Answer: The correct answer is Option a.
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
We are given:
Mass of solute (carbon dioxide) = 44 g
Molar mass of carbon dioxide = 44 g/mol
Volume of solution = 0.5 L
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{44g}{44g/mol\times 0.5L}\\\\\text{Molarity of solution}=2mol/L[/tex]
Hence, the correct answer is Option a.
Use the References to access imp A sample of hydrogen gas has a mass of 48.3 grams. How many H, molecules are there in the sample? molecules Submit Answer & Next
Answer: The number of hydrogen molecules in the given amount is [tex]1.454\times 10^{25}[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of hydrogen gas = 48.3 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of hydrogen gas}=\frac{48.3g}{2g/mol}=24.15mol[/tex]
According to mole concept:
1 mole of a compound contains [tex]6.022\time 10^{23}[/tex] number of molecules
So, 24.15 moles of hydrogen gas will contain = [tex]24.15\times 6.022\times 10^{23}=1.454\times 10^{25}[/tex] number of molecules.
Hence, the number of hydrogen molecules in the given amount is [tex]1.454\times 10^{25}[/tex]
For the following pair, indicate which element has the lower first ionization energy: Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.
1. Given the elements Cl and Ge ,_________ has the smaller first ionization energy.
2. Given the elements Te and Se , __________ has the smaller first ionization energy.
3. Given the elements Ba and Ti , __________ has the smaller first ionization energy.
4. Given the elements Cu and Ag ,___________ has the smaller first ionization energy.
to fill in the blank:
(A) Ag
(B) Ge
(C) Cl
(D) Ti
(E) Cu
(F) Ba
(G) Se
(H) Te
Answer:
For 1: The correct answer is Ge.
For 2: The correct answer is Te.
For 3: The correct answer is Ba.
For 4: The correct answer is Ag.
Explanation:
Ionization energy is defined as the energy required to remove an electron from the outermost shell of an isolated gaseous atom. It is represented as [tex]E_i[/tex]
[tex]X(g)\rightarrow X^+(g)+1e^-;E_i[/tex]
Ionization energy increases as we move from left to right in a period. This happens because the atomic radius of an element decreases moving across a period, which increases the effective attraction between the negatively charged electrons and positively-charged nucleus. Hence, the removal of electron from the outermost shell becomes difficult and requires more energy.
Ionization energy decreases on moving from top to bottom in a group. This happens because the number of shells increases as we move down the group. The electrons get added in the new shell. So, the shielding of outermost electrons from the inner ones is more which decreases the attraction between the electrons and the nucleus. Hence, the removal of electron from the outermost shell becomes easy and requires less energy.
For the given options:
Option 1:Chlorine is the 17th element of the periodic table belonging to Period 3 and Group 17.
Germanium is the 32nd element of the periodic table belonging to Period 4 and Group 14.
Hence, germanium will have smaller first ionization energy.
Option 2:Tellurium is the 52nd element of the periodic table belonging to Period 5 and Group 16.
Selenium is the 34th element of the periodic table belonging to Period 4 and Group 16.
Hence, tellurium will have smaller first ionization energy.
Option 3:Barium is the 56th element of the periodic table belonging to Period 6 and Group 2.
Titanium is the 22nd element of the periodic table belonging to Period 4 and Group 4.
Hence, barium will have smaller first ionization energy.
Option 4:Copper is the 29th element of the periodic table belonging to Period 4 and Group 11.
Silver is the 47th element of the periodic table belonging to Period 5 and Group 11.
Hence, silver will have smaller first ionization energy.
Ionization energy increases across the period but decreases down the group.
First ionization energy of an element refers to the energy required to remove an electron from the atom. Ionization energy is a periodic trend that increases across the period but decreases down the group.
The following are true regarding the statements in the question;
Given the elements Cl and Ge, Ge has the smaller first ionization energyGiven the elements Te and Se, Te has the smaller first ionization energy.Given the elements Ba and Ti, Ba has the smaller first ionization energy. Given the elements Cu and Ag, Ag has the smaller first ionization energy.Learn more: https://brainly.com/question/11969651
2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), and lead (d = 11.3 g/mL). If all of the samples have the same mass, which one occupies the largest volume? Why?
