Answer:
oxidation occurs at the cathode.
Explanation:
In a voltaic cell electrons move from anode to cathode. At the anode, species give up electrons. This is an oxidation reaction depicted by the oxidation half equation. At the cathode, species accept electrons and become reduced. This is depicted by the reduction half equation. In summary; in a Voltaic cell, oxidation occurs at the anode while reduction occurs at the cathode.
In a voltaic cell, oxidation does not occur at the cathode.
A voltaic cell is an electrochemical cell that converts chemical energy to electrical energy, hence it can be used as a source of energy. It consists of two separate half-cells. There is a salt bridge between the two half cells allowing for the free flow of ions from one cell to another.
There are two electrodes, the anode undergoes oxidation and the cathode undergoes reduction. Electrons flow from the anode to the cathode.
Find out more at: https://brainly.com/question/1370699
Calculation of Molar Ratios of Conjugate Base to Weak Acid from pH For a weak acid with a pKa of 6.0, calculate the ratio of conjugate base to acid at a pH of 5.0.
Choice of Weak Acid for a Buffer Which of these com-pounds would be the best buffer at pH 5.0: formic acid (pKa 5 3.8), acetic acid (pKa 5 4.76), or ethylamine (pKa 5 9.0)? Briefly justify your answer.
Explanation:
According to the Handerson equation,
pH = [tex]pK_{a} + log \frac{\text{salt}}{\text{acid}}[/tex]
or, pH = [tex]pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}[/tex]
Putting the given values into the above equation as follows.
pH = [tex]pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}[/tex]
5.0 = 6.0 + log \frac{\text{conjugate base}}{\text{acid}}[/tex]
[tex]log \frac{\text{conjugate base}}{\text{acid}}[/tex] = -1.0
or, [tex]\frac{\text{conjugate base}}{\text{acid}} = 10^{-1.0}[/tex]
= 0.1
Therefore, we can conclude that molar ratios of conjugate base to weak acid for given solution is 0.1.
The molar ratio of conjugate base to acid for a weak acid at pH 5.0 with a pKa of 6.0 is 0.1. Among formic acid, acetic acid, and ethylamine, the best buffer at pH 5.0 would be acetic acid due to its pKa being closest to 5.0.
Explanation:The calculation of molar ratios of conjugate base to weak acid from pH and the choice of a weak acid for a buffer can be accomplished by using the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]).
Firstly, for a weak acid with a pKa of 6.0 at a pH of 5.0, the ratio of conjugate base ([A-]) to acid ([HA]) can be calculated by rearranging this equation to [A-]/[HA] = 10^(pH-pKa) = 10^(5.0-6.0) = 0.1. So the ratio of conjugate base to acid is 0.1.
Secondly, the choice of weak acid for a buffer at pH 5.0 is ideally the one with a pKa closest to the desired pH. In this case, acetic acid with a pKa of 4.76 would be the best buffer since its pKa is closest to the pH of 5.
https://brainly.com/question/32396139
#SPJ3
A solution containing 500,000 units of polymyxin B sulfate in 10 mL of sterile water for injection is added to 250 mL of 5% dextrose injection. T he infusion is to be administered over 2 hours. If the administration set delivers 15 drops/mL, at what rate, in drops per minute, should the flow be adjusted to administer the infusion over the designated time interval?
Explanation:
It is given that the total units of polymixn B silfate are present as follows.
(10 + 250) ml = 260 ml
Therefore, total volume necessary to inject the whole volume is as follows.
[tex]260 \times 15[/tex] drops
= 3900
It is given that total time required to inject the solution is 2 hours. Converting it into minutes as follows.
[tex]2 hrs \times \frac{60 min}{1 hr}[/tex]
= 1200 minutes
Hence, rate of injection is calculated as follows.
Rate of injection (r) = [tex]\frac{\text{total drops}}{\text{total time}}[/tex]
r = [tex]\frac{3900}{120}[/tex]
= 32.5 drops/min
Thus, we can conclude that the rate of injection is 32.5 drops/min.
To infuse 260 mL of solution over 2 hours with a drop factor of 15 drops/mL, the flow rate should be adjusted to 32.5 drops per minute.
Explanation:We need to calculate the rate at which the solution should be administered in order to complete the infusion over 2 hours. First, we look at the total volume of the solution to be infused which is 250 mL of 5% dextrose injection plus the initial 10 mL, making it 260 mL in total. To calculate the rate, we'll use the formula:
Rate (drops/min) = Total volume (mL) × Administration set drop factor (drops/mL) ÷ Time (min)
Since we are told that the administration set delivers 15 drops/mL and that the infusion needs to be administered over 2 hours (which is 120 minutes), we can substitute the values into the formula:
Rate (drops/min) = (260 mL) × (15 drops/mL) ÷ (120 min)
After calculating, we find:
Rate (drops/min) = 32.5 drops/min
Note: This calculation is based on the student's given values and intended to demonstrate the process of determining the flow rate for an IV infusion.
Three different metabolic pathways are responsible for the production of ATP. What are they
Answer:
Explanation:
1) Glycolysis which occurs in the cytoplasm yield 2 ATP 2) Kreb cycle (critic acid cycle) yields 2 ATP per glucose molecule 3)electron transport chain yields the highest with 34 ATP both of which occurs in the mitochondria.
Calcium carbide can be made by heating calcium oxide with carbon. How many kj of heat are absorbed in the reaction?
Answer:
179Kj/mol
Explanation:
The equation for the reaction is:
CaO + 3C ⇒ CaC₂ + CO ₂
The formula for calculating the heat absorbed is:
∆Hf (Products) - ∆Hf (Reactants)
Using the heat of formation (∆Hf) for each component in the reaction
∆Hf Cao = -635kj/mol
∆Hf C = 0kj/mol
∆Hf CaC2 = -63kj/mol
∆Hf CO2 = -393.5
The heat absorbed = ( -63 + -393.5) – (-635)
= - 456 + 635
= 179kj/mol
Calculate Δ G for ATP hydrolysis in liver at 18 °C. Use the liver concentrations from the first question.
This question is incomplete and the full question can be seen below:
b) Calculate ΔG for ATP hydrolysis in liver at 18° C. Use the liver concentration From part a.
