Answer:
FALSE
Explanation:
Grunge refers to the genre of rock music and the fashion inspired by it. It originated in the mid-1980s in Seattle, Washington State.
Grunge was described as the fusion of punk rock and heavy metal.
This genre of music became popular in the early mid-1990s and included lyrics based on the theme of emotional and social alienation, betrayal, abuse, trauma etc.
The statement that grunge is a rock style from Detroit is false; grunge originated in Seattle, Washington. This music genre is characterized by a stripped-down aesthetic, distorted guitars, and socially conscious lyrics, gaining popularity in the late 1980s and early 1990s.
Grunge is indeed a style of rock music, but it is not from Detroit; that statement is false. Grunge, often referred to as the Seattle sound, originated in Seattle, Washington and became widely popular in the late 1980s and early 1990s. It represented a departure from the extravagant stage productions and fashion of contemporary rock bands, favoring a more stripped-down sound with an emphasis on distorted electric guitars and direct, often socially conscious lyrics.
The grunge movement produced several iconic bands, including Nirvana, Pearl Jam, and Soundgarden, which significantly shaped the alternative rock landscape. Seattle's geographic isolation played a pivotal role in the development of grunge music. Local bands evolved their unique sound and aesthetic away from the influence of mainstream music industry expectations, which at the time were centered around Los Angeles. This isolation contributed to grunge's authenticity and appeal when it eventually burst onto the national and international stages.
Contrary to the glamorous rock scene in Los Angeles, grunge musicians typically wore simple, unremarkable clothing, often sourced from second hand stores, and their lyrics addressed themes like political issues, mental health, and substance abuse, resonating with the disaffected Generation X.
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.
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Write 0.0251089 in Scientific Notation with 4 significant figures.
Answer: The given number in scientific notation is [tex]2.511\times 10^{-2}[/tex]
Explanation:
Scientific notation is defined as the notation in which a number is expressed in the decimal form. This means that the number is always written in the power of 10 form. The numerical digit lies between 0.1.... to 9.9.....
If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.
We are given:
A number having value = 0.0251089
Converting this into scientific notation, we get:
[tex]\Rightarrow 0.0251089=2.511\times 10^{-2}[/tex]
Hence, the given number in scientific notation is [tex]2.511\times 10^{-2}[/tex]
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.
The pOH of an aqueous solution at 25°C was found to be 2.40. The pH of this solution is The hydronium ion concentration is The hydroxide ion concentration is TEM. Submit Answer Retry Entire Group 9 more group attempts rema
Answer:
The hydroxide ion concentration is 0.003981 M.
The hydronium ion concentration is[tex] 2.5119\times 10^{-12} M[/tex].
Explanation:
The pOH of an aqueous solution at 25°C = 2.40
[tex]pOH=-\log[OH^-][/tex]
[tex]2.40=-\log[OH^-][/tex]
[tex][OH^-]=0.003981 M[/tex]
The hydroxide ion concentration is 0.003981 M.
The pH of an aqueous solution at 25°C = ?
The relationship between pH and pOH :
pH + pOH = 14
[tex]pH=14- pOh = 14-2.40 =11.6 [/tex]
[tex]pH=-\log[H_3O^+][/tex]
[tex]11.6=-\log[H_3O^+][/tex]
[tex][H_3O+]=2.5119\times 10^{-12} M[/tex]
The hydronium ion concentration is[tex] 2.5119\times 10^{-12} M[/tex].
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.
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!
The reaction described by the equation CH 3 Cl + NaOH → CH 3 OH + NaCl follows the second-order rate law, rate = k [ CH 3 Cl ] [ NaOH ] . When this reaction is carried out with starting concentrations [ CH 3 Cl ] = 0.2 M and [ NaOH ] = 1.0 M , the measured rate is 1 × 10 − 4 mol L − 1 s − 1 . What is the rate after one-half of the CH 3 Cl has been consumed? (Caution: The initial concentrations of the starting materials are not identical in this experiment. Hint: Determine how much of the NaOH has been consumed at this point and what its new concentration is, compared with its initial concentration.)
Answer:
The rate is [tex] 4,5 \times 10^{-5}\frac{mole}{Ls}[/tex]
Explanation:
Stoichiometry
[tex]CH_{3}Cl+NaOH \rightarrow CH_{3}OH+NaCl [/tex]
Kinetics
[tex]-r_{A}=k \times [CH_{3}Cl] \times [NaOH] [/tex]
The rate constant K can be calculated by replacing with the initial data
[tex] 1 \times 10^{-4}\frac{mole}{Ls}=k \times [0,2M] \times [1,0M] =5 \times 10^{-4}\frac{L}{mole s}[/tex]
Taking as a base of calculus 1L, when half of the [tex] CH_{3}Cl [/tex] is consumed the mixture is composed by
[tex] 0,1 mole CH_{3}Cl [/tex] (half is consumed)
[tex] 0,9 mole NaOH [/tex] (by stoicheometry)
[tex] 0,1 mole CH_{3}OH [/tex]
[tex] 0,1 mole NaCl [/tex]
Then, the rate is
[tex]-r_{A}=5 \times 10^{-4} \frac{L}{mole s}\times 0,1M \times 0,9 M=4,5 \times 10^{-5}\frac{mole}{Ls}[/tex]
The reaction rate decreases because there’s a smaller concentration of reactives.
How many grams of sodium fluoride should be added to 300. mL of 0.0310 M of hydrofluoric acid to produce a buffer solution with a pH of 2.60?
Answer : The mass of sodium fluoride added should be 0.105 grams.
Explanation : Given,
The dissociation constant for HF = [tex]K_a=6.8\times 10^{-4}[/tex]
Concentration of HF (weak acid)= 0.0310 M
First we have to calculate the value of [tex]pK_a[/tex].