Answer:
The sample of lithium occupies the largest volume.
Explanation:
Given the densities for the four elements, we have the expression [tex]d=\frac{m}{V}[/tex] that shows the relationship between mass and Volume to express the density of an element.
For each element we have:
[tex]d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL[/tex]
[tex]d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL[/tex]
[tex]d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL[/tex]
[tex]d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL[/tex]
The problem says that all the samples have the same mass, so:
[tex]m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m[/tex]
it means that m is a constant
Now, solving for the Volume in each element and with m as a constant, we have:
[tex]V_{lithium}=\frac{m}{d_{lithium}}[/tex]
[tex]V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m[/tex]
[tex]V_{lithium}=1.88\frac{mL}{g}*m[/tex]
[tex]V_{gold}=\frac{m}{d_{gold}}[/tex]
[tex]V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m[/tex]
[tex]V_{gold}=5.18*10^{-2}\frac{mL}{g}*m[/tex]
[tex]V_{aluminum}=\frac{m}{d_{aluminum}}[/tex]
[tex]V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m[/tex]
[tex]V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m[/tex]
[tex]V_{lead}=\frac{m}{d_{lead}}[/tex]
[tex]V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m[/tex]
[tex]V_{lead}=8.85*10^{-2}\frac{mL}{g}*m[/tex]
If we assume m = 1g, we find that:
[tex]V_{lithium}=1.88mL[/tex]
[tex]V_{gold}=5.18*10^{-2}mL[/tex]
[tex]V_{aluminum}=3.70*10^{-1}mL[/tex]
[tex]V_{lead}=8.85*10^{-2}mL[/tex]
So we can see that the sample of lithium occupies the largest volume with 1.88mL
Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.
What is an isochoric process? (1 mark) b) Can heat be exchanged in an isochoric process? (1 mark) c) A 100L container holding an ideal gas at an initial pressure of 10MPa is raised to a pressure of 15MPa. How much work is done?
Explanation:
(a) An isochoric process is defined as the process in which volume of a system remains constant.
For example, an air conditioner follows an isochoric process.
(b) As pressure and temperature can change in an isochoric process but volume will remain the same.
Hence, heat can also be exchanged in an isochoric process. Whereas in an adiabatic process heat remains 0. Therefore, it means the in adiabatic process there occurs no exchange of heat between the system and its surrounding.
(c) It is given that V = 100 L = 0.1 [tex]m^{3}[/tex] (as 1 L = 0.001 [tex]m^{3}[/tex])
[tex]P_{1}[/tex] = [tex]10 \times 10^{6} Pa[/tex]
[tex]P_{2}[/tex] = [tex]15 \times 10^{6} Pa[/tex]
Relation between work, pressure and volume is as follows.
W = [tex]\Delta PV[/tex]
= PdV + VdP
= 0 + 0.1 [tex]m^{3} (15 - 10) \times 10^{6}[/tex]
(Since V = constant so, dV = 0)
= [tex]5 \times 10^{5} J[/tex]
Thus, we can conclude that work done is [tex]5 \times 10^{5} J[/tex].
Ethyl acetate can be prepared by an SN2 reaction. Draw the alkylbromide and nucleophile used in the reaction. Remember to include formal charges (do not include counterions).
Answer:
Few important points related to [tex]S_N2[/tex] reaction:
1. [tex]S_N2[/tex] is a one-step reaction that follows second order kinetics.
2. In [tex]S_N2[/tex] reaction, a transition state is formed in situ.
3. Strong nucleophiles like [tex]OH^- \ or\ CN^-[/tex] are used in case of bi-molecular nucleophilic substitution reaction.
Ethyl acetate can be prepared by a second-order nucleophilic substitution reaction between acetic acid and ethyl bromide.
The reaction between acetic acid and ethyl bromide is drawn below:
In the SN2 reaction to create ethyl acetate, the nucleophile acetate ion attacks the alkyl bromide, leading to the formation of ethyl acetate and bromide ions.
