Answer:
Sodium chloride has a crystalline face-centered cubic structure whereas metallic sodium body-centered cubic structure.
Explanation:
Hello, atomic arrangements provide the molecules different features and behaviors, since the sodium metal has a body-centered cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-body-centered.svg) the lack of inner atoms, allows the material to be soft and bendable. On the other hand the compacted sodium chloride's face-centered cubic structure (https://en.wikipedia.org/wiki/Cubic_crystal_system#/media/File:Cubic-face-centered.svg), provides it a crystalline structure which is hard and brittle since the atoms are closer.
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Liquid dimethyl disulfide (CH3SSCH3) flows through a pipe with a mass flow rate of 86.0 g's. Given that the density of dimethyl disulfide is 1.0625 g/cm, find: The molar flow rate in molimin: Number mol/min The volumetric flow rate in L/hr.
Explanation:
Molar mass of [tex]CH_{3}SSCH_{3}[/tex] is 94 g/mol. As it is known that number of moles is equal to mass of a substance divided by its molar mass.
Then, calculate the number of moles as follows.
No. of moles = [tex]\frac{86.0 g}{94 g/mol}[/tex] in 1 s
= 0.914 mol
So, in 60 sec number of moles will be equal to 0.914 x 60 = 54.89 mol/min.
Hence, the molar flow rate = 54.89 mol/min
Also, density is equal to mass of a substance divided by its volume.
Density = [tex]\frac{mass}{volume}[/tex]
Volume = [tex]\frac{mass}{Density}[/tex]
= [tex]\frac{86.0 g}{1.0625 g/cm^{3}}[/tex]
= 80.941 [tex]cm^{3}[/tex]
As, 80.941 [tex]cm^{3}[/tex] of volume flows in 1 s . Therefore, flow of volume in 1 hour will be calculated as follows.
In 1 hr = 80.941 [tex]cm^{3} \times 3600[/tex]
= 291388.24 [tex]cm^{3}/hr[/tex]
Since, 1 [tex]cm^{3}[/tex] = 0.001 L.
So, 291388.24 [tex]cm^{3}/hr \times 0.001 L/cm^{3}[/tex]
= 291.38824 L/hr
Thus, we can conclude that molar flow rate in mol/min is 54.89 mol/min and the volumetric flow rate in L/hr is 291.38824 L/hr.
Chlorine is used to disinfect swimming pools. The accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 g of chlorine per million grams of water. Calculate the volume of a chlorine solution (in milliliters) a homeowner should add to her swimming pool if the solution contains 5.50 percent chlorine by mass and there are 6.52 × 104 gallons (gal) of water in the pool (1 gal = 3.79 L; density of liquids = 1.00 g/mL). Enter your answer in scientific notation.
Answer:
The volume of a chlorine solution a homeowner should add to her swimming pool is [tex]4.480\times 10^3 mL[/tex].
Explanation:
The solution contains 5.50 percent chlorine by mass.
Mass of the chlorine required to added in the pool= x
Mass of the water in the pool = y
Volume of the water ,V= [tex]6.52\times 10^4 gal=2.464\times 10^5 L[/tex]
(1 gal = 3.78 L)
V = [tex]2.464\times 10^5 L=2.464\times 10^8 mL[/tex]
( 1 L = 1000 mL)
Density of the water ,d= 1.00 g/mL
[tex]y=d\times V=1.00 g/mL\times 2.464\times 10^8 mL=2.464\times 10^8 g[/tex]
y = [tex]2.464\times 10^8 g[/tex]
1.00 g of chlorine per million grams of water .
Then for [tex]2.464\times 10^8 g[/tex] of water, we required x amount of chlorine:
[tex]z=\frac{2.464\times 10^8 g}{10^6 g}=246.4 g[/tex]
x = 246.4 grams of chlorine
Mass of the chlorine solution = M'
[tex](w/w)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100[/tex]
[tex]5.5\%=\frac{z}{M'}\times 100[/tex]
[tex]M'=\frac{246.4}{5.5}\times 100=4,480 g[/tex]
Density of the chlorine solution = D = 1 g/mL
Volume of the chlorine solution to be added in the pool : v
[tex]v=D\times M'=1 g/mL\times 4,480 g=4480 mL[/tex]
[tex]v=4.480\times 10^3 mL[/tex]
The volume of a chlorine solution a homeowner should add to her swimming pool is [tex]4.480\times 10^3 mL[/tex].
Answer:
4.45 L
Explanation:
The concentration of chlorine in the pool must be 1.00g/10⁶g, and the solution has 5.50g/100g, so it will be diluted in the pool.
The product of the concentration by the volume is always constant in a solution, thus, we can use the equation:
C1V1 = C2V2
Where C is the concentration, V is the volume, 1 represents the solution, and 2 the pool. So, C1 = 5.50g/100g, C2 = 1.00g/10⁶g, V2 = 6.52x10⁴ gal = 247108 L. Thus,
(5.50/100)*V1 = (1.00/10⁶)*247108
0.055V1 = 0.247108
V1 = 4.5 L
One kilogram is slightly more than A) 2 B) 5 U.S. pounds. C) 0.5 D) 1 11) The mass of a proton is 1.67 A) 1.67. 10-30 G B 10-27 kg.