The equation for the ATP hydrolysis is:
H₂0
ATP ------------> ADP + P₁ ΔG = -30.5 [tex]\frac{KJ}{mol}[/tex]
a) calculate ΔG for ATP hydrolysis to rank the following conditions from most favorable to least favorable. Assume a temperature of 37.0° C, R = 8.315 [tex]\frac{J}{mol.k}[/tex]
muscle: [ATP]= 8.1mM; [ADP]= 0.9mM; [P₁]= 8.1mM
liver: [ATP]= 3.4mM; [ADP]= 1.3mM; [P₁]= 4.8mM
brain: [ATP}= 2.6mM; [ADP}= 0.7mM; [P₁]= 2.7mM
b) Calculate ΔG for ATP hydrolysis in liver at 18° C. Use the liver concentration From part a.
Answer:
−45.8 KJ/mol
Explanation:
Equilibrium constant (k) is defined as a measure of the ratio of the equilibrium concentration of the products of a reaction, to the Equilibrium concentration of the reactants with each concentration raised to the power corresponding to the coefficient in the balanced equation of the reaction.
In the reaction in the question given above;
[tex]K=\frac{[ADP][P_1]}{ATP}[/tex]
For muscle;
ADP= 0.9 × 10⁻³
P₁= 8.1 × 10⁻³
ATP= 8.1 × 10⁻³
∴ K = [tex]\frac{(0.9*10^-3)(8.1*10^-3)}{(8.1*10^-3)}[/tex]
K = 0.9 × 10⁻³
For liver;
ADP= 1.3 × 10⁻³
P₁= 4.8 × 10⁻³
ATP= 3.4 × 10⁻³
∴ K = [tex]\frac{(1.3*10^-3)(4.8*10^-3)}{(3.4*10^-3)}[/tex]
K = 1.8 × 10⁻³
For brain;
ADP= 0.7 × 10⁻³
P₁= 2.7 × 10⁻³
ATP= 2.6 × 10⁻³
∴ K = [tex]\frac{(0.7*10^-3)(2.7*10^-3)}{(2.6*10^-3)}[/tex]
K = 7.3 × 10⁻⁴
b) Since we are concerned about calculating ΔG for ATP hydrolysis in liver at 18° C and we've already obtained the liver concentration from part a; we can therefore calculate ΔG as:
ΔG = ΔG° + RTInK
ΔG° = -30.5
R= 8.315 [tex]\frac{J}{mol.k}[/tex]
T= 18° C = 18 + 273.15k = 291.15k
K= 1.8 × 10⁻³
InK = In(1.8 × 10⁻³ )
≅ -6.32
∴ ΔG = -30.5 + 8.315 [tex]\frac{J}{mol.k}[/tex] × 291.15k × (-6.32)
ΔG = -30.5 + (−15.30016542)
ΔG = −45.80016542
ΔG ≅ −45.8 KJ/mol
Change in free energy for ATP hydrolysis at 18 degree Celsius is -45.8 kJ.
The equation for the ATP hydrolysis is,
[tex]\bold {ATP \rightarrow ADP + Pi }[/tex] ΔG = -30.5 kJ/mol at STP
Change in free energy for ATP hydrolysis in liver at 18° C can be calculated by the formula,
[tex]\bold {\Delta G = \Delta G^o + RT \times lnK}[/tex]
Where,
ΔG° - Free energy at STP = -30.5
R - gas constant = 8.315
T - temperature in Kelvin = 18° C = 18 + 273.15k = 291.15k
K - Equilibrium constant = 1.8 × 10⁻³
In K = In(1.8 × 10⁻³ ) = -6.32
Put the values in the formula,
[tex]\bold {\Delta G = -30.5 + 8.315 \times 291.15 \times (-6.32)}\\\\\bold {\Delta G = -30.5 + (- 15.3)}\\\\\bold {\Delta G = - 45.8 KJ}\\\\[/tex]
Therefore, change in free energy for ATP hydrolysis at 18 degree Celsius is -45.8 kJ.
To know more about Free energy,
https://brainly.com/question/15319033
Considering only electron density, will the following reaction occur?
327d1 + 327d2 → yes no
Answer:
The question is not so clear, but there is a similar question that talks about reaction of alkanes with ammonia
Explanation:
The underlying factor dependent on whether alkane will react with ammonia or not is in their bond and reactivity. basically, alkanes are saturated hydrocarbons with a single bond existing between their chains. They are single bonded hydrocarbons and they majorly undergo SUBSTITUTION REACTION WITH HALOGENS
Alkanes are less reactive in this case, if other functional groups are attached or takes place alongside the reaction, the reaction may have a high chances of taking place.
The provided reaction notation is incorrect and does not correspond to known chemical species. Electron density alone does not determine a reaction's viability; conservation laws, such as those of mass-energy, charge, and nucleon number, are crucial in nuclear reactions.
Explanation:The reaction provided in the question seems to be incorrectly or incompletely specified, as '327d1 + 327d2' does not correspond to any known chemical species. However, we can discuss the general rules that determine whether a nuclear reaction can occur, based on electron density and other factors.
In nuclear reactions, electron density alone is not sufficient to determine if a reaction will occur. Instead, factors such as conservation of mass-energy, conservation of charge, and conservation of nucleon number (protons and neutrons) are critical. Furthermore, the reaction must not violate any fundamental forces or principles, such as the conservation of lepton number in beta decay processes, where a neutron (n) can decay into a proton (p), an electron (e−), and an electron antineutrino (ve).
For example, in nuclear equations such as 238U → 234Ra + 4He, we can confirm the reaction is possible by checking these conservation laws. This is an alpha decay process, where the uranium nucleus emits an alpha particle (identical to a helium nucleus) and thus produces radium while observing all conservation laws.
Suppose an EPA chemist tests a 250.0 ml sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
NiCl_2(aq) + 2AgNO_3 (aq) -------> AgCl (s) + Ni(NO_3)_2 (aq)
The chemist adds 58.0m silver nitrate solution to the sample until silver chloride stops forming. He then washes, dries, and weighs the precipitate. He finds he has collected of 3.6 mg of silver chloride.
1. Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits.
Final answer:
The concentration of the nickel(II) chloride contaminant in the groundwater sample is 5.0 x 10^-5 M, calculated by determining the moles of AgCl precipitate and its relation to NiCl2 through the stoichiometry of the reaction.