Explanation:Ethyl acetate can be prepared by an SN2 reaction. In this context, an alkylbromide and nucleophile initiate the reaction. Ethyl bromide (CH3CH2Br) would be the alkylbromide used in the reaction. The nucleophile, in this case, would be acetate ion (CH3COO-).
Ethyl bromide is a good leaving group and acetate ion is a strong nucleophile, which comes from acetic acid (ethanoic acid). In SN2 reactions, the nucleophile attacks the substrate, and the leaving group (Br- in this case) leaves, leading to a reverse configuration. In this reaction, the nucleophile (acetate ion) will attack the alkyl bromide, resulting in the formation of ethyl acetate, CH3COOCH2CH3, and bromide ions, which is a good leaving group.
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What type of chemical bond would form between an atom of carbon (C) and an atom of nitrogen (N). Explain specifically why this type of bond would form.
Answer:
Due to the low difference of electronegativity in carbon and nitrogen, these will form a covalent bond, and this covalent union of C-N is one of the most common bonds in the organic chemistry and biological systems. This type of bonds can be found in amines, amides, imines, etc. Also, a single atom of C and a single atom of N can form a cyanide, that is a triple covalent bond between the atoms.
When a compound is dissolved in hot ethanol during a recrystallization, what is changing on a molecular level? The intermolecular forces between the solute and solvent The covalent bonds of the solute The compound slowly melts into the hot solvent The covalent bonds between the solute and solvent
Answer:
The intermolecular forces between the solute and solvent.
Explanation:
When you are heating a solvent, the intermolecular forces are reduced because the distances between molecules are large. Thus, in a solution where solvent is hot the intermolecular forces between solute and solvent are lower than those solutions where solvent is in room temperature.
The covalent bonds do not change because this mean a chemical reaction that doesn't occur in a solution.
Usually solid solutes melts in a higher temperature than boiling point in solvents. Thus, a compound normally doesn't melt in a hot solvent.
I hope it helps!
When a compound is dissolved in hot ethanol during recrystallization, the intermolecular forces between the solute and solvent are changing on a molecular level. The process of solution formation involves the formation of new intermolecular forces between the solute and solvent molecules, allowing the compound to dissolve in the solvent.
Explanation:Intermolecular forces are attractive or repulsive forces between molecules. They include London dispersion forces, dipole-dipole interactions, and hydrogen bonding. These forces influence physical properties like boiling points and solubility in molecular substances. The intermolecular forces between the solute and solvent are changing on a molecular level when a compound is dissolved in hot ethanol during recrystallization.
When the compound is dissolved in the hot solvent, the solute-solvent interactions which is also known as solvatio then naturally takes place. In addition to this, these interactions involve the formation of new intermolecular forces between the solute and solvent molecules, which are nearly as strong as the intermolecular forces within the solute and solvent alone. This favorable solution formation process allows the compound to dissolve in the solvent.
how many grams of solid NaOH are required to prepare a 400ml of a 5N solution? show your work!
Answer: The mass of solid NaOH required is 80 g
Explanation:
Equivalent weight is calculated by dividing the molecular weight by n factor. The equation used is:
[tex]\text{Equivalent weight}=\frac{\text{Molecular weight}}{n}[/tex]
where,
n = acidity for bases = 1 (For NaOH)
Molar mass of NaOH = 40 g/mol
Putting values in above equation, we get:
[tex]\text{Equivalent weight}=\frac{40g/mol}{1eq/mol}=40g/eq[/tex]
Normality is defined as the umber of gram equivalents dissolved per liter of the solution.
Mathematically,
[tex]\text{Normality of solution}=\frac{\text{Number of gram equivalents} \times 1000}{\text{Volume of solution (in mL)}}[/tex]
Or,
[tex]\text{Normality of solution}=\frac{\text{Given mass}\times 1000}{\text{Equivalent mass}\times \text{Volume of solution (in mL)}}[/tex] ......(1)
We are given:
Given mass of NaOH = ?