Answer:
The correct option is: a) 2 U.S. pounds;
B) 1.672622×10⁻²⁷ kg
Explanation
Pound is a United States customary unit for mass and is abbreviated as lb. One pound is defined as being equal to 0.45359237 kilograms (kg), exactly.
Kilogram is the SI unit of mass and is abbreviated as kg.
Since, 1 lb = 0.45359237 kg
Therefore, 1 kg = 1 ÷ 0.45359237 lb ≈ 2.20462 lb
Therefore, we can say that 1 kilogram is slightly more than 2 pounds.
A proton is a positively charged subatomic particle, that possesses +1e electric charge and has mass approximately equal to 1.672622×10⁻²⁷ kg.
In AG calculations, temperature is expressed in a. degrees Celsius. b. kelvins. c. degrees Fahrenheit, d. kilojoules.
Answer:
The correct option is: a. degrees Celsius
Explanation:
The anion gap is the difference in the cations and anions in plasma, serum or urine, calculated from medical lab test results. It can be calculated by measuring the concentration of the anions or cations, which are expressed in millimoles/litre (mmol/L) or milliequivalents/liter (mEq/L).
The temperature in this test is expressed in degrees Celsius (°C).
In a combustion furnace, 2094 standard ft3 per hour
of natural gas (Methane) is burned with 6% excess air.
How many standard ft3 of air are drawn from outside
per hour by the fan that supplies the air?
Explanation:
The chemical reaction is as follows.
[tex]CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O[/tex]
It is given that 2094 [tex]ft^{3}/hr[/tex]. And, it is known that 1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]
Hence, convert 2094 [tex]ft^{3}/hr[/tex] into [tex]m^{3}/s[/tex] as follows.
[tex]\frac{2094 ft^{3}/hr}{127133 ft^{3}/hr} \times 1 m^{3}/s[/tex]
= [tex]0.0165 m^{3}/s[/tex]
As ideal gas equation is PV = nRT. So, calculate the number of moles as follows.
n = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{1 atm \times 0.0165 m^{3}}{0.0821 \times 298 K}[/tex]
= 0.673 mol/sec
According to the stoichiometry of the given reaction, 1 mol of methane reacts with 2 mol of oxygen.
So, 1 mol [tex]CH_{4}[/tex] = 2 mol [tex]O_{2}[\tex]
Hence, [tex]O_{2}[\tex] required theoretically = [tex]2 \times 0.673 mol/s[/tex] = 1.346 mol/s
Hence, air required theoretically = [tex]\frac{1.346}{0.21}[/tex] = 6.4095 mol/s.
Since, 6% of excess air is being supplied. Therefore, total air supplied will be calculated as follows.
Total air supplied = [tex]6.4095 mol/s [1 + \frac{6}{100}][/tex]
= 6.794 mol/s
Now, calculate the volume using ideal gas law equation as follows.
PV = nRT
[tex]1 atm \times V = 6.794 mol/s \times 8.21 \times 10^{-5} Latm/K mol \6 times 298 K[/tex]
V = 0.166229 [tex]m^{3}/s[/tex]
Converting calculated volume into [tex]ft^{3}/hr[/tex] as follows.
1 [tex]m^{3}/s[/tex] = 127133 [tex]ft^{3}/hr[/tex]
So, 0.166229 [tex]m^{3}/s[/tex] = [tex]0.166229 m^{3}/s \times 127133 \frac{ft^{3}/hr}{1 m^{3}/s}[/tex]
= 21133.191 [tex]ft^{3}/hr[/tex]
Thus, we can conclude that 21133.191 [tex]ft^{3}/hr[/tex] of air are drawn from outside per hour by the fan that supplies the air.
how could you easily and accurately determine the glass
transition and melting temperature of a polymer?
Answer:
You can use a thermal analysis called Differential Scanning Calorimetry (DSC)
Explanation:
The glass transition temperature (Tg) and melting temperature (Tm) are thermal properties of substances, such as polymers. You can determine them with a very low sample quantity by differential scanning calorimetry (DSC).
In DSC, there are two aluminium pans. In one pan, you put the sample, and the other pan is empty because is the reference pan. The pans are placed in a chamber of a DSC calorimeter and they are heated at a temperature range you select at a given speed (usually in ºC/min). The system register the difference of heat between the pans (sample pan and reference pan) and it shows you a plot of heat flow (in J/s) vs temperature.
Tg is commonly detected as a step in the plot
Tm is usually an endothermic peak (upwards in the plot), because is a process in which heat is absorbed.
Differential scanning calorimetry (DSC) under quasi-isothermal conditions is utilized to accurately determine the glass transition temperature and melting temperature of a polymer, reflecting its chain interactions and molecular weight.