Explanation:
To calculate the concentration of nickel(II) chloride in the original groundwater sample, we need to use the information about the precipitate of silver chloride formed during the reaction. First, we must determine the moles of AgCl produced using its molar mass. The molar mass of AgCl is 143.32 g/mol, and we have 3.6 mg, or 0.0036 g of AgCl:
Moles of AgCl = mass (g) / molar mass (g/mol) = 0.0036 g / 143.32 g/mol = 2.51 x 10-5 moles
From the balanced chemical equation, we know that 1 mole of NiCl2 reacts with 2 moles of AgNO3 to produce 2 moles of AgCl. Therefore, the moles of NiCl2 present in the reaction will be half the moles of AgCl:
Moles of NiCl2 = 1/2 * Moles of AgCl = 1/2 * 2.51 x 10-5 moles = 1.255 x 10-5 moles
To find the concentration of NiCl2, we divide the moles of NiCl2 by the volume in liters:
Concentration of NiCl2 = moles / volume (L) = 1.255 x 10-5 moles / 0.250 L = 5.02 x 10-5 M
Therefore, the concentration of the nickel(II) chloride contaminant in the groundwater sample is 5.0 x 10-5 M, rounded to two significant digits.
Final answer:
The concentration of nickel(II) chloride in the original groundwater sample is calculated to be approximately 1.63 mg/L, based on the mass of the AgCl precipitate produced in the reaction with silver nitrate.
Explanation:
Calculating the Concentration of Nickel(II) Chloride
To calculate the concentration of nickel(II) chloride in the original groundwater sample, we first need to understand the reaction that takes place between nickel(II) chloride and silver nitrate to form silver chloride (AgCl) precipitate:
NiCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + Ni(NO3)2(aq).
From the balanced equation, we see that one mole of NiCl2 reacts with two moles of AgNO3 to yield two moles of AgCl. The mass of the collected AgCl precipitate is given as 3.6 mg.
The molar mass of AgCl is 143.32 g/mol. Since we have 3.6 mg (or 0.0036 g) of AgCl, we can calculate the number of moles:
Moles of AgCl = mass (g) / molar mass (g/mol) = 0.0036 g / 143.32 g/mol ≈ 2.51 x 10-5 mol AgCl.
Since the ratio of AgCl to NiCl2 in the reaction is 2:1, there were half as many moles of NiCl2 as there were of AgCl in the original sample:
Moles of NiCl2 = 0.5 x Moles of AgCl = 0.5 x 2.51 x 10-5 mol ≈ 1.255 x 10-5 mol NiCl2.
The volme of the groundwater sample was 250.0 mL, or 0.250 L. The concentration (C) of NiCl2 in moles per liter (mol/L) is:
Concentration of NiCl2 = Moles of NiCl2 / Volume (L) = 1.255 x 10-5 mol / 0.250 L = 5.02 x 10-5 mol/L.
To convert this into mg/L (knowing that the molar mass of NiCl2 is 129.6 g/mol), we have
Mass (mg) = Moles x molar mass (g/mol) x 1000 mg/g = 1.255 x 10-5 mol x 129.6 g/mol x 1000 mg/g ≈ 1.63 mg.
Therefore, the concentration of NiCl2 in the groundwater sample is roughly 1.63 mg/L.
Calculate the volume in liters of a potassium iodide solution that contains of potassium iodide . Be sure your answer has the correct number of significant digits.
The volume of a potassium iodide solution with a concentration of 0.0380 M and containing 150 g of potassium iodide is 3.95 L, rounded to three significant digits.
To calculate the volume V of a solution, you can use the formula:
[tex]\[ V = \frac{\text{mass of solute}}{\text{molarity}} \][/tex]
Given that the molarity M is [tex]\(0.0380 \, \text{M}\)[/tex] and the mass of potassium iodide is [tex]\(150 \, \text{g}\)[/tex], substitute these values into the formula:
[tex]\[ V = \frac{150 \, \text{g}}{0.0380 \, \text{M}} \]\[ V = 3947 \, \text{mL} \][/tex]
Since the answer should have the correct number of significant digits, the volume is [tex]\(3.95 \, \text{L}\)[/tex] (rounded to three significant digits).
For more questions on potassium iodide:
https://brainly.com/question/2913015
#SPJ3
The complete question is:
Calculate the volume in liters of a 0.0380M potassium iodide solution that contains 150 g of potassium iodide. Be sure your answer has the correct number of significant digits.
A solution is made by mixing equal masses of methanol, CH 4 O , and ethanol, C 2 H 6 O . Determine the mole fraction of each component to at least three significant figures.
Answer:
Mole fraction of [tex]CH_4O[/tex] = 0.58
Mole fraction of [tex]C_2H_6O[/tex] = 0.42
Explanation:
Let the mass of [tex]CH_4O[/tex] and [tex]C_2H_6O[/tex] = x g
Molar mass of [tex]CH_4O[/tex] = 33.035 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles_{CH_4O}= \frac{x\ g}{33.035\ g/mol}[/tex]
[tex]Moles_{CH_4O}=\frac{x}{33.035}\ mol[/tex]
Molar mass of [tex]C_2H_6O[/tex] = 46.07 g/mol
Thus,
[tex]Moles= \frac{x\ g}{46.07\ g/mol}[/tex]
[tex]Moles_{C_2H_6O}=\frac{x}{46.07}\ mol[/tex]
So, according to definition of mole fraction:
[tex]Mole\ fraction\ of\ CH_4O=\frac {n_{CH_4O}}{n_{CH_4O}+n_{C_2H_6O}}[/tex]
[tex]Mole\ fraction\ of\ CH_4O=\frac{\frac{x}{33.035}}{\frac{x}{33.035}+\frac{x}{46.07}}=0.58[/tex]
Mole fraction of [tex]C_2H_6O[/tex] = 1 - 0.58 = 0.42
The enzyme urease enhances the rate of urea hydrolysis at pH 8.0 and 20C by a factor of 1014 . If a given quantity of urease can completely hydrolyze a given quantity of urea in 5.0 min at 20 C and pH 8.0, how long would it take for this amount of urea to be hydrolyzed under the same conditions in the absence of urease? Assume that both reactions take place in sterile systems so that bacteria cannot attack the urea.
Answer:
It will take 950 million years for this amount of urea to be hydrolyzed under the same conditions in the absence of urease.
