Explanation:
STP means standard temperature and pressure where values are as follows.
T = 273.15 K, P = 1 atm
According to ideal gas equation, PV = nRT. Since, it is given that mass is 48.3 g and we have to find the volume as follows.
n = [tex]\frac{mass}{\text{molar mass}}[/tex]
So, PV = nRT
V = [tex]\frac{mass}{\text{molar mass}} \times \frac{RT}{P}[/tex]
= [tex]\frac{48.3 g}{28 g/mol} \times \frac{0.0821 L atm/mol K \times 273.15 K}{1 atm}[/tex] (molar mass of CO = 28 g/mol)
= 38.68 L
Thus, we can conclude that volume of 48.3 g of carbon monoxide at STP is 38.68 L.
Calculate the freezing point of the solution.After mixing these 2 bottles together, set Kf of water = 1.86 ° C / m.
Bottle 1 contained 0.3 grams of glucose in 1000 grams of water.
The 2nd bottle contains 0.5 mol fructose in 1000 grams of water.
Answer : The freezing point of solution is 273.467 K
Explanation : Given,
Mass of glucose (solute) = 0.3 g
Mass of water (solvent) = 1000 g = 1 kg
Moles of fructose (solute) = 0.5 mol
Mass of water (solvent) = 1000 g = 1 kg
Molar mass of glucose = 180 g/mole
First we have to calculate the moles of glucose.
[tex]\text{Moles of glucose}=\frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}=\frac{0.3g}{180g/mole}=0.00167mole[/tex]
Now we have to calculate the total moles after mixing.
[tex]\text{Total moles}=\text{Moles of glucose}+\text{Moles of fructose}[/tex]
[tex]\text{Total moles}=0.00167+0.5=0.502moles[/tex]
Now we have to calculate the molality.
[tex]\text{Molality}=\frac{\text{Total moles}}{\text{Mass of water(solvent in kg)}}[/tex]
[tex]\text{Molality}=\frac{0.502mole}{(1+1)kg}=0.251mole/kg[/tex]
Now we have to calculate the freezing point of solution.
As we know that the depression in freezing point is a colligative property that means it depends on the amount of solute.
Formula used :
[tex]\Delta T_f=K_f\times m[/tex]
[tex]T^o_f-T_f=K_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]T_f^o[/tex] = temperature of pure water = [tex]0^oC[/tex]
[tex]T_f[/tex] = temperature of solution = ?
[tex]K_f[/tex] = freezing point constant of water = [tex]1.86^oC/m[/tex]
m = molality = 0.251 mole/kg
Now put all the given values in this formula, we get
[tex]0^oC-T_f=1.86^oC/m\times 0.251mole/kg[/tex]
[tex]T_f=-0.467^oC=273.467K[/tex]
conversion used : [tex]K=273+^oC[/tex]
Therefore, the freezing point of solution is 273.467 K
If the caffeine concentration in a particular brand of soda is 2.13 mg/oz, drinking how many cans of soda would be lethal? Assume that 10.0 g of caffeine is a lethal dose, and there are 12 oz in a can. cans of soda:
Answer:
400 cans of soda would be lethal.
Explanation:
In a can of soda, there is (2.13 mg/oz * 12 oz) 25 mg caffeine.
25 mg * (1g / 1000 mg) = 0.025 g
If in a can of soda there is 0.025 g of caffeine, a lethal dose of caffeine will be ingested after drinking (10.0 g * (1 can / 0.025 g)) 400 cans of soda.
The two methods in the lab for measuring the volume of a geometric solid object are weighing it on the balance and measuring it with a ruler measuring it with a ruler and then noting its volume in water using final volume minus initial volume in the graduated cylinder measuring it with a ruler and then again with a different ruler noting its volume in water using final volume minus initial volume in the graduated cylinder and then weighing it.
Answer:
Measuring with a ruler and using final volume minus initial volume
Explanation:
You can measure the volume of a geometric object by measuring its sides with a ruler and calculating the volume according to the corresponding formula for each object. For example, for a rectangular prism it would be
[tex]volume=length*width*height[/tex]
You can also measure the volume of an object by measuring how much water it displaces. To do this you have to fill a measuring cylinder with enough water for the object to be completely submerged and take note of the volume. Then, add the object and note again the volume of the water+object. The difference between both is the volume of the object.
[tex]Volume of the object= volume of water and object - volume of water[/tex]
The advantage of the second method is that it can be used for objects with irregular shapes as long as they do not float.
In an x-ray tube with 70kV, what is the velocity of the electrons upon reaching the anode? And what is the shortest wavelength of the x-rays emitted from this tube?