Determining Glass Transition and Melting Temperature
To easily and accurately determine the glass transition temperature (Tg) and melting temperature (Tm) of a polymer, differential scanning calorimetry (DSC) is the most commonly used technique. The glass transition is a measure of the change in the polymer's heat capacity (Cp) as it crosses certain heat energy thresholds, and it can be more accurately determined under quasi-isothermal conditions as these conditions produce precise Cp measurements as a function of time. When a polymer is heated to its Tg, it enters a molten state, and upon cooling, becomes brittle. If the heating continues beyond Tg, the polymer will soften, allowing it to be molded or deformed.
The glass transition temperature can also provide insight into a polymer's chain length and structure, as it reflects the number of van der Waals or entangling chain interactions. Additionally, the molecular weight of a polymer influences Tg, typically following the Flory-Fox equation. The melting temperature (Tm) is the temperature at which a polymer transitions from a crystalline to a molten state and is detectable as a peak on a DSC trace above Tg.
Overall, by using DSC and applying quasi-isothermal steps across the apparent glass transition range, we can observe changing Cp rates that indicate the exact Tg range. Similarly, for Tm, we can detect the transition between crystalline and molten states as the temperature rises above Tg.
A computer depreciates at a rate of 15% per year. If the computer is worth $324.50 when it is 5 years old, how much was it worth when it was new? Round your answer to two decimal places
Answer:
the price of the new computer was [tex]\$ 731.34[/tex]
Explanation:
let's assume that the price of the computer was [tex]\$ x[/tex] when it was new.
As price depreciates at a rate of 15% therefore we can mathematically express this situation when the computer is 5 yrs old as-
[tex]x\times (1-0.15)^{5}=324.50[/tex]
or, [tex]x\times (0.85)^{5}=324.50[/tex]
or, [tex]x=\frac{324.50}{(0.85)^{5}}[/tex]
or, [tex]x=731.34[/tex]
So, the price of the new computer was [tex]\$ 731.34[/tex]
A 1 m ID spherical cooler made of 2 cm thick foam insulation (k = 0.15 Wm-1K-1) is used to store iced water at 0°C. The tank is located outdoors at 25 °C and is subjected to horizontal cross winds at 1 m/s. Determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0 °C that melts during a 24 h period. Assume the inner surface of the cooler is at 0 °C and take the thermal resistance of the cooler and heat transfer by radiation into consideration. Assume the average surrounding surface temperature for radiation exchange to be 20 °C and the outer surface of the cooler to have an emissivity of 0.85. The heat of fusion of water at atmospheric pressure is hif = 333.7 kJ/kg.
The answer is unknown
A chemist determines by measurements that 0.0200 moles of iodine solid participate in a chemical reaction. Calculate the mass of iodine solid that participates. Round your answer to 3 significant digits. $ x 6 ?
To calculate the mass of iodine solid that participates in a chemical reaction, multiply the number of moles of iodine by its molar mass.
Explanation:To calculate the mass of iodine solid that participates in a chemical reaction, we need to use the concept of moles and the molar mass of iodine. Given that 0.0200 moles of iodine solid participate in the reaction, we can convert moles to grams by multiplying the number of moles by the molar mass of iodine. The molar mass of iodine (I2) is 253.8 grams per mole. Multiplying 0.0200 moles by 253.8 grams/mole gives us a mass of 5.08 grams of iodine solid that participates in the chemical reaction.
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A water treatment plant has a total flow of 45 MGD (mega gallons per day). The plant was designed with five sedimentation basins, each with a total volume of 2500 m3.
Originally, the plant was operating the basins in series: the entire flow goes through one tank, then the second, and so on. What is the retention time in each tank?
The engineer realizes that the sedimentation would improve if they changed the operation to parallel flow. Under this new configuration, the flow is evenly split between the four tanks. What is the retention time in each tank?
Answer:
a) 21.133 minutes for series
b) 84.54 minutes when split into 4
Explanation:
Data provided:
Total flow, V₀ = 45 MGD
total volume of each basin, V = 2500 m³
Now,
1 MGD = 3785.4118 m³/day
also,
1 day = 1440 minutes
thus,
45 MGD = 45 × 3785.4118 m³/day
or
45 MGD = 170343.5305 m³/day
and,
170343.5305 m³/day in 'm³/min'
= 170343.5305 / 1440
= 118.2941 m³/min.
Therefore,
The retention time, t = [tex]\frac{\textup{Volume of tank}}{\textup{Volumetric Flowrate }}[/tex]
or
The retention time, t = [tex]\frac{\textup{2500}}{\textup{118.2941 }}[/tex]
or
The retention time, t = 21.13 min
Hence,
for series of tanks the retention time is 21.133 min.
Now,
on splitting the tanks in 4
the volumetric flow rate will be
= [tex]\frac{\textup{118.2941}}{\textup{4}}[/tex]
= 29.57 m³/min.