Explanation:
Given that:
In the presence of urease:Urease enhances the rate of hydrolysis by = 10^14
Time taken in hydrolysis is 5 min
In the absence of urease:Time taken in hydrolysis will be = 5 min x 10^14
Now, converting minutes into years
Time = (5 min x 10^14) / (60 min/hr x 24 hr/day x 365 days/year)
Time = 9.50 x 10^8 years
Time = 950 X 10^6 years
Time = 950 million years
Answer:
It would take [tex]5*10^{14}min[/tex] for this amount of of urea to be hydrolysed under the same conditions in the absence of urease.Explanation:
The enzyme increases the rate of urea hydrolysis by a factor of [tex]10^{14}[/tex]
So in absence of enzyme, reaction will take [tex]10^{14}[/tex] times more time for completion.
Therefore time taken,
[tex]\frac{rate catalysed}{rate uncatalysed} = 10^{14}[/tex]
[tex]\frac{t uncatalysed}{t catalysed} = 10^{14}\\\\t uncatalysed = 10^{14} * 5min\\\\t uncatalysed = 5*10^{14}min[/tex]
For more information, visit
https://brainly.com/question/14555858
Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.11 m FeBr3 A. Highest boiling point 2. 0.15 m CuBr2 B. Second highest boiling point 3. 0.24 m AgNO3 C. Third highest boiling point d 4. 0.51 m Glucose(nonelectrolyte) D. Lowest boiling point
Answer:
1. For 0.11 m [tex]FeBr_3[/tex] : Lowest boiling point
2. For 0.15 [tex]CuBr_2[/tex]
: Second highest boiling point
3. For 0.24 [tex]AgNO_3[/tex]
: Third highest boiling point
4. 0.51 m glucose : Highest boiling point
Explanation:
Elevation in boiling point:
[tex]\Delta T_b=ik_b\times m[/tex]
where,
[tex]T_b[/tex] = change in boiling point
i= vant hoff factor
[tex]k_b[/tex] = boiling point constant
m = molality
1. For 0.11 m [tex]FeBr_3[/tex]
[tex]FeBr_3\rightarrow Fe^{3+}+3Br^{-}[/tex]
, i= 4 as it is a electrolyte and dissociate to give 4 ions and concentration of ions will be [tex]1\times 0.11+3\times 0.11=0.44[/tex]
2. For 0.15 [tex]CuBr_2[/tex]
[tex]CuBr_2\rightarrow Cu^{2+}+2Br^{-}[/tex]
, i= 3 as it is a electrolyte and dissociate to give 3 ions, concentration of ions will be [tex]1\times 0.15+2\times 0.15=0.45[/tex]
3. For 0.24 [tex]AgNO_3[/tex]
[tex]AgNO_3\rightarrow Ag^{+}+NO_3^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be [tex]1\times 0.24+1\times 0.24=0.48[/tex]
4. 0.51 m glucose
i= 1 as it is a non electrolyte and does not dissociate to give ions, concentration will be [tex]1\times 0.51=0.51[/tex]
Thus as boiling point depends on the concentration of solutes, the solution having highest concentration will have highest boiling point.
The production capacity for acrylonitrile (C3H3N) in the United States exceeds 2 million pounds per year. Acrylonitrile, the building block for polyacrylonitrile fibers and a variety of plastics, is produced from gaseous propylene, ammonia, and oxygen.
2C3H6(g) + 2NH3(g) + 3O2(g) → 2C3H3N(g) + 6H2O(g)
(a) What mass of acrylonitrile can be produced from a mixture of 1.14 kg of propylene (C3H6), 1.65 kg of ammonia, and 1.85 kg of oxygen, assuming 100% yield?
(b) What mass of water is produced?
Answer:
a) 1.44 kg
b) 1.47 kg
Explanation:
a) By the reaction given, the stoichiometry, the molar ratio, is:
2 moles of C3H6 : 2 moles of NH3 : 3 moles of O2 : 2 moles of C3H3N : 6 moles of H2O, or 2:2:3:2:6.
By the mixture given, one or two of the reactants may be in excess, so, we must found which of them is the limiting reactant, the reactant that will be totally consumed.
The molar masses of the compounds are:
C3H6 = 42.0 g/mol
NH3 = 17.0 g/mol
O2 = 32.0 g/mol
C3H3N = 53.1 g/mol
H2O = 18.0 g/mol
Thus, the mass ratio between the reactants will be the molar ratio multiplied by the molar mass:
2 moles of C3H6*42.0 g/mol --- 2 moles of NH3*17.0 g/mol -- 3 moles of O2* 32.0 g/mol
84.0 g of C3H6 -- 34.0 g of NH3 -- 96.0 g of O2
Thus, assuming C3H6 as limiting, let's use the rule of three to find out which would be the masses of the other reactants needed:
84.0 g of C3H6 -- 34.0 g of NH3
1.14 kg -- x
84.0x = 38.76 kg
x = 0.46 kg of NH3
Because there's more NH3 than it's required, NH3 is in excess, and C3H6 is limiting. Let's test using O2 as reference:
84.0 g of C3H6 -- 96.0 f of O2
1.14 kg -- x
84.0x = 109.44
x = 1.30 kg of O2
So, O2 is also in excess, and the limiting reactant is C3H6.
Thus, the molar ratio between C3H6 and C3H3N is 2:2, which is simplified by 1:1, and so the mass ratio is 42 g of C3H6 -- 53.1 g of C3H3N, so the mass of acrylonitrile can be found by a rule of three:
42 g of C3H6 -- 53.1 f of C3H3N
1.14 kg -- x
42x = 60.534
x = 1.44 kg
b) The molar ratio between C3H6 and water is 2:6, which can be simplified by 1:3, thus the mass ratio is 1*42g/mol of C3H6 -- 3*18 g/mol of H2O
42 g of C3H6 -- 54 g of H2O
By a rule of three the mass of water is:
42 g of C3H6 -- 54 g of H2O
1.14 kg -- x
42x = 61.56
x = 1.47 kg
Final answer:
The mass of acrylonitrile that can be produced from the given mixture of reactants is approximately 1437.73 g, and the mass of water produced is 1464.68 g, assuming a 100% yield and propylene as the limiting reactant.