Equivalent mass of NaOH = 40 g/eq
Volume of solution = 400 mL
Normality of solution = 5 eq/L
Putting values in equation 1, we get:
[tex]5eq/L=\frac{\text{Mass of NaOH}\times 1000}{40g/eq\times 400mL}\\\\\text{Mass of NaOH}=80g[/tex]
Hence, the mass of solid NaOH required is 80 g
A certain insulation has a thermal conductivity of 0.2 W/m °C What thickness is necessary to effect a temperature drop of 400°C for a heat flux of 200 W/m2?
Answer:
It is necessary 1 meter of thickness.
Explanation:
You need to know the equation (P/A)=(k*T)/(L), where P/A is the heat flux, k is the thermal conductivity, T the temperature drop, and L the unknown thickness. So if rearrange the variables you will have that L = (k*T)/(P/A), and substituting the terms L = (0.2*400)/(200) = 1 meter
Final answer:
The necessary thickness of insulation to achieve a temperature drop of 400°C with a heat flux of 200 W/m^2, given a thermal conductivity of 0.2 W/m°C, is 0.4 meters.
Explanation:
The necessary thickness of insulation for a specific temperature drop and heat flux. By using the equation of thermal conduction Q = (kAΔT)/d, where Q is the heat flux, k is the thermal conductivity, A is the area, ΔT is the temperature difference, and d is the thickness, we can solve for the thickness d.
To start, we know that k (thermal conductivity) is 0.2 W/m°C, the temperature drop ΔT is 400°C, and the heat flux Q is 200 W/m^2. Plugging these values into the equation gives us d = (kAΔT)/Q. Since the area A is not specified, it will cancel out in this case, allowing us to solve for d directly. Rearranging the equation yields d = (kΔT)/Q, which equals (0.2 W/m°C * 400°C)/200 W/m^2, resulting in d = 0.4 m. Therefore, the thickness required for this insulation is 0.4 meters.
List and discuss two major sources of errors in numerical methods
Answer:
There are three major sources of errors in numerical method, namely human error, truncation error and round off error, but here we have to discuss only two sources-
Human error: This type of error occurs when the experiment is being performed but student isn't attentive during the experiment, or noted down the wrong observation, this error is very difficult to detect.There are rules for significant figures and rounding off, but if we round off the values incorrectly it will lead to error in final answer.Answer:
sdfsdfsdf
Explanation:dsf
Alkynes are hydrocarbons that have.. O a.... at least one triple bond between carbon atoms O b. None of the above. O c. ...all single bonds between carbon atoms. O d....at least one double bond between carbon atoms
Answer: Option (a) is the correct answer.
Explanation:
Alkynes are the hydrocarbons, that is, they contain only atoms of carbon and hydrogen. Their general chemical formula is [tex]C_{n}H_{2n-2}[/tex], where n is a whole number.
A compound that contains a double or triple bond is known as an unsaturated compound. An alkyne contains a triple bond and it is also an unsaturated compound.
Thus, we can conclude that alkynes are hydrocarbons that have at least one triple bond between carbon atoms.
As the volume of confined gas decreases at the constant temperature, the pressure exerted by the gas________________.
Fluctuates
decreases
Increases
stays the same
As the volume of confined gas decreases at the constant temperature, the pressure exerted by the gas increases . The correct option is C.
What is the Boyle's law?Boyle–Mariotte law is a gas law, shows the relation between pressure and volume. With the increase in the volume, the pressure decreases.
The pressure exerted by the mass, is inversely proportional to the volume of the gas.
Thus, the correct option is C, increases.
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At 14,000 ft elevation the air pressure drops to 0.59 atm. Assume you take a 1L sample of air at this altitude and compare it to 1 L of air taken at sea level. How much less O2 (in g) is available in 1 L of air at 14,000 ft (assume temperature of 298 K and that relative gas percentages are constant in both locations).
Answer:
There are 0.1125 g of O₂ less in 1 L of air at 14,000 ft than in 1 L of air at sea level.