Therefore,
The retention time = [tex]\frac{\textup{Total Volume of tank}}{\textup{Volumetric Flowrate }}[/tex]
or
The retention time, t = [tex]\frac{\textup{2500}}{\textup{29.57 }}[/tex]
or
The retention time, t = 84.54 min
When is Raoult's law not a good approximation for a particular compound? When the compound is very dilute in the liquid phase. When the compound has a high mole fraction in the liquid phase. When the compound has a high mole fraction in the vapor phase.
Answer:
Raoult's law is not a good approximation when the compound is very dilute in the liquid phase.
Explanation:
Thermodynamics defines ideal mixture as a mixture in which all the molecules of the different species involved are so similar between each other that intermolecular forces between different molecules are the same to those of similar molecules. Thus, Raoult's law takes this definition and stablish that when a compound is highly concentrated in the solution behaves as if it is in a pure solution.
Also, the Raoult's law stablish that the partial pressure ([tex]P_{i}[/tex]) of every component in an ideal mixture of liquids is equal to the pressure of the pure component ([tex]P_{i}^{*}[/tex]) multiplicated by the molar fraction in the solution of that component ([tex]x_{i}[/tex])
[tex]P_{i} = P_{i}^{*}. x_{i}[/tex]
The answer "When the compound has a high mole fraction in the liquid phase" is wrong because this is the case that follows the Raoult's law. And the answer "When the compound has a high mole fraction in the vapor phase." talks about the fraction in the vapor, and Raoult's law use the molar fraction in the solution.
The reaction rate low is independent of which of the following charactetics • Reactor type] • Temperature • Concentration of components • Pressure
Answer: In the reaction rate law the rate is expressed in terms of concentrations of species. It is important to know how much time a reaction will take to complete itself. It depends on some factors. Temperature, concentration of component, catalyst and pressure. On increasing these factors the rate of reaction of a respective reaction increases. It doesn't depends upon reactor type.
If 5.00 g helium gas is added to a 1.00 L balloon containing 1.00 g of helium gas, what is the new volume of the balloon? Assume no change in temperature or pressure. Enter your answer in the box provided. L
Explanation:
The given data is as follows.
[tex]mass_{1}[/tex] = 5 g, [tex]Volume_{1}[/tex] = 1 L
[tex]mass_{2}[/tex] = 1 g, [tex]Volume_{1}[/tex] = ?
No. of moles of helium present in 5 g helium gas are as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{5 g}{4 g/mol}[/tex]
= 1.25 mol
No. of moles of helium present in 1 g helium gas are as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{1 g}{4 g/mol}[/tex]
= 0.25 mol
According to the ideal gas equation, PV = nRT. And, since temperature and pressure are held constant. Therefore,
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]
[tex]\frac{1 L}{1.25 mol} = \frac{V_{2}}{0.25 mol}[/tex]
[tex]V_{2}[/tex] = 0.2 L
Thus, we can conclude that the new volume of the balloon is 0.2 L.
The final volume of the balloon after adding 5.00 g of helium to the already present 1.00 g of helium, under constant temperature and pressure, is 6.00 L.
Explanation:The subject of your question pertains to the ideal gas law which is represented by the formula PV = nRT, where P is pressure, V is volume, n is the number of moles, R refers to the gas constant, and T is temperature. Since the question states that there is no change in temperature or pressure, you can use Avogadro's law which states that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules.
In your problem, you are adding more helium to the balloon. Therefore, the volume of the balloon will increase proportionally. Since the initial volume of 1.00 L corresponds to 1.00 g of Helium, 5.00 g of Helium will correspond to 5.00 L. This is because the amount of Helium has increased by a factor of 5, and thus, the volume will also increase by a factor of 5.
Therefore, the final volume of the balloon will be the sum of the initial volume and the increase, which equals 1.00 L + 5.00 L = 6.00 L.
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A 1.92 mol sample of hydrogen gas occupies 22.2 L. How many mol of hydrogen are needed to fill a 102.1 L container at the same pressure and temperature? Enter your answer in the provided box. I mol H2
Answer:
8.83 moles of Hydrogen
Explanation:
We are told that 1.92 moles of H2 occupies 22.2 L at a certain pressure and temperature, and we are asked to calculate the quantity of moles of H2 are necessary to fill a container of 102.1 L. So, if the pressure and temperature are constant, then the equality expressed before 1.92 H2 moles = 22.2 L maintains. Therefore, we can calculate what is being asked as follows:
22.2 L ----- 1.92 H2 moles
102.1 L ---- x = (102.1 L × 1.92 H2 moles)/22.2 L = 8.83 H2 moles
This means that 8.83 H2 moles are necessary to fill a 102.1 L container at a certain pressure and temperature.
The energy transferred between samples of matter because of a difference in their temperatures is called a. heat. b. thermochemistry c. chemical kinetics. d. temperature.
Answer:
The energy transferred between samples of matter because of a difference in their temperatures is called a. heat.
Explanation:
The first law of thermodynamics establishes that when two bodies with different temperatures are put in contact they will find thermic equilibrium to a final temperature by transferring heat. Thus the correct answer is (a).