Explanation:
Calculating Mass of Acrylonitrile and Water Produced
To answer your questions about the production of acrylonitrile and water, we first need to use the provided stoichiometry and the masses of the reactants. The balanced chemical reaction is:
2C₃H₆(g) + 2NH₃(g) + 3O₂(g) → 2C₃H₃N(g) + 6H₂O(g)
This informs us that two moles of propylene (C₃H₆) react with two moles of ammonia (NH₃) and three moles of oxygen (O₂) to produce two moles of acrylonitrile (C₃H₃N) and six moles of water (H₂O).
Firstly, we determine the number of moles of each reactant:
Propylene (C₃H₆): Molar mass = 42.08 g/mol, Moles = 1140 g / 42.08 g/mol = 27.11 moles
Ammonia (NH₃): Molar mass = 17.031 g/mol, Moles = 1650 g / 17.031 g/mol = 96.91 moles
Oxygen (O₂): Molar mass = 32.00 g/mol, Moles = 1850 g / 32.00 g/mol = 57.81 moles
Since the limiting reactant determines the amount of product formed, we can find it by comparing the mole ratios. In this case, propylene is the limiting reactant. Using the stoichiometry, we calculate the mass of acrylonitrile produced:
Moles of acrylonitrile produced = Moles of propylene used = 27.11 moles
Mass of acrylonitrile produced = Moles of acrylonitrile produced x Molar mass of acrylonitrile = 27.11 moles x 53.06 g/mol = 1437.73 g
For the mass of water produced:
Moles of water produced = 3 x Moles of propylene used = 3 x 27.11 mol = 81.33 moles
Mass of water produced = Moles of water produced x Molar mass of water = 81.33 moles x 18.015 g/mol = 1464.68 g
Assuming 100% yield, we can produce approximately 1437.73 g of acrylonitrile and 1464.68 g of water from the given reactants.
When NaHCO3 is heated above 270 °C, it decomposes to Na2CO3(s), H2O(g), and CO2(g). Calculate the mass (in grams) of CO2 produced from the decomposition of 25.0 g NaHCO3.
When heated, 25 grams of sodium bicarbonate (NaHCO3) will decompose and produce approximately 13.1 grams of carbon dioxide (CO2) gas.
Explanation:The decomposition of sodium bicarbonate (NaHCO3) can be represented by the equation: NaHCO3 → Na2CO3 + H2O + CO2. The ratio of NaHCO3 to CO2 is 1:1, so in theory, one mole of NaHCO3 will produce one mole of CO2. Based on molecular weights, 84 grams of NaHCO3 (the weight of one mole) will produce 44 grams of CO2 (the weight of one mole).
First, we need to find how many moles are in 25 grams of NaHCO3. Using the formula number of moles=(given mass/molar mass), we get approximately 0.297 moles.
Now, applying the molar relationship between NaHCO3 and CO2, we will also produce 0.297 moles of CO2. Converting this to grams: number of CO2 grams = number of moles * molecular weight, we get approximately 13.1 grams of CO2 gas.
Learn more about Chemical Decomposition here:https://brainly.com/question/32272087
#SPJ3
To determine the mass of CO2 produced from the decomposition of 25.0 g NaHCO3, we use stoichiometry based on the chemical equation, resulting in 6.55 grams of CO2.
To calculate the mass of CO2 produced from the decomposition of 25.0 g NaHCO3, we need to use the stoichiometry of the balanced chemical equation:
2 NaHCO3(s) ightarrow Na2CO3(s) + H2O(g) + CO2(g)
First, we calculate the molar mass of NaHCO3, which is (23+1+12+3(16)) = 84 g/mol. Next, we determine the moles of NaHCO3 in 25.0 g:
Moles of NaHCO3 = 25.0 g \/ 84 g/mol = 0.2976 moles
According to the balanced equation, 2 moles of NaHCO3 produce 1 mole of CO2. Thus, 0.2976 moles of NaHCO3 would produce:
Moles of CO2 = 0.2976 moles NaHCO3 \/ 2 = 0.1488 moles CO2
The molar mass of CO2 is (12 + 2(16)) = 44 g/mol, so the mass of CO2 produced is:
Mass of CO2 = 0.1488 moles \/ 44 g/mol = 6.5472 g
Therefore, 6.55 grams of CO2 are produced from the decomposition of 25.0 g NaHCO3.
Ultraviolet radiation and radiation of shorter wavelengths can damage biological molecules because they carry enough energy to break bonds within the molecules. A carbon-carbon bond requires 348 kJ/mol to break.What is the longest wavelength of radiation with enough energy to break carbon-carbon bonds?
Answer:
344 nm is the longest wavelength of radiation with enough energy to break carbon-carbon bonds.
Explanation:
[tex]C-C(g)\rightarrow 2C(g)[/tex] ,ΔH = 348 kJ/mol
Energy required to break 1 mole of C-C bond = 348 kJ
Energy required to break 1 C-C bond = E
[tex]E = \frac{348,000J}{6.022\times 10^{23}}=5.779\times 10^{-19} J[/tex]
Energy related with the wavelength of light is given by Planck's equation:
[tex]E=\frac{hc}{\lambda }[/tex]
[tex]\lambda =\frac{hc}{E}[/tex]
[tex]=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{5.779\times 10^{-19} J}[/tex]
[tex]\lambda =3.44\times 10^{-7} m = 344 nm[/tex]
[tex]1 m =10^9 nm[/tex]
344 nm is the longest wavelength of radiation with enough energy to break carbon-carbon bonds.
Ultraviolet radiation and radiation of shorter wavelengths can break carbon-carbon bonds within biological molecules. The longest wavelength that can break carbon-carbon bonds is approximately 1.72 x 10^-6 meters or 1720 nm.
Explanation:Ultraviolet radiation and radiation of shorter wavelengths can damage biological molecules because they carry enough energy to break bonds within the molecules. A carbon-carbon bond requires 348 kJ/mol to break. To find the longest wavelength of radiation with enough energy to break carbon-carbon bonds, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength. Rearranging the equation to solve for λ, we get λ = hc/E. Substituting the given energy of 348 kJ/mol (which is equivalent to 348,000 J/mol) and solving for λ, we find that the longest wavelength is approximately 1.72 x 10^-6 meters or 1720 nm.
Learn more about radiation here:https://brainly.com/question/4075566
#SPJ12
a) Show the alkyl bromide and alcohol used to make methyl t-butyl ether using the Williamson ether synthesis to the right of the retrosynthetic arrow.