The expression used for the calculation of [tex]pK_a[/tex] is,
[tex]pK_a=-\log (K_a)[/tex]
Now put the value of [tex]K_a[/tex] in this expression, we get:
[tex]pK_a=-\log (6.8\times 10^{-4})[/tex]
[tex]pK_a=4-\log (6.8)[/tex]
[tex]pK_a=3.17[/tex]
Now we have to calculate the concentration of NaF.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[NaF]}{[HF]}[/tex]
Now put all the given values in this expression, we get:
[tex]2.60=3.17+\log (\frac{[NaF]}{0.0310})[/tex]
[tex][NaF]=0.00834M[/tex]
Now we have to calculate the moles of NaF.
[tex]\text{Moles of NaF}=\text{Concentration of NaF}\times \text{Volume of solution}=0.00834M\times 0.300L=0.0025mole[/tex]
Now we have to calculate the mass of NaF.
[tex]\text{Mass of }NaF=\text{Moles of }NaF\times \text{Molar mass of }NaF=0.0025mole\times 42g/mole=0.105g[/tex]
Therefore, the mass of sodium fluoride added should be 0.105 grams.
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.
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).
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 .
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.
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]
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.
Be sure to answer all parts. Calculate the molality, molarity, and mole fraction of FeCl3 in a 24.0 mass % aqueous solution (d = 1.280 g/mL).
Answer:
m= 1.84 m
M= 1.79 M
mole fraction (X) =
Xsolute= 0.032
Xsolvent = 0.967
Explanation:
1. Find the grams of FeCl3 in the solution: when we have a mass % we assume that there is 100 g of solution so 24% means 24 g of FeCl3 in the solution. The rest 76 g are water.
2. For molality we have the formula m= moles of solute / Kg solvent
so first we pass the grams of FeCl3 to moles of FeCl3:
24 g of FeCl3x(1 mol FeCl3/162.2 g FeCl3) = 0.14 moles FeCl3
If we had 76 g of water we convert it to Kg:
76 g water x(1 Kg of water/1000 g of water) = 0.076 Kg of water
now we divide m = 0.14 moles FeCl3/0.076 Kg of water
m= 1.84 m
3. For molarity we have the formula M= moles of solute /L of solution
the moles we already have 0.14 moles FeCl3
the (L) of solution we need to use the density of the solution to find the volume value. For this purpose we have: 100 g of solution and the density d= 1.280 g/mL
The density formula is d = (m) mass/(V) volume if we clear the unknown value that is the volume we have that (V) volume = m/d
so V = 100 g / 1.280 g/mL = 78.12 mL = 0.078 L
We replace the values in the M formula
M= 0.14 moles of FeCl3/0.078 L
M= 1.79 M
3. Finally the mole fraction (x) has the formula
X(solute) = moles of solute /moles of solution
X(solvent) moles of solvent /moles of solution
X(solute) + X(solvent) = 1
we need to find the moles of the solvent and we add the moles of the solute like this we have the moles of the solution:
76 g of water x(1 mol of water /18 g of water) = 4.2 moles of water
moles of solution = 0.14 moles of FeCl3 + 4.2 moles of water = 4.34 moles of solution
X(solute) = 0.14 moles of FeCl3/4.34 moles of solution = 0.032
1 - X(solute) = 1 - 0.032 = 0.967
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.
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
Hafnium has six naturally occurring isotopes: 0.16% of 174Hf, with an atomic weight of 173.940 amu; 5.26% of 176Hf, with an atomic weight of 175.941 amu; 18.60% of 177Hf, with an atomic weight of 176.943 amu; 27.28% of 178Hf, with an atomic weight of 177.944 amu; 13.62% of 179Hf, with an atomic weight of 178.946 amu;. and 35.08% of 180Hf, with an atomic weight of 179.947 amu. Calculate the average atomic weight of Hf. Give your answer to three decimal places.
The average atomic weight of Hafnium is calculated by multiplying the atomic weight of each isotope by its relative abundance and summing the products. The result is an average atomic weight of 178.433 amu.
Explanation:To calculate the average atomic weight of Hafnium (Hf), we multiply the atomic weight of each isotope by its natural abundance, expressed as a fraction, and then sum the results. Here are the calculations:
(0.0016 × 173.940 amu) for 174Hf(0.0526 × 175.941 amu) for 176Hf(0.1860 × 176.943 amu) for 177Hf(0.2728 × 177.944 amu) for 178Hf(0.1362 × 178.946 amu) for 179Hf(0.3508 × 179.947 amu) for 180HfAdding these products together we get the average atomic weight of Hafnium:
(0.0016 × 173.940) + (0.0526 × 175.941) + (0.1860 × 176.943) + (0.2728 × 177.944) + (0.1362 × 178.946) + (0.3508 × 179.947) = 0.278304 + 9.255366 + 32.922978 + 48.523072 + 24.373172 + 63.079956 = 178.432848 amu
The average atomic weight of Hafnium truncated to three decimal places is 178.433 amu.
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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.
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 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.
One of the significant achievements of Fahrenheit was to: O make thermometers smaller using air O make thermometers compact using alcohol O make thermometers more accurate using alcohol O make thermometers smaller using mercury
Answer:
make thermometers smaller using mercury
Explanation:
Daniel Fahrenheit invented first accurate thermometer which used mercury instead of the alcohol and the water mixtures. In laboratory, he used this invention of him to develop first temperature scale which was enough precise to become the worldwide standard.
The key to the Fahrenheit's thermometer was that the mercury is able to rise and fall within tube without sticking to sides. It was ideal substance for the reading temperatures since mercury expanded at more constant rate than the alcohol and is also able to be read the temperature at much higher and also lower temperatures.
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.
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
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
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
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]
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.
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.
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