Thermochemistry is the study of the transformations of heat energy on the chemical reactions. Chemical kinetics is the study of the rate of chemical reactions. And temperature is the measure of the heat.
Calculate the volume of 8,410 g of a substance whose density is 0.71 g/mL. Express your answer in milliliters using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
12 mL
Explanation:
Density is expressed as mass divided by volume, so the expression can be rearranged to solve for volume:
D = m/V ⇒ V = m/D
V = m/D = (8.410 g)/(0.71g/mL) = 12 mL
Final answer:
The volume of a substance with a mass of 8,410 g and a density of 0.71 g/mL is 12,000 mL, rounded to two significant figures to match the precision of the given density.
Explanation:
To calculate the volume of a substance, you can use the density formula: Density = Mass / Volume. Given the density of the substance is 0.71 g/mL and the mass is 8,410 g, the volume can be found by rearranging the formula to Volume = Mass / Density.
Volume = 8,410 g / 0.71 g/mL = 11,845.07042 mL. This raw calculation has more significant figures than is justified by the precision of the given data. Since the density value has two significant figures, the volume should also be reported with two significant figures, resulting in 12,000 mL.
The molecular mass of magnesium sulfate heptahydrate is 246.46 g/mol. What weight is needed to obtain 0.150 moles?
Answer : The mass of magnesium sulfate heptahydrate needed are, 36.969 grams.
Explanation : Given,
Moles of magnesium sulfate heptahydrate = 0.150 mole
Molar mass of magnesium sulfate heptahydrate = 246.46 g/mole
Formula used :
[tex]\text{Mass of magnesium sulfate heptahydrate}=\text{Moles of magnesium sulfate heptahydrate}\times {\text{Molar mass of magnesium sulfate heptahydrate}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Mass of magnesium sulfate heptahydrate}=0.150mole\times 246.46g/mole[/tex]
[tex]\text{Mass of magnesium sulfate heptahydrate}=36.969g[/tex]
Therefore, the mass of magnesium sulfate heptahydrate needed are, 36.969 grams.
To obtain 0.150 moles of magnesium sulfate heptahydrate, you will need 36.969 grams.
To determine the weight needed to obtain 0.150 moles of magnesium sulfate heptahydrate (MgSO₄·7H₂O), you can use the molar mass and the number of moles given.
The molar mass of MgSO₄·7H₂O is given as 246.46 g/mol.The number of moles needed is 0.150 moles.To find the mass needed, use the formula:Mass = Number of Moles x Molar Mass
Substitute the values into the formula:
Mass = 0.150 moles x 246.46 g/mol
Mass = 36.969 g
Therefore, to obtain 0.150 moles of magnesium sulfate heptahydrate, you need 36.969 grams.
Consider the following reaction: A+B C+D According to Le Chatelier's Principle what would happen if you added more D to the reaction? From the point of view of the molecules and atoms why would the reaction respond this way? Your answer may use a maximum of 15 words. Using more than 15 words will result in a 0 score
Explanation:
According to the Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the direction of equilibrium where there is less stress or disturbance.
For example, [tex]A + B \rightleftharpoons C + D[/tex]
So, when we increase the concentration of D in this reaction then disturbance will increase on the product side.
And, frequency of successful collisions between the reactant atoms will decrease. Hence, there will be less stress on reactant side.
As a result, equilibrium will shift in the backward direction that is, on the reactant side it will shift.
You have a 0.5 L of a 1.0 M buffer solution with pka = 5.14 and current pH = 5.23. Calculate the new pH when 1.50 mL of 5.00 M HCl is added. Be mindful of units. (How can you tell if your answer makes sense?) pH = pka + 109 U base 5.23 - 5.14+ log ( )
Answer:
pH = 5,22
Explanation:
A buffer consist in a solution with both conjugate acid (HA) and conjugate base (A⁻). To know the concentration of both A⁻ and HA you use Henderson-Hasselbalch equation:
pH = pka + log₁₀ [tex]\frac{A^{-} }{HA}[/tex]
In the problem:
5,23 = 5,14 + log₁₀ [tex]\frac{A^{-} }{HA}[/tex]
1,23 = [tex]\frac{A^{-} }{HA}[/tex] (1)
Also, you know buffer concentration is 1,0 M:
1,0 M = A⁻ + HA (2)
Replacing (2) in (1):
HA = 0,45 M, thus:
A⁻ = 0,55 M.
The addition of 1.50 mL of 5.00 M HCl represents a concentration of:
1,50x10⁻³ L×[tex]\frac{5,00 mol}{L}[/tex] ÷ 0,5 L = 0,015 M of HCl
The buffer equilibrium is:
HA ⇄ A⁻ + H⁺ ka = [tex]10^{-5,14}[/tex]
The concentrations in equilibrium are:
[HA] = 0,45 M + x
[A⁻] = 0,55 M - x
[H⁺] = 0,015 M -x
Because the equilibrium is displaced to the left.