The alkyl bromide used is Tertiary-Butyl Bromide and the alcohol used is Methanol. The reaction involves reacting Tertiary-Butyl Bromide with Sodium Methoxide to form methyl t-butyl ether and Sodium Bromide as a byproduct.
The alkyl bromide and alcohol used to make methyl t-butyl ether using the Williamson ether synthesis are:
Alkyl Bromide: Tertiary-Butyl Bromide (C4H9Br)
Alcohol: Methanol (CH3OH)
Step 1: React Tertiary-Butyl Bromide (C4H9Br) with Sodium Methoxide (CH3ONa).
Step 2: Heating the reaction mixture will result in the formation of methyl t-butyl ether (MTBE) and Sodium Bromide (NaBr) as byproduct.
For more such questions on methyl t-butyl, click on:
https://brainly.com/question/33650659
#SPJ6
Calculate the theoretical yield of aspirin to be obtained when 2.0 g of salicylic acid and 5.0 mL of acetic anhydride (density = 1.08 g/mL) are mixed together. What is the limiting reagent?
Answer:
The theoretical yield is 2.61 grams of aspirin
salicylic acid is the limiting reactant.
Explanation:
Step 1: Data given
Mass of salicylic acid = 2.0 grams
Volume of acetic acid = 5.0 mL
Density of acetic acid = 1.08 g/mL
Molar mass of acetic anhydride = 102.09 g/mol
Molar mass of salicylic acid = 138.12 g/mol
Step 2: The balanced equation
C7H6O3 + C4H6O3 → C9H8O4 + CH3COOH
Step 3: Calculate mass of acetic acid
Mass acid acid = density * volume
Mass acetic acid = 1.08 g/mL * 5.0 mL
Mass acetic acid = 5.5 grams
Step 4: Calculate moles salicylic acid
Moles salicylic acid = mass salicylic acid / molar mass salicylic acid
Moles salicylic acid = 2.00 grams / 138.12 g/mol
Moles salicylic acid = 0.0145 moles
Step 5: Calculate moles acetic anhydride
Moles acetic anhydride= 5.5 grams / 102.09 g/mol
Moles acetic anhydride = 0.0538 moles
Step 6: Calculate the limiting reactant
For 1 mol salicylic acid we need 1 mol acetic anhydride to produce 1 mol aspirine.
The limiting reactant is salicylic acid. It will completely be consumed. (0.0145 moles). Acetic anhydride will be in excess. There will react 0.0145 moles . There will remain 0.0538 - 0.0145 = 0.0393 moles
Step 7: Calculate moles aspirine
For 1 mol salicylic acid we need 1 mol acetic anhydride to produce 1 mol aspirine.
For 0.0145 moles salicylic acid, we'll have 0.0145 moles aspirine
Step 8: Calculate mass of aspirin
Mass aspirin = moles aspirin * molar mass aspirin
Mass aspirin = 0.0145 moles * 180.158 g/mol
Mass aspirin = 2.61 grams = Theoretical yield
The theoretical yield is 2.61 grams of aspirin
salicylic acid is the limiting reactant.
A chemist prepares a solution of aluminum chloride by measuring out of aluminum chloride into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's aluminum chloride solution.
The question is incomplete, here is the complete question:
A chemist prepares a solution of aluminum chloride by measuring out of 11. g aluminum chloride into a 50. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's aluminum chloride solution.
Answer: The concentration of aluminium chloride solution is 1.65 mol/L
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Given mass of aluminium chloride = 11. g
Molar mass of aluminium chloride = 133.34 g/mol
Volume of solution = 50. mL
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{11\times 1000}{133.34\times 50}\\\\\text{Molarity of solution}=1.65mol/L[/tex]
Hence, the concentration of aluminium chloride solution is 1.65 mol/L
How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?
Complete Question:
To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?
Answer:
2.23x10⁶ g
Explanation:
The concentration of the fluoride (F⁻) must be 0.800 ppm, which is 0.800 parts per million, so the water must have 0.800 g of F⁻/ 1000000 g of the solution. The density of the water at room temperature is 997 kg/m³ = 997x10³ g/m³. So, the concentration of the fluoride will be:
0.800 g of F⁻/ 1000000 g of the solution * 997x10³ g/m³
0.7976 g/m³
The volume of the reservoir is the volume of the cylinder: area of the base * depth. The base is a circumference, which has an area:
A = πR², where R is the radius = 1.01x10² m (half of the diameter)
A = π*(1.01x10²)²
A = 32047 m²
The volume is then:
V = 32047 * 87.32
V = 2.7983x10⁶ m³
The mass of the F⁻ is the concentration multiplied by the volume:
m = 0.7976 * 2.7983x10⁶
m = 2.23x10⁶ g
When determining rock composition, describe the test that can be used to determine whether it contains the mineral calcite.
Answer:
Check explanation
Explanation:
There are two ways to test for calcite in rocks. Calcite is popularly known as calcium carbonate and it is found almost everywhere. There are many minerals that looks like calcite/calcium carbonates, so we there is a need for a good test for proper identification.
The tests are:
(1). ACID TEST: the calcite/calcium carbonate reacts vigorously with Hydrogen Chloride acid,HCl with the evolution of carbondioxide,CO2. The equation of reaction is given below;
CaCO3 + 2 HCl ---------> CaCl2 + CO2 + H2O.
(2). DOUBLE REFRACTION METHOD: There are other minerals in the rock that can react with Hydrochloric acid,HCl to evolve Carbondioxide,CO2(although their response to HCl differs). There is a need for a confirmatory test. This double refraction method is a confirmatory test. In this test, one will pass light through the calcite, if the light splits into two rays and reflected twice then, it is a calcite.
The acid test for calcite involves applying dilute hydrochloric acid to a rock sample. If the rock contains calcite, effervescence (the release of bubbles) occurs due to the reaction between the acid and calcite, forming calcium chloride, water, and carbon dioxide. This test is a useful tool in determining the presence of calcite, an essential mineral in the identification and classification of various types of rocks and minerals.
To determine whether a rock contains the mineral calcite, one common test that geologists use is the acid test, specifically the reaction of calcite with dilute hydrochloric acid (HCl). This test helps identify calcite due to its distinctive reaction with the acid. Here's a description of the test:
Acid Test for Calcite in Rocks:
Materials Needed:
Dilute hydrochloric acid (HCl)
The rock sample in question
Procedure:
Safety Precautions: Before conducting the test, ensure that you are wearing appropriate safety gear, including gloves and eye protection, as HCl is a corrosive acid.
