LAfter rounding to two significant digits, the concentration is 0.069 mmol/L.
A gas obeys the ideal-gas equation of state P V = N k T , where N = n NA is the number of molecules in the volume V at pressure P and temperature T and n is the number of gm-moles of the gas. Calculate the volume occupied by 1 gm·mole of the gas at atmospheric pressure and a temperature of 313.7 K . Avogadro’s constant is 6.02214 × 1023 mol−1 , Boltzmann’s constant is 1.38065 × 10−23 N · m/K , and 1 atm = 1.013 × 105 N/m2 . Answer in units of L.
Answer:
The volume occupied is 25.7 L
Explanation:
Let's replace all the data in the formula
P . V = N . k . T
N = nNA
1 gm.mole . 6.02x10²³
k = Boltzmann's contant
T = T° in K
1 atm = 1.013 × 10⁵ N/m2
1.013 × 10⁵ N/m2 . Volume = 6.02x10²³ . 1.38065 × 10⁻²³ N · m/K . 313.7K
Volume = (6.02x10²³ . 1.38065 × 10⁻²³ N · m/K . 313.7K) / 1.013 × 10⁵ m2/N
Volume = 2607.32 N.m / 1.013 × 10⁵ m2/N = 0.0257 m³
1 dm³ = 1 L
1m³ = 1000 dm³
25.7 L
The volume of 1 gm·mole of gas at atmospheric pressure and a temperature of 313.7 K is approximately 25.7 Liters, as calculated using the Ideal Gas Law PV=nRT, with n=1, P=1 atm, T=313.7K, and R=0.08206 L·atm/mol·K.
Explanation:To calculate the volume occupied by 1 gm·mole of the gas at atmospheric pressure and a temperature of 313.7 K, you'll want to use the Ideal Gas Law (PV=nRT), where P is the pressure of a gas, V is the volume it occupies, n is the number of moles of the gas, T is the temperature, and R is the universal gas constant. We'll use the given values: n=1 gm·mole, P=1 atm, T=313.7K, and also utilize the known universal gas constant for these units R=0.08206 L·atm/mol·K.
The ideal gas equation PV=nRT can be rearranged to solve for the volume of the gas: V=nRT/P. Substituting the values, V=(1 gm·mole * 0.08206 L·atm/mol·K * 313.7K) / 1 atm. By performing the calculation, we find that the volume V is approximately 25.7 Liters.
Learn more about Ideal Gas Law here:https://brainly.com/question/30458409
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Fructose-1-P is hydrolyzed according to: Fructose-1-P + H2O → Fructose + Pi If a 0.2 M aqueous solution of Froctose-1-P is allowed to reach equilibrium, its final concentration is 6.52 × 10-5 M.
What is the standard free energy of Froctose-1-P hydrolysis?
Answer:
[tex]\Delta G^{\circ}=-15902 J/mol[/tex]
Explanation:
In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):
[tex]\Delta G^{\circ}=-R*T*ln(K_{eq})[/tex]
Being Keq:
[tex]K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}[/tex]
Initial conditions:
[tex][Fructose-1-P]=0.2M[/tex]
[tex][Fructose]=0M[/tex]
[tex][Pi]=0M[/tex]
Equilibrium conditions:
[tex][Fructose-1-P]=6.52*10^{-5}M[/tex]
[tex][Fructose]=0.2M-6.52*10^{-5}M[/tex]
[tex][Pi]=0.2M-6.52*10^{-5}M[/tex]
[tex]K_{eq}=\frac{(0.2M-6.52*10^{-5}M)*(0.2M-6.52*10^{-5}M)}{6.52*10^{-5}M}[/tex]
[tex]K_{eq}=613.1[/tex]
Free-energy for T=298K (standard):
[tex]\Delta G^{\circ}=-8.314\frac{J}{mol*K}*298K*ln(613.1)[/tex]
[tex]\Delta G^{\circ}=-15902 J/mol[/tex]
The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning the formation and decomposition of ozone in the stratosphere. Rowland and Molina hypothesized that chlorofluorocarbons (CFCs) in the stratosphere break down upon exposure to UV radiation, producing chlorine atoms. Chlorine was previously identified as a catalyst in the breakdown of ozone into oxygen gas. Using the enthalpy of reaction for two reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine with ozone. ( 1 ) ClO ( g ) + O 3 ( g ) ⟶ Cl ( g ) + 2 O 2 ( g ) Δ H ∘ rxn = − 122.8 kJ ( 2 ) 2 O 3 ( g ) ⟶ 3 O 2 ( g ) Δ H ∘ rxn = − 285.3 kJ ( 3 ) O 3 ( g ) + Cl ( g ) ⟶ ClO ( g ) + O 2 ( g ) Δ H ∘ rxn = ?
Answer:
The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.
Explanation:
[tex]ClO ( g ) + O_3 ( g )\rightarrow Cl ( g ) + 2 O_2 ( g ),\Delta H^o_{1,rxn} =-122.8 kJ [/tex]..[1]
[tex]2 O_3 ( g )\rightarrow 3O_2 ( g ),\Delta H^o_{2,rxn} = -285.3 kJ [/tex]..[2]
[tex]O_3(g) + Cl(g)\rightarrow ClO (g)+O_2(g),\Delta H^o_{3,rxn}=?[/tex]..[3]
The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:
[2] - [1] = [3]
[tex]\Delta H^o_{3,rxn}=\Delta H^o_{2,rxn}-\Delta H^o_{1,rxn}[/tex]
[tex]=-285.3 kJ-(-122.8 kJ)=162.5 kJ[/tex]
The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.