The equation of equilibrium is:
[tex]10^{5,14}[/tex] = [tex]\frac{[0,015M -x][0,55M-x]}{[0,45M-x]}[/tex]
You will obtain:
x² - 0,565x + 8,24674x10⁻³ = 0
Solving:
x = 0,5500061 ⇒ No physical sense
x = 0,01499391
Thus, [H⁺] = 0,015 - 0,01499391 = 6,09x10⁻⁶
As pH = -log₁₀ [H⁺]
pH = 5,22
I hope it helps!
How does a buffer resist change in pH upon addition of a strong acid? The strong acid reacts with the strong base in the buffer to form a salt, which produces few H+ ions in solution and therefore only a little change in pH. The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH. The strong acid reacts with the weak acid in the buffer to form a weak base, which produces few H+ ions in solution and therefore only a little change in pH.
Answer:
The strong acid reacts with the weak base in the buffer to form a weak acid, which produces few H+ ions in solution and therefore only a little change in pH.
Explanation:
When a strong acid is added to the buffer, the acid dissociates and furnish hydrogen ions which combine with the conjugate of the weak acid, forming weak acid. The weak acid dissociates to only some extent and can furnish only some protons and there is no significant change in the pH.
Hence, option B is correct.
In a buffer, a strong acid reacts with the weak base to form a weak acid, yielding fewer H+ ions, so only a slight change in pH is noted. Buffer solutions, including weak conjugate acid-base pairs, resist pH changes. However, the buffering ability may be disrupted if large amounts of acid or base are added, a concept known as buffer capacity.
Explanation:A buffer essentially resists changes in pH when an acid or base is added to the solution. This is due to the presence of a weak conjugate acid-base pair. In the buffer, a strong acid will react with the weak base to form a weak acid, thus producing fewer H+ ions. This, in essence, leads to only a slight change in pH.
A typical example of a buffer might include a solution of acetic acid and sodium acetate. This buffer consists of a weak acid and its salt. Similarly, a solution of ammonia and ammonium chloride could serve as a buffer consisting of a weak base and its salt.
It's crucial to note, however, that buffers do not have infinite capacity to resist pH changes. This is known as the buffer capacity. If a large amount of acid or base is added to the buffer solution, potentially lowering the concentration of the conjugate acid or base pair, the ability of the buffer to regulate pH may be disrupted.
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A worker is told her chances of being killed by a particular process are 1 in every 300 years. Should the worker be satisfied or alarmed? What is the FAR (assuming normal working hours) and the deaths per person per year? What should her chances be, assuming an average chemical plant?
Answer:
(a) Yes, he should be worried. The Fatal accident rate (FAR) is too high according to standars of the industry. This chemical plant has a FAR of 167, where in average chemical plants the FAR is about 4.
(b) FAR=167 and Death poer person per year = 0.0033 deaths/year.
(c) The expected number of fatalities on a average chemical plant are one in 12500 years.
Explanation:
Asumming 50 weeks of work, with 40 hours/week, we have 2000 work hours a year.
In 300 years we have 600,000 hours.
With these estimations, we have (1/600,000)=1.67*10^(-6) deaths/hour.
If we have 2000 work hours a year, it is expected 0.0033 deaths/year.
[tex]1.67*10^{-6} \frac{deaths}{hour}*2000 \frac{hours}{year}=0.0033 deaths/year[/tex]
The Fatal accident rate (FAR) can be expressed as the expected number of fatalities in 100 millions hours (10^(8) hours).
In these case we have calculated 1.67*10^(-6) deaths/hour, so we can estimate FAR as:
[tex]FAR=1.67*10^{-6} \frac{deaths}{hour}*10^{8} hours=1.67*10^{2} =167[/tex]
A FAR of 167 is very high compared to the typical chemical plants (FAR=4), so the worker has reasons to be worried.
If we assume FAR=4, as in an average chemical plant, we expect
[tex]4\frac{deaths}{10^{8} hour} *2000\frac{hours}{year}=8*10^{-5} \frac{deaths}{year}[/tex]
This is equivalent to say
[tex]\frac{1}{8*10^{-5} } \frac{years}{death}=1.25*10^{4} \frac{years}{death} =12500 \, \frac{years}{death}[/tex]
The expected number of fatalities on a average chemical plant are one in 12500 years.
calcualte pressure at STP in 10.0 L vessel after reaction of 1.0 L hydrochloride acid (concentration 35% and density 1.28 g/cub.cm) with 1.0 kg of limestone. yield of reaction 97%
Answer:
The pressure in the vessel is 13,3 atm.
Explanation:
The reaction that occurs in vessel (where limestone is 96% of CaCO₃) is:
2 HCl (aq)+ CaCO₃ (s) → CaCl₂(aq)+ H₂O(l)+ CO₂(g)
The increase in the pressure of the vessel after the reaction is by formation of a gas (CO₂). So we have to find the produced moles of this gas and apply the gas ideal law to find the pressure.