Select a Representative Sample: Choose a representative sample of the rock you want to test. It's essential to select an area that appears to contain calcite or where you suspect its presence.
Apply the Acid: Place a small drop of dilute hydrochloric acid (typically 10% HCl solution) onto the surface of the rock. You can also use a dropper or a small pipette for precision.
Observe the Reaction: Watch the reaction between the acid and the rock. If the rock contains calcite, you will observe effervescence or the release of bubbles of carbon dioxide (CO₂). This effervescence is a result of the chemical reaction between the HCl and the calcite, producing soluble calcium chloride (CaCl₂), water (H₂O), and carbon dioxide:
CaCO₃ (calcite) + 2HCl → CaCl₂ + H₂O + CO₂↑
For more such information on: acid test
https://brainly.com/question/25814739
#SPJ3
How much sodium chloride (NaCl) would need to be added to 750 milliliters of water to make a 15% solution?
To create a 15% solution of sodium chloride in 750 milliliters of water, you would need to add 112.5 grams of sodium chloride.
Explanation:To create a 15% solution of sodium chloride (NaCl) in 750 milliliters (ml) of water, you need to know the concept of weight/volume percentage. A 15% w/v solution means 15g of solute (in this case, NaCl) is present in 100 ml of solution. To find how much NaCl is needed for 750ml, we simply multiply the percentage by the volume of solution (in this case, 750ml) and divide by 100.
So, to answer your question, the calculation would be: (15g/100ml)*750ml = 112.5g
Therefore, you would need to add 112.5 grams of sodium chloride to 750 ml of water to create a 15% solution.
Learn more about Solution Preparation here:https://brainly.com/question/32551796
#SPJ3
371. mg of an unknown protein are dissolved in enough solvent to make 5.00 ml of solution. The osmotic pressure of this solution is measured to be at 0.118 atm at 25 C
Calculate the molar mass of the protein.
Answer:
mm protein = 15365.8183 g/mol
Explanation:
∴ molar mass (mm) ≡ g/mol
osmotic pressure (π) = C RT∴ π = 0.118 atm
∴ T = 25°C ≅ 298 K
∴ concentration (C ) [=] mol/L
∴ mass protein = 371 mg = 0.371 g
∴ volume sln = 5.00 mL = 5 E-3 L
⇒ C = π / RT = (0.118 atm)/((0.082 atm.L/K.mol)(298 K))
⇒ C = 4.8289 E-3 mol/L
⇒ mol protein = (4.8289 E-3 mol/L)×(5 E-3 L) = 2.4145 E-5 mol
⇒ mm protein = (0.371 g)/(2.4145 E-5 mol) = 15365.8183 g/mol
Final answer:
To find the molar mass of the protein, we begin by converting mass to grams and volume to liters, and then apply the van 't Hoff equation for osmotic pressure. After rearranging the equation, we calculate the molar mass using the given osmotic pressure, volume, temperature, and ideal gas constant.
Explanation:
To calculate the molar mass of the protein using the osmotic pressure, we can use the van 't Hoff equation for osmotic pressure, Π = MRT, where Π is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin. In this scenario, the student provided the mass of the protein (371 mg), the volume of the solution (5.00 mL), the osmotic pressure (0.118 atm), and the temperature at which the osmotic pressure was measured (25 °C which is 298.15 K).
First, convert the mass of the protein to grams:
371 mg = 0.371 g
Next, convert the volume from mL to liters:
5.00 mL = 0.005 L
Now, calculate the molarity (M) of the protein:
Number of moles = mass (g) / molar mass (g/mol)
As we do not know the molar mass yet, let's call it 'Mm'. The molarity (concentration) of the solution is:
Molarity (M) = Number of moles / Volume (L) = (mass (g) / Mm (g/mol)) / Volume (L)
Convert the temperature to Kelvin:
25 °C = 298.15 K
Using the van 't Hoff equation, we solve for the molar mass (Mm):
Π = (mass (g) / Mm (g/mol)) / Volume (L) × R × T
0.118 atm = (0.371 g / Mm (g/mol)) / 0.005 L × 0.0821 (L·atm/K·mol) × 298.15 K
Rearranging the equation and solving for Mm gives us:
Mm = (0.371 g / 0.005 L) / (0.118 atm / (0.0821 L·atm/K·mol × 298.15 K))
After calculating, we find the molar mass of the protein.
Which concepts can be used to explain the difference in acidicty between ethanol and 2-fluoroethanol?
Answer:
The answer is Inductive effect
Explanation:
To determine the acidity or alkalinity of an organic compound. We have to keep in mind that the whole analysis is based on the comparison between the compounds, and we must work with the conjugated base of the molecule. Keeping in mind, the more unstable the base, the less acidic the molecule is. Thus, to determine instability, the Inductive Effect of the molecule can be used.
This type of effect occurs when atoms of different electronegativities are linked or very close in the compound. The most electronegative atom has a tendency to bring electrons close to it, thus creating a dipole. This dipole can have a stabilizing effect on the molecule, as it “relieves” the excessive charge on some occasions, better accommodating the charges.
However, in some cases, instead of chains with chlorine radicals, we may have chains with methyl radicals. This has a major impact on the inductive effect, keeping in mind that alkyl groups are electron donors.
Compare and contrast the positive and negative effects that result from the use of the Haber process.
The Haber process increased agricultural productivity and human support through the large-scale production of fertilizers, but it also enabled the creation of harmful chemical weapons and resulted in environmental issues such as algal blooms and dead zones.
Explanation:The Haber process, developed by Fritz Haber in the early 20th century, had both positive and negative effects. The positive effects of this process include the mass production of nitrogen fertilizers, which increased agricultural productivity and the sustainable support of human populations. An example of this is the jump in the number of humans per hectare that arable land could support, increasing from 1.9 in 1908 to 4.3 by 2008.
However, there are also negative consequences associated with the Haber process. It enabled the creation of harmful chemical weapons during the World War, marking a dark turn in its utilization. Also, the excessive use of nitrogen-based fertilizers, majorly produced through this process, has led to detrimental environmental consequences such as algal blooms, red tides, and dead zones in water bodies. These effects end up destroying marine life and disrupting ecosystems.