Answer:
v=3.92*10¹⁷ m/s, and l=2.93*10⁻³⁰m
Explanation:
The velocity of the electrons in the x-ray tube is related to the mass of electrons and the energy applied, the equation is E=0.5*m*v², where E is the energy in kV, m is the mass of the electrons (9.11x10-31kg), and v the velocity. Substituting, 70000 = 0.5*(9.11x10-31kg)*v², and rearranging the terms, sqrt((70000)/(0.5)(9.11x10-31)) = v = 3.92*10¹⁷ m/s. Then the energy is related to the constant h (6.62607015×10−34 Js), the speed of light 299792458 m/s, and "l" that is the wavelength. So the equation is l=(h*c)/E, and l = 2.93*10⁻³⁰m.
Silver Nitrate.
Hi experts, can someone give me some real word applications beyond the chemistry laboratory about silver nitrate?
what is it used for in real life?
Explanation:
Except for the use in the chemistry laboratory , were it is used to synthesize many useful products, silver nitrate is also has biolofical and medical relevance.
Silver nitrate is commonly used for silver staining, for demonstrating the reticular fibers, the proteins and the nucleic acids. It is also used as stain in the scanning electron microscopy.
Silver Nitrate is also used for the bone ulcers as well as the burns and the acute wounds.
People tend to speak more quietly in restaurants than they do when they are having an ordinary conversation Restaurant conversation is about 45 dB. If ordinary conversation is 100 times greater than restaurant conversation, how loud is ordinary conversation?
Answer:
A loud ordinary conversation following the supplied information in the question is about 4500 dB. But, in the official decibel system measure a loud conversation does not overcome 100 dB.
Explanation:
Using the supplied data of the exercise, we say that in a restaurant conversation the value is 45 dB. If we multiply this by 100 we will have a value for a laud ordinary conversation.
45×100 = 4500 dB.
but as I mentioned in the answer, in the official decibel system measure a loud conversation between 2 man reaches a maximal of 100 dB.
A wall is constructed of a section of stainless steel (k = 16 W/m -°C) 40 mm thick with identical layers of plastic on both sides of the steel The overall heat-transfer coefficient, considering convection on both sides of the plastic, is 120 W/m2 °C If the overall temperature difference across the arrangement is 60°C, calculate the temperature difference across the stainless steel The area of the wall is one meter square
Answer:
The temperature difference across the stainless steel is 18°C
Explanation:
The heat flows through the plastic layer by convection and then through the steel layer by conduction and then through the plastic layer on the other side.
The heat flux q/A can be expressed as:
[tex]q/A = h*\Delta T[/tex]
It can also be expressed as
[tex]q/A=h*(T_A-T_B) = h_1(T_A-T_1)=k/\Delta x*(T_1-T_2)=h_2(T_2-T_B)[/tex]
being Δx/k*(T1-T2) the conductive heat flux through the steel.
If we want to know the temperatur difference across the stainless steel (T1-T2) we can write:
[tex](k/\Delta x)*(T_1-T_2)=h*(T_A-T_B)\\\\(T_1-T_2)=(\Delta x/k)*h*(T_A-T_B)\\\\(T_1-T_2)=(0.04/16)*120*60=18 \, ^{\circ} C[/tex]
Verona dissolves 20. grams of NaCl with water to make a 100 ml solution. What is the molarity of the solution? There are 1,000 mL in 1 L O a. 3.4 M O b.0.34 M O G.58 M O d. 20. M
Answer:
b.0.34 M
Explanation:
Given that:
Mass of NaCl = 20 grams
Molar mass of NaCl = 58.44 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{20\ g}{58.44\ g/mol}[/tex]
[tex]Moles= 0.3422\ mol[/tex]
Given that volume = 100 mL
Also,
[tex]1\ mL=10^{-3}\ L[/tex]
So, Volume = 100 / 1000 L = 0.1 L
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity=\frac{0.3422}{0.1}[/tex]
Molarity = 0.34 M
What volume (in microlitres) of a 200M stock solution of a primer (molecular weight = 7.3 kDa) would you need to include in a 100ul PCR reaction to achieve a final concentration of primer of 300nM?