We have to find the limit reactant, to do so, we have to calculate the moles of each reactant in the reaction, the one that have the less moles will be the limit reactant:
HCl:
1,0L × (35/100) × (1000 cm³/1L) × (1,28 g/ 1cm³) × (1mol HCl/ 36,46 g) ÷ 2mol
(Concentration) (L to cm³) (cm³ to g) (g to mol) (moles of reaction)
moles of HCl= 6,14 mol
CaCo₃:
1,0 kg × (96/100) × (1000 g/1kg) × (1 mol/100,09g)
(Limestone) (CaCo₃ in limestone) (kg to g) (g to mol)
moles of CaCo₃= 9,59 mol
So, reactant limit is HCl
This reaction have a yield of 97%. So, the CO₂ moles are:
6,14 mol × 97÷ = 5,96 mol CO₂
The ideal gas formula to obtain pressure is:
P = nRT/V
Where: n = 5,96mol; R= 0,082 atm×L/mol×K; T = 273,15 (until STP conditions) and V= 10,0 L
Replacing this values in the equation the pressure is
P = 13,3 atm
I hope it helps!
You have 3.27g of O2. How many grams of CO2 can be made?
Answer:
Amount of Carbon dioxide equals 4.49625 grams.
Explanation:
From the basic stichometric reaction between carbon and oxygen we know that 1 mole of carbon combines with 1 mole of oxygen to form 1 mole of carbon dioxide.
Thus we can say that 12 grams of carbon combines with 32 grams of oxygen to form 44 grams of carbon dioxide.
In the given question assuming that there is no limited supply of carbon we can find the find the amount of carbon dioxide formed from 3.27 grams of Oxygen using ratio and proportion method.
As we can see that 32 grams of oxygen form 44 grams of carbon dioxide thus we can say 1 gram of oxygen yields [tex]\frac{44}{32}grams[/tex] Carbon Dioxide
Thus the carbon dioxide formed by 3.27 grams of Oxygen equals
[tex]3.27\times \frac{44}{32}=4.49625grams[/tex]
Write 274,541,005,000 in Engineering Notation with 3 significant figures.
Answer : The correct answer is, [tex]0.274\times 10^{12}[/tex]
Explanation :
Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power. It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.
The engineering notation written in the form:
[tex]a\times 10^b[/tex]
where,
a = the number which is greater than 0 and less than 999
b = an integer multiple of 3
For example : [tex]45.89\times 10^3[/tex] or [tex]56.45\times 10^{-6}[/tex]
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.
As we are given the 274,541,005,000 in standard notation.
Now converting this into engineering notation, we get:
[tex]\Rightarrow 274,541,005,000=0.274\times 10^{12}[/tex]
As, the decimal point is shifting to left side, thus the power of 10 is positive.
Hence, the correct answer is, [tex]0.274\times 10^{12}[/tex]
Final answer:
To write 274,541,005,000 in Engineering Notation with 3 significant figures, we need to move the decimal point so that we have a number between 1 and 10 multiplied by a power of 10. The number becomes 2.75 × 10¹¹.
Explanation:
Engineering Notation is a way of expressing numbers in scientific notation but with the power of 10 always being a multiple of 3. To write 274,541,005,000 in Engineering Notation with 3 significant figures, we need to move the decimal point so that we have a number between 1 and 10 multiplied by a power of 10. The number becomes 2.75 × 10¹¹.
At constant temperature 10.0L of N, at 0.983 atm is compressed to 2.88L.What is the final pressure of N,? 0.283 atm b) 1.51 atm c) 3.41 atm d 29.3 atm a)
Answer:
c) 3.41 atm
Explanation:
We can calculate the final pressure using Boyles Law, P₁V₁ = P₂V₂.
P₁V₁ = P₂V₂
(0.983 atm)(10.0 L) = P₂(2.88 L)
9.83 ÷ 2.88 = P₂
P₂ = 3.41 atm
The final pressure can be determined by using Boyles law and will be 3.41 atm.
What is pressure?The force delivered perpendicularly to a surface of the structure per unit area throughout whom that force would be dispersed is known as pressure.
What is Boyles law?
According to Boyle's law, the relationship between a gas's pressure as well as volume seems to be inverse.
Given data:
[tex]P_{1} = 0.983 atm\\V_{1} = 10 L\\V_{2} = 2.88 L[/tex]
Calculation of pressure.
The formula of Boyles law:
[tex]P_{1} V_{1} = P_{2} V_{2}[/tex]
Put the value of given data:
[tex](0.983) (10 ) = P_{2} (2.88)\\P_{2} = 3.41 atm[/tex]
Therefore, the final pressure will be 3.41 atm.
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Which situation is an indication that a chemical change has occurred? No color change occurs. A gas is formed. No heat is released. The temperature does not change.