Learn more about Haber Process here:https://brainly.com/question/26667299
#SPJ3
In the previous question, you calculated the amount of CO2 that was required to heat the air in your room. Which of the following are true statements?
Explanation:
In the previous question, you calculated the amount of CO2 that was required to heat the air in your room. Which of the following are true statements?
You didn't complete the above question, please complete the question and reupload, thanks for your anticipated cooperation.
Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 9.0 g. Silver has 47 electrons per atom, and its molar mass is 107.87 g/mol.
Answer: The number of electrons in the given amount of silver are [tex]2.36\times 10^{24}[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of silver = 9.0 g
Molar mass of silver = 107.87 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of silver}=\frac{9.0g}{107.87g/mol}=0.0834mol[/tex]
We are given:
Number of electrons in one atom of silver = 47
According to mole concept:
1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of particles
So, 0.0834 moles of silver contains [tex](47\times 0.0834\times 6.022\times 10^{23})=2.36\times 10^{24}[/tex] number of electrons
Hence, the number of electrons in the given amount of silver are [tex]2.36\times 10^{24}[/tex]
How many grams of perchloric acid, HClO4, are contained in 39.1 g of 74.9 wt% aqueous perchloric acid
Answer:
29.3 g HClO₄
Explanation:
We have 39.1 grams of 74.9 wt% aqueous perchloric acid solution, that is, there are 74.9 grams of perchloric acid in 100 grams of perchloric acid solution. The mass, in grams, of perchloric acid contained in 39.1 grams of perchloric acid solution is:
39.1 g Solution × (74.9 g HClO₄/100 g Solution) = 29.3 g HClO₄
Calculate the mass of oxygen (in mg) dissolved in a 4.97 L bucket of water exposed to a pressure of 1.08 atm of air. Assume the mole fraction of oxygen in air to be 0.21 and the Henry's law constant for oxygen in water at this temperature to be 1.3 × 10-3 M/atm O2. (Enter your value using three significant figures.)
Answer:
46.9mg of oxygen
Explanation:
From Henry's law,
Concentration of oxygen (C) = Henry's constant (K) × partial pressure of oxygen in air (p)
K = 1.3×10^-3M/atm O2, p = mole fraction of oxygen in air × pressure of air = 0.21×1.08atm = 0.2268atm
C = K×p = 1.3×10^-3 × 0.2268 = 0.00029484M of O2
Concentration (C) = number of moles of oxygen (n)/volume of water (V)
Volume of water (V) = 4.97L
n = CV = 0.00029484 × 4.97 = 0.001465mole
number of moles (n) = mass of O2/MW of O2
mass of O2 = number of moles of O2 × MW of O2 = 0.001465mole × 32g/mole = 0.0469g = 0.0469×1000mg = 46.9mg (to three significant figures)
What is the strongest intermolecular force present in SO2? (EN Values: S = 2.5; O = 3.5). Please explain!
Ion-Dipole force
Hydrogen-Bond
Dipole-Dipole force
Dispersion forces (London dispersion)
Covalent Bond
Answer:
Dipole-Dipole force
Explanation:
Dipole - Dipole force -
These are the force of attraction , that occurs between two dipole , i.e. ,a species with two poles , hence , the attraction between the delta positive charge of first species with the delta negative charge of the second species , arises to a dipole - dipole force of attraction.
Hence, from the question,
SO₂ , is a polar compound , where O is more electronegative in comparison to S , thus , O attains a delta negative charge and S attains a delta positive charge and therefore , generates a dipole , and interacts with the dipole of the second molecule of SO₂ , arising a dipole - dipole force of attraction .
At 25 oC, the vapor pressure of water is 23.8 torr and the heat of vaporization is 43.9 kJ/mol. Calculate the vapor pressure of water at 50. oC
Answer: The vapor pressure of water at [tex]50^0C[/tex] is 93.8 torr
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
[tex]ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]
where,
[tex]P_1[/tex] = initial pressure at [tex]25^oC[/tex] = 23.8 torr
[tex]P_2[/tex] = final pressure at [tex]50^oC[/tex] = ?
[tex]\Delta H_{vap}[/tex] = enthalpy of vaporisation = 43.9 kJ/mol = 43900 J/mol
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]
[tex]T_2[/tex] = final temperature = [tex]50^oC=273+50=323K[/tex]
Now put all the given values in this formula, we get
[tex]\log (\frac{P_2}{23.8}=\frac{43900}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{323K}][/tex]
[tex]\frac{P_2}{23.8}=antilog(0.5955)[/tex]
[tex]P_2=93.8torr[/tex]
Therefore, the vapor pressure of water at [tex]50^0C[/tex] is 93.8 torr
Estimate the surface-to-volume ratio of a C60 fullerene by treating the molecule as a hollow sphere and using 77pm for the atomic radius of carbon.
To estimate the surface-to-volume ratio of a C60 fullerene, treat the molecule as a hollow sphere and calculate its surface area and volume. The surface area is found using the formula A = 4πr² and the volume is found using the formula V = (4/3)πr³, where r is the radius of the sphere. Dividing the surface area by the volume gives the surface-to-volume ratio.
Explanation:To estimate the surface-to-volume ratio of a C60 fullerene, we can treat the molecule as a hollow sphere and calculate its surface area and volume.
The surface area of a sphere is given by the formula A = 4πr^2, where r is the radius of the sphere.
The volume of a sphere is given by the formula V = (4/3)πr^3.
Since the C60 fullerene is composed of 60 carbon atoms, we can divide the atomic radius of carbon by 2 to get the radius of the molecule. Using the given atomic radius of 77pm, the radius of the C60 fullerene is 38.5pm (or 0.385nm).
Using these values, we can calculate the surface area and volume of the C60 fullerene:
Surface Area (A) = 4π(0.385nm)^2Volume (V) = (4/3)π(0.385nm)^3Now we can calculate the surface-to-volume ratio by dividing the surface area by the volume:
Surface-to-Volume Ratio = A / V
Substituting the calculated values into the formula, we get:
Surface-to-Volume Ratio = (4π(0.385nm)^2) / ((4/3)π(0.385nm)^3)
Simplifying the equation, we find that the surface-to-volume ratio of the C60 fullerene is approximately 0.649 nm-1.