Answer:
150 × 10⁻⁹ μL
Explanation:
Data provided in the question:
Molarilty of the stock solution, M₁ = 200 M
Final Volume of the solution, V₂ = 100 μL = 100 × 10⁻⁶ L
Final concentration, M₂ = 300 nM = 300 × 10⁻⁹ M
Now,
M₁V₁ = M₂V₂
where,
V₁ is volume of the stock solution
Thus,
200 × V₁ = 100 × 10⁻⁶ × 300 × 10⁻⁹
or
V₁ = 150 × 10⁻⁹ μL
The Prandtl number, Pr, is a dimensionless group important in heat transfer. It is defined as Pr Cp*mu/k = where Cp is the heat capacity of a fluid, mu is the fluid viscosity, and k is the fluid thermal conductivity. For a given fluid, Cp 0.5 J/(g * deg C), k 0.2 W/(m * deg C), and mu 2200 lbm (ft* h}. Determine the value of the Prandtl number for this fluid. Please keep two significant figures in your final answer
Answer:
Pr = 2273.58
Explanation:
Pr = Cp*μ/κ∴ Cp = 0.5 J/g.°C
∴ κ = 0.2 W/m.°C * ( J/s / W ) = 0.2 J/s.m.°C
∴ μ = 2200 Lbm/ft.h * ( 453.592 g/Lbm ) * ( ft / 0.3048 m ) * ( h/3600 s )
⇒ μ = 909.433 g/m.s
⇒ Pr = ((0.5 J/g.°C )*( 909.433 g/m.s )) / 0.2 J/s.m.°C
⇒ Pr = 2273.58
If you mix 10 mL of a 0.1 M HCl solution with 8 mL of
a0.2 M NaOH solution, what will be the resulting pH?
Answer:
The pH of the resulting solution is 12.52.
Explanation:
[tex]Molarity=\frac{n}{V}[/tex]
n = number of moles
V = volume of the solution in Liters
1)1 mol of HCl gives 1 mol of hydrogen ion.
[tex][HCl]=[H^+]=0.1 m[/tex]
Concentration of the hydrogen ion = 0.1 M
Volume of the solution = 10 mL = 0.010 L
[tex]0.1 M=\frac{n}{0.010L}[/tex]
Moles of hydrogen ions = = 0.001 mol
2) 1 mol of NaOH gives 1 mol of hydroxide ion.
[tex][NaOH]=[OH^-]=0.2 M m[/tex]
Concentration of the Hydroxide ions = 0.2 M
Volume of the solution ,V'= 8 mL = 0.008 L
[tex]0.2=\frac{n'}{V'}[/tex]
Moles of hydroxide ions ,n ' = 0.0016
1 mol of HCl neutralizes 1 mol of NaOH ,then 0.001 mol of HCl will neutralize 0.001 mol NaOH.
So left over moles of hydroxide ions in the solution will effect the pH of the solution:
Left over moles of hydroxide ions in the solution = 0.0016 mol - 0.0010 mol = 0.0006 mol
Left over concentration of hydroxide ions:
[tex][OH^-]'=\frac{0.0006 mol}{0.010 L+0.008 L}=0.0333 mol/L[/tex]
[tex]pOH=-\log[OH^-]=-\log[0.03333 M]=1.48[/tex]
pH +pOH = 14
pH = 14 - 1.48 = 12.52
The pH of the resulting solution is 12.52.
How many total electrons does the P^3- ion have? O a. 3 O b. 31 O c.1 O d. 15 O e. 18
Answer:
e. 18
Explanation:
A neutral P atom has an atomic number of 15, which means there are 15 protons in the atom. In order to be neutral, the P atom must also have 15 electrons.
The P³⁻ anion has 3 electrons more than the neutral P atom since it has a charge of -3.
Thus, the total number of electrons are 15 + 3 = 18 electrons.
If the half-life of 37Rb is 4.7x101 years, how long would it take for 0.5 grams of a 2 gram sample to radioactively decay?
Answer: The time required will be 19.18 years
Explanation:
All the radioactive reactions follows first order kinetics.
The equation used to calculate half life for first order kinetics:
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
We are given:
[tex]t_{1/2}=4.7\times 10^1yrs[/tex]
Putting values in above equation, we get:
[tex]k=\frac{0.693}{4.7\times 10^1yr}=0.015yr^{-1}[/tex]
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]0.015yr^{-1}[/tex]
t = time taken for decay process = ?
[tex][A_o][/tex] = initial amount of the reactant = 2 g
[A] = amount left after decay process = (2 - 0.5) = 1.5 g
Putting values in above equation, we get:
[tex]0.015yr^{-1}=\frac{2.303}{t}\log\frac{2}{1.5}\\\\t=19.18yrs[/tex]
Hence, the time required will be 19.18 years
In the reaction C zH4 + H2 - e) +4 CHo the carbon atoms are a) oxidized b) reduced c) cannot be determined
Answer: Option (b) is the correct answer.