Answer:
Its A
Explanation:
Its A
The correct answer is B. A gas is formed
Explanation:
In chemistry, changes in substances or elements occur as a product of chemical reactions in which a substance reacts with others or decomposes. In any of these cases, the original substance or element changes into new or new substances, which does not occur in physical changes. This change in composition can be identified through different changes and signals that include a change in odor, color or temperature; the production of bubbles or gas; and the formation of a precipitate. According to this, the situation that is an indication of a chemical change is the formation of a gas, because only if the original substance has changed there would be a new substance such as a gas.
Balance the equation and show the calculation of the number of moles and grams of Ca3(PO42 formed from 10.8 grams of Ca(OH)2. Show your answers to 3 significant figures Ca(OH)2 H3PO4-- Ca3(PO4)2 + H2O]
Answer:
The number of moles of calcium phosphate is 0.0485 mol. The mass is 15.1 g.
Explanation:
The balanced equation is:
[tex]3 Ca(OH)_{2} + 2 H_{3}PO_{4} => Ca_{3}(PO_{4})_{2} + 6H_{2}O[/tex]
As we can see, 3 moles of calcium hidroxide produced 1 mol of calcium phosphate. So the quantity of moles of Calcium posphate is:
[tex]10.8g Ca(OH)_{2}*\frac{1 molCa(OH)_{2}}{74.093gCa(OH)_{2}} *\frac{1mol Ca_{3}(PO_{4})_{2}}{3molCa(OH)_{2}} =0.0485 molCa_{3}(PO_{4})_{2} [/tex]
The mass of calcium phosphate in grams is:
[tex]0.0485 molCa_{3}(PO_{4})_{2}*\frac{310.176g}{mol} =15.1 g[/tex]
Why is a cis-1,3-disubstituted cyclohexane more stable thanits
trans isomer?
Answer:
Explanation has been given below
Explanation:
In diaxial conformation of cis-1,3-disubstituted cyclohexane, 4 gauche-butane interactions along with syn-diaxial interaction are present. Hence it readily gets converted to diequitorial conformation where no such gauche-butane interaction is presentIn two possible conformations of trans-1,3-disubstituted cyclohexane, 2 gauche-butane interactions are present in each of them.Hence cis-1,3-disubstituted cyclohexane exists almost exclusively in diequitorial form. But trans-1,3-disubstituted cyclohexane has no such option. Trans-1,3-disubstituted cyclohexane experiences gauche butane interaction in each of the two conformations.Therefore cis-1,3-disubstituted cyclohexane is more stable than trans conformationA cis-1,3-disubstituted cyclohexane is more stable than its trans isomer due to lower steric hindrance in the cis configuration as compared to the trans.
Explanation:A cis-1,3-disubstituted cyclohexane is more stable than its trans isomer because of a property known as steric hindrance. In organic chemistry, steric hindrance is a phenomenon where the size of groups within a molecule prevents chemical reactions from taking place. In a cis isomer, the substituents are on the same side of the ring, specifically on the ring's axial position. This arrangement minimizes the steric hindrance, which makes it more stable. On the other hand, the trans isomer has the substituents on opposite sides of the ring, causing more steric hindrance and thus it is less stable.
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What is the energy in joules of one photon of
microwaveradiation with a wavelength 0.122m?
Final answer:
One photon of microwave radiation with a wavelength of 0.122m has an energy of approximately 1.63 x 10⁻²⁴ joules, calculated using Planck's equation (E=hc/λ).
Explanation:
The energy of a photon can be calculated using the equation E = hc/λ, where E is the energy of the photon in joules, h is Planck's constant (6.626 x 10⁻³⁴ J·s), c is the speed of light in a vacuum (3.00 x 10⁸ m/s), and λ is the wavelength of the photon in meters. Using the given wavelength of 0.122 meters for the microwave radiation, we can calculate the energy of one photon by substituting the values into the equation:
E = (6.626 x 10⁻³⁴ J·s)(3.00 x 10⁸ m/s) / 0.122 m
E = 1.63 x 10⁻²⁴ Joules per photon. Therefore, one photon of microwave radiation with a wavelength of 0.122m has an energy of approximately 1.63 x 10⁻²⁴ joules.
Which of the following is a definition of a base? O a. behaves as a catalyst. O b. accepts H ions from acids. Oc. turns cloudy when dissolved in water O d. donates H ions when dissolved in water O e. releases heat when dissolved in water
Answer: b. accepts [tex]H^+[/tex] ions from acids
Explanation:
According to Arrhenius concept, a base is defined as a substance which donates hydroxide ions [tex](OH^-)[/tex] when dissolved in water and an acid is defined as a substance which donates hydronium ions [tex](H_3O^+)[/tex] in water.
According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates [tex]H^+[/tex] ions and a base is defined as a substance which accepts [tex]H^+[/tex] ions.
[tex]NH_3+H_2O\rightarrow NH_4^++OH^-[/tex]
Here water is donating [tex]H^+[/tex] ions, and thus act as acid and ammoia [tex]NH_3[/tex] is accepting [tex]H^+[/tex] ions from water and thus is a base.
According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.
Thus base is a substance which accepts [tex]H^+[/tex] ions from acids.