Explanation:
Reduction is defined as the process in which there occurs gain of hydrogen. Whereas oxidation is defined as the process in which there occurs loss of hydrogen.
As the given reaction is as follows.
[tex]C_{2}H_{4} + H_{2} \rightarrow C_{2}H_{6}[/tex]
Since, hydrogen is being added in this chemical reaction. It means that reduction is taking place and carbon atom is reduced.
Thus, we can conclude that in the given reaction carbon atoms are reduced.
Valence electrons are: O a. Electrons that have been lost in an atom O b. Electrons in the nucleus of an atom O c. electrons in the outer-most shell of an atom O d. electrons in the inner-most shell of an atom
Copper sulfate is a blue solid that is used to control algae growth. Solutions of copper sulfate that come in contact with the surface of galvanized (zinc-plated) steel pails undergo the following reaction that forms copper metal on the zinc surface. How many grams of zinc would react with 454 g (1 lb) of copper sulfate (160 g/mol)?
Answer:
185.5156 g
Explanation:
The reaction of copper sulfate with zinc is shown below as:
[tex]CuSO_4+Zn\rightarrow ZnSO_4+Cu[/tex]
Given that :
Amount of copper sulfate = 454 g
Molar mass of copper sulfate = 160 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus, moles are:
[tex]moles= \frac{454\ g}{160\ g/mol}[/tex]
[tex]moles\ of\ copper\ sulfate= 2.8375\ mol[/tex]
From the reaction,
1 mole of copper sulfate reacts with 1 mole of zinc
Thus,
2.8375 moles of copper sulfate reacts with 2.8375 moles of zinc
Moles of Zinc that should react = 2.8375 moles
Mass of zinc= moles×Molar mass
Molar mass of zinc = 65.38 g/mol
Mass of zinc = 2.8375 ×65.38 g = 185.5156 g
Final answer:
185.595 grams of zinc would react with 454 g of copper sulfate, based on the one-to-one mole ratio between copper sulfate and zinc and the molar mass of zinc (65.38 g/mol).
Explanation:
The student is asking how many grams of zinc would react with 454 g of copper sulfate, which has a molar mass of 160 g/mol. According to the provided reaction, CuSO4 (aq) + Zn(s) → Cu(s) + ZnSO4 (aq), there is a one-to-one mole ratio between copper sulfate and zinc. Given the molar mass of copper sulfate (160 g/mol), we first find the number of moles of copper sulfate in 454 g.
Number of moles of CuSO4 = (454 g) / (160 g/mol) = 2.8375 mol
Since the mole ratio between copper sulfate and zinc is 1:1, an equal number of moles of zinc will react with the copper sulfate. We then need to find the molar mass of zinc to convert moles of zinc to grams. Zinc has a molar mass of approximately 65.38 g/mol.
Mass of zinc = Number of moles of Zn × molar mass of Zn = 2.8375 mol × 65.38 g/mol = 185.595 g
Therefore, 185.595 grams of zinc would react with 454 g of copper sulfate.
Which is the largest scale? Subatomic Miniscopic Atomic Macroscopic Microscopic
Answer:
Macroscopic scale
Explanation:
Subatomic scale is the scale at which atomic constituents, such as nucleus which contains protons and neutrons, and electrons, which orbit in the elliptical paths around nucleus exists.
Miniscopic scale is a reference scale and is not a standard scale for measurement. Usually, this refers to minute objects.
Atomic scale is the scale which is at size of the atoms.
Macroscopic scale is length scale on which the objects or the phenomena are enough large to be visible with naked eye, without magnifying the optical instruments. It is the largest scale.
Microscopic scale is scale of the objects that require microscope to see them.
Assume a pill has a dosage of 350 mg of medication. How much medication is this in grams?
Answer:
0.350 grams
Explanation:
The given mass of the pill that has to be taken as a dosage for the medication = 350 mg
The medication has to be determined in grams which means that milligram has to be converted to grams.
The conversion of mg to g is shown below:
[tex]1\ milligram=10^{-3}\ gram[/tex]
So,
[tex]350\ milligram=350\times 10^{-3}\ gram[/tex]
[tex]350\ milligram=0.350\ gram[/tex]
The medication required in grams = 0.350 grams
Final answer:
To convert 350 mg to grams, divide by 1000, resulting in 0.35 grams of medication.
Explanation:
To convert the dosage of medication from milligrams to grams, you need to know the conversion factor between these two units.
There are 1000 milligrams in one gram.
Therefore, you can find the amount in grams by dividing the milligram dosage by 1000.
For the pill with a dosage of 350 mg of medication, the conversion to grams would be:
Total given grams/1000
= 350 mg ÷ 1000
= 0.35 grams
A liquid of mass 10 kg is enclosed in a cylinder of radius 1 m and length 5 m, what is the density of liquid? a) 0.63 kg/m3 b) 0.44 kg/m3 c) 0.54 kg/m3 d) 0.83 kg/m3
Answer:
a) [tex]0.63\frac{kg}{m^{3}}[/tex]
Explanation:
Density is given by the expression [tex]d=\frac{m}{V}[/tex], where m is the mass of the substance and V is the volume occupied by the substance.
As the problem says that the liquid is enclosed in a cylinder, you should find the volume of that cylinder that will be the same volume of the liquid, so:
For a cylinder the volume is given by V=[tex]2\pi r^{2}h[/tex]
Replacing the values given, we have:
[tex]V=2\pi (1m)^{2}(5m)[/tex]
[tex]V=15.708m^{3}[/tex]
Replacing the values of m and V in the equation of density, we have:
[tex]d=\frac{10kg}{15.708m^{3}}[/tex]
[tex]d=0.63\frac{kg}{m^{3}}[/tex]
As accurately as possible find the residual molar volume of saturated liquid water at 83 bar.
Answer:
v = 2.512 E-5 m³/mol
Explanation:
∴ P = 80 bar → V = 0.001384 m³/Kg......sat. liq water table
∴ P = 85 bar → V = 0.0014013 m³/Kg
⇒ P = 83 bar → V = ?
specific volume ( V ):
⇒ V = 0.001384 + (( 83 - 80 ) / ( 85 - 80 ))*( 0.0014013 - 0.001384 )
⇒ V = 0.00139438 m³/Kg
molar volume ( v ):
∴ Mw water = 18.01528 g/mol
⇒ v = 0.00139438 m³/Kg * ( Kg/1000g ) * ( 18.01528 g/mol )
⇒ v = 2.512 E-5 m³/mol
If heat flows into a system and the system does work on the surroundings, what will be the signs on q and w? Select the correct answer below O positive q, positive w O positive q, negative w O negative q, positive w O negative q, negative w
Answer:
q = Positive
w = Negative
Explanation:
As per first law of thermodymanics,
ΔE = q + w
Where,
ΔE = Change in internal energy
q = Heat absorbed or heat released by the system
w = Work done
Sign conventions are used for heat transfer and work done during a thermodynamics process.
Sign convention for Heat transfer
q is positive when heat is added to the system or heat absorbed by the system this is because energy of the system is increased.q is negative when heat is withdrawn from the system or heat released by the system.Sign convention for Work done
w is positive if work is done on the system or work is done by the surroundingsw is negative if work is done on the system or work is done on the surrounding.In the given question, work is done on the surroundings so, w is negative.
Heat flows into a system or in other word heat is added to the system,
So q is positive.
What typically occurs in a substance where hydrogen bonding exists when compared to the same substance without H-bonds?
Question options:
A) Decrease in boiling point and decrease in vapor pressure
B) Increase in boiling point and decrease in vapor pressure
C) Increase in boiling point and increase in vapor pressure
E) Decrease in boiling point and increase in vapor pressure
F) There is no difference
Answer:
B) Increase in boiling point and decrease in vapor pressure
Explanation:
Vapor pressure is inversely related to the Boiling point , as
higher the boiling point, lower the vapor pressure. and
Lower the boiling point, higher the vapor pressure.
Hydrogen bonding.
The electrostatic attraction between Hydrogen , bonded to electronegative atom like F, O, N and the more electronegative atom is called as Hydrogen bonding.
For example -
In alcohols, - OH group has Hydrogen that is bonded to more electronegative atom O.
As ,
Extra energy is required to break Hydrogen bonds.
because the substance which exhibits Hydrogen bonding have lower vapor pressure than that of the substance with out Hydrogen bonding.
Hence , the substance with Hydrogen bonding , has higher boiling point,.
Hence , the correct option is Increase in boiling point and decrease in vapor pressure .
Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. The chemical reaction is H2001) H2(g) + 0.502(g) Data (at 298°K and 1 atm): AH = 286 kJ for this reaction, Suzo = 70 JK, SH2 = 131 JIK, and Soz = 205 J/ºK.
Explanation:
The given data is as follows.
[tex]\Delta H[/tex] = 286 kJ = [tex]286 kJ \times \frac{1000 J}{1 kJ}[/tex]
= 286000 J
[tex]S_{H_{2}O} = 70 J/^{o}K[/tex], [tex]S_{H_{2}} = 131 J/^{o}K[/tex]
[tex]S_{O_{2}} = 205 J/^{o}K[/tex]
Hence, formula to calculate entropy change of the reaction is as follows.
[tex]\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)[/tex]
= [tex][(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}][/tex]
= [tex][(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)][/tex]
= 163.5 J/K
Therefore, formula to calculate electric work energy required is as follows.
[tex]\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}[/tex]
= [tex]286000 J - (163.5 J/K \times 298 K)[/tex]
= 237.277 kJ
Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.
Final answer:
To find the electrical work required for the electrolysis of water to produce 1 mole of hydrogen, calculate the Gibbs Free Energy (ΔG) for the reaction. Using the given thermodynamic data, ΔG at 298 K is 236.4 kJ, representing the minimum electrical work needed.
Explanation:
The student has asked how to determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298 K and 1 atm. The question involves understanding the thermodynamics of the reaction: H2O(l) → H2(g) + 0.5O2(g), with given data points including ΔH and standard entropies (S°) for the reactants and products. To find the electrical work required, you first calculate the ΔG (Gibbs Free Energy) of the reaction using the formula ΔG = ΔH - TΔS. Knowing ΔG allows you to determine the maximum work that can be extracted from the reaction, which, in the case of electrolysis, corresponds to the minimum work required to drive the reaction in reverse.
ΔS for the reaction can be calculated using the given entropies: ΔS = ⅛∑S°(products) - ⅛∑S°(reactants) = (131 + 0.5×205) - 70 = 166.5 J/K. Therefore, ΔG at 298 K can be calculated as ΔG = 286,000 J - (298K × 166.5 J/K) = 286,000 J - 49,617 J = 236,383 J or 236.4 kJ. This value represents the minimum electrical work required to produce one mole of hydrogen gas via electrolysis of water under the specified conditions.
What is the ground state electron configuration of a
calciumatom?
Answer:
[tex]1s^22s^22p^63s^23p^64s^2[/tex]
Explanation:
Calcium is the chemical element with symbol Ca and the atomic number equal to 20. As alkaline earth metal, the element, calcium is reactive metal. IT lies in the second group and forth period of the periodic table.
The number of the valence electrons of the calcium element is 2 and thus primarily denotes these electrons and forms ionic bond.
The ground state- electron configuration for calcium is: [tex]1s^22s^22p^63s^23p^64s^2[/tex]
What is the transition interval for phenol red? 24 a. pH 3.1-4.4 b. pH 6.4-8.0 c. pH 6.2-7.6 d. pH 8.0-10.0
Answer:
The correct option is: b. pH 6.4-8.0
Explanation:
Phenol red is a weak acid that is used as a pH indicator and exists in the form of stable red crystals.
The color of the phenol red solution changes from yellow to red when the change in pH is observed. The color of phenol red transitions from yellow to red when the pH is 6.8 - 8.2 or 6.4 - 8.0
Above the pH of 8.2, the phenol red solution turns a bright pink in color.
An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (report your answer to three significant figures)?
Answer:
The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0. It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it
Explanation:
This is the begin:
Q1 = Q which is gained from the ice to be melted
Q2 = Q which is lost from the water to melt the ice
Q1 + Q2 = 0
We are informed that the ice is at 0 ° so we have to start calculating how many J, do we need to melt it completely. If the ice had been at a lower temperature, it should be brought to 0 ° with the formula
Q = mass. specific heat. (ΔT)
and then make the change of state by the latent heat of fusion.
The heat of fusion for water at 0 °C is approximately 334 joules per gram.
So Q = Hf . mass
Q1 = 334 J/g . 8.32 g = 2778,88 J
For water we should use this:
Q = mass. specific heat. (ΔT)
Q2 = 55g . 4180 J/kg. K (Tfinal -T initial)
Q2 = 55g . 4180 J/kg. K (Tfinal -T initial)
(notice we have kg, so we have to convert 55 g, to kg, 0,055kg so units can be cancelled)
Q2 = 0,055kg . 4180 J/kg. K (Tfinal (The unknown) -25°)
T° should be in K for the units of Specific heat but it is the same. The difference is the same, in K either in °C
25°C = 298K
Q2 = 0,055kg . 4180 J/kg. K (Tfinal -298K)
Now the end
Q1 + Q2 = 0
334 J/g . 8.32 g + 0,055kg . 4180 J/kg. K (Tfinal -298K)
2778,88 J + 229,9 J/K (Tfinal - 298 K) = 0
2778,88 J + 229,9 J/K x Tfinal - 68510,2 J = 0
229,9 J/K x Tfinal = 68510,2 J - 2778,88 J
229,9 J/K x Tfinal = 65731,4 J
Tfinal = 65731,4 J / 229,9 K/J
Tfinal = 285,9 K
Tfinal = 285,9 K - 273K = 12,9 °C
The final temperature of the water after an 8.32 g ice cube at 0.00 °C is placed into 55 g of water at 25 °C and all the ice melts is 12.959 °C, using the conservation of energy and assuming no heat loss to the surroundings.
When an 8.32 g ice cube at 0.00 °C is placed into 55 g of water at 25 °C, we can find the final temperature after all the ice is melted by applying the principle of conservation of energy. The heat gained by the ice melting must be equal to the heat lost by the water cooling down. We will assume that no heat is lost to the surroundings in this perfectly insulated system.
We need to calculate the heat required to melt the ice cube (Qmelt) using the heat of fusion of water (which is 79.9 cal/g or 334 J/g), and the heat lost by the water as it cools down (Qwater) using the specific heat of water (which is 4.184 J/g°C).
First, calculate the heat necessary to melt the ice:
Qmelt = mass of ice * heat of fusion
Qmelt = 8.32 g * 334 J/g
Qmelt = 2778.08 J
Next, set up the equation based on the conservation of energy, where the heat lost by the water (Qwater) equals the heat gained by the ice (Qmelt):
Qwater = mass of water * specific heat of water * temperature change
Qwater = Qmelt
55 g * 4.184 J/g°C * (25 °C - final temperature) = 2778.08 J
Solving for the final temperature:
230.62 J/°C * (25 °C - final temperature) = 2778.08 J
25 °C - final temperature = 2778.08 J / 230.62 J/°C
25 °C - final temperature = 12.041 °C
Final temperature = 25 °C - 12.041 °C
Final temperature = 12.959 °C
Therefore, the final temperature of the water after all the ice has melted is 12.959 °C (rounded to three significant figures).
A sealed can with an internal pressure of 721 mmHg at
25degrees C is thrown into an incinerator operating at 755 degrees
C.What will be the pressure inside the heated can, assuming
thecontainer remains intact during incineration?
Answer:
2486 mmHg
Explanation:
Gay-Lussac's Law states the pressure varies directly with temperature when volume remains constant:
P₁/T₁ = P₂/T₂
Where P₁ and T₁ are initial pressure and temperature and P₂ and T₂ are final pressure and temperatue.
The problem says initial pressure is 721 mmHg, initial temperature is 25°C and final temperature is 755°C. The question is final pressure.
°C must be converted to absolute temperature (K), thus:
25°C + 273,15 = 298,15 K
755°C + 273,15 = 1028,15 K
Thus, pressure P₂ is:
(T₂·P₁) / T₁ = P₂
1028,15K · 721mmHg / 298,15 K = 2486 mmHg
I hope it helps!
Air containing 0.06% carbon dioxide is pumped into a room whose volume is 12,000 ft3. The air is pumped in at a rate of 3,000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.3% carbon dioxide, determine the subsequent amount A(t), in ft3, in the room at time t.
Answer:
[tex]A(t)=1.8+34.2*e^{-t}[/tex]
Explanation:
The concentration of CO2 in the room will be the amount of CO2 in the room at time t, divided by the volume of the room.
Let A(t) be the amount of CO2 in the room, in ft3 CO2.
The air entering the room is 3000 ft3/min with 0.06% concentrarion of CO2. That can be expressed as (3000*0.06/100)=1.8 ft3 CO2/min.
The mixture leaves at 3000 ft3/min but with concentration A(t)/V. We can express the amount of CO2 leaving the room at any time is A(t).
We can write this as a differential equation
[tex]dA/dt=v_i-v_o=1.8-A[/tex]
We can rearrange and integrate
[tex]dA/dt=v_i-v_o=1.8-A\\\\dA/(A-1.8)=-dt\\\\\int(dA/(A-1.8) = -\int dt\\\\ln(A-1.8)=-t+C\\\\A-1.8=e^{-t}* e^{C}=C*e^{-t}\\\\A=1.8+C*e^{-t}[/tex]
We also know that A(0) = 12000*(0.3/100)=36 ft3 CO2.
[tex]A(0)=1.8+C*e^{-0}\\36=1.8+C*1\\C=34.2[/tex]
Then we have the amount A(t) as
[tex]A(t)=1.8+34.2*e^{-t}[/tex]
The subsequent amount A(t) of carbon dioxide in the room at time t is determined by solving the differential equation that models the problem as a tank mixing problem, accounting for the rate of air being pumped in and out of the room.
Explanation:To determine the subsequent amount A(t), in ft³, of carbon dioxide in the room at any given time t, we need to set up a differential equation that accounts for the rate of air being pumped in and out of the room. This situation is analogous to a classic problem in differential equations and mathematical modeling called the tank mixing problem.
The rate at which carbon dioxide enters the room is constant at 0.06% of the 3,000 ft³/min being pumped in, which equals 1.8 ft³/min of CO₂. The rate at which it leaves the room is proportional to the concentration of carbon dioxide in the room at time t. The change in the amount of carbon dioxide at any moment is then given by the rate in minus the rate out.
Let C0 be the initial concentration of CO₂ (0.3%), CV be the volume of the room (12,000 ft³), and r be the rate of air exchange (3,000 ft³/min). The differential equation modeling this situation is:
dA(t)/dt = r * (0.06% * V) - r * (A(t) / V)with initial condition A(0) = C0 * V. This equation can be solved using separation of variables and integrating both sides,
A(t) = V * (C0 - 0.0006) * e^(-rt/V) + 0.0006VUpon inserting the values for V, C0, and r, we can find the expression for A(t).
What is the pH if 1mL of 0.1M HCl is added to 99mL of pure water?
Now if instead of pure water a buffer is used: HPO4-2/H2PO4- pKa = 7.2 Assume the initial pH of this buffer is 7 (like the pure water example). -First you must you must use the Henderson-Hasselbalch equation to determine the ratio of HPO4-2/H2PO4- , which is 0.063M to .1M. Using the same amount of HCl added (.001M), determine the change in pH that occurs to the buffer when the HCl is added.
(i already answered the first part, I just need the second part. Show and explain your work please!)
Answer:
pH of buffer after addition of 1 mL of 0,1 M HCl = 7,0
Explanation:
It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:
pH = pka + log₁₀
Where A⁻ is conjugate base and HA is conjugate acid
The equilibrium of phosphate buffer is:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,2
Thus, Henderson–Hasselbalch equation for 7,00 phosphate buffer is:
7,0 = 7,2 + log₁₀ [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex]
Ratio obtained is:
0,63 = [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex]
As the problem said you can assume [H₂PO₄⁻] = 0,1 M and [HPO4²⁻] = 0,063M
As the amount added of HCl is 0,001 M the concentrations in equilibrium are:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺
0,1 M +x 0,063M -x 0,001M -x -because the addition of H⁺ displaces the equilibrium to the left-
Knowing the equation of equilibrium is:
[tex]K_{a} = \frac{[HPO_{4}^{2-}][H^{+}]}{[H_{2} PO_{4}^{-}]}[/tex]
Replacing:
6,20x10⁻⁸ = [tex]\frac{[0,063-x][0,001-x]}{[0,1+x]}[/tex]
You will obtain:
x² -0,064 x + 6,29938x10⁻⁵ = 0
Thus:
x = 0,063 → No physical sense
x = 0,00099990
Thus, [H⁺] in equilibrium is:
0,001 M - 0,00099990 = 1x10⁻⁷
Thus, pH of buffer after addition of 1 mL of 0,1 M HCl =
-log₁₀ [1x10⁻⁷] = 7,0
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!
I hope it helps!
Drugs.com contains information on all the following categories except: a. Pill Identification b. Product d. Manufacturers C. Approval date b. Drugs A-Z c. Interaction Checker e. Calculators
Answer: Drugs.com is the site which have all we need regarding medicines and uses of medicines. It give the facility to search A-Z drugs, once you have searched the desired medicine by this feature of site, it further gives information about identification of pills, approval date, and interaction between drug-drug and drug- the food you eating, the manufacturing date of medicines. It doesn't give information about the calculators.
Therefore, (e) is the correct option here.
Drugs.com is a resourceful website for information about various drugs. It provides details like Pill Identification, Drugs A-Z, Interaction Checker, and more. However, it does not provide any calculator-type feature.
Explanation:Drugs.com provides comprehensive information related to different types of medications. It covers areas such as Pill Identification, Product, Manufacturers, Drug Approval Date, Drugs A-Z, and Interaction Checker. However, Drugs.com does not offer any feature or functionality related to 'Calculators' pertaining to drug calculations or dosage calculations. Use of the site could enable someone to determine whether they have a substance use disorder, but it does not replace professional medical advice.
Learn more about Drugs.com here:https://brainly.com/question/13264475
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