Explanation:
To solve this problem we use the ideal gas law:
PV=nRT
Where P is pressure (in atm), V is volume (in L), n is the number of moles, T is temperature (in K), and R is a constant (0.082 atm·L·mol⁻¹·K⁻¹)
Now we calculate the number of moles of air in 1 L at sea level (this means with P=1atm):
1 atm * 1 L = n₁ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K
n₁=0.04092 moles
Now we calculate n₂, the number of moles of air in L at an 14,000 ft elevation, this means with P = 0.59 atm:
0.59 atm * 1 L = n₂ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K
n₂=0.02414 moles
In order to calculate the difference in O₂, we substract n₂ from n₁:
0.04092 mol - 0.02414 mol = 0.01678 mol
Keep in mind that these 0.01678 moles are of air, which means that we have to look up in literature the content of O₂ in air (20.95%), and then use the molecular weight to calculate the grams of O₂ in 20.95% of 0.01678 moles:
[tex]0.01678mol*\frac{20.95}{100} *32\frac{g}{mol} =0.1125 gO_{2}[/tex]
A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. After the sizzling subsides, what is the final equilibrium temperature? (Make the reasonable assumptions that any steam produced condenses into liquid water during the process of equilibration and that the evaporation and condensation don’t affect the outcome, as we’ll see in the next section.)
Answer:
[tex]T_{2}=16,97^{\circ}C[/tex]
Explanation:
The specific heats of water and steel are
[tex]Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}[/tex]
[tex]Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}[/tex]
Assuming that the water and steel are into an adiabatic calorimeter (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium [tex]T_{2}_{w}= T_{2}_{s}[/tex]
An energy balance can be written as
[tex] m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s} [/tex]
Replacing
[tex] 1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)[/tex]
Then, the temperature [tex]T_{2}=16,97^{\circ}C[/tex]
The equilibrium temperature given that You drop a piece of steel at 215 °C into the water at 10.0 °C in the calorimeter is 16 °C
How to calculate the equilibrium temperature?The equilibrium temperature given that You drop a piece of steel at 215 °C into the water at 10.0 °C can be calculated as follow:
Mass of steel (M) = 0.385 KgTemperature of steel (T) = 215 °CSpecific heat capacity of steel (C) = 420 J/kg°CMass of water (Mᵥᵥ) = 1.28 KgTemperature of water (Tᵥᵥ) = 10.0 °CSpecific heat capacity of water (Cᵥᵥ) = 4184 J/Kg°CEquilibrium temperature of steel and water mixture (Tₑ) =?[tex]MC(T - T_e) = M_wC_w(T_e - T_w)\\\\0.385\ \times\ 420(215 - T_e) = 1.28\ \times\ 4184(T_e - 10)[/tex]
[tex]34765.5 - 161.7T_e = 5355.52T_e - 53555.2\\\\34765.5\ +\ 53555.2 = 5355.52T_e\ +\ 161.7T_e\\\\88320.7 = 5517.22T_e\\\\T_e = \frac{88320.7}{5517.22} \\\\T_e = 16\ \textdegree C[/tex]
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What volume of concentrated (10.2 M) HCl would be required to prepare 1.11 x 104 mL of 1.5 M HC1? Enter your answer in scientific notation. Be sure to answer all parts. x 10 (select) L
Answer:
The required volume is 1.6 x 10³mL.
Explanation:
When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:
C₁ . V₁ = C₂ . V₂
where,
C₁ and V₁ are the concentration and volume of the concentrated solution
C₂ and V₂ are the concentration and volume of the dilute solution
In this case, we want to find out V₁:
C₁ . V₁ = C₂ . V₂
[tex]V_{1} = \frac{C_{2}.V_{2}}{C_{1}} = \frac{1.5M \times1.11.10^{4}mL }{10.2M} =1.6\times10^{3} mL[/tex]
Pressure drop in packed column ..... a tray column
a) More than
b) Less than
c) Equal to
d) No relation to
Answer:
The answer is b)
Explanation:
In packed column the mass transfer area is higher and the packing can be random, that could provoque an increase in the pressure, because obstruction of the mass flux, While in a tray column the sieves can have specific porosity and the flux of mass can be facilitated, thus the pressure drop.
If a substance increases the OH- concentration of pure water by 10^4. what is the pH of the solution?
Answer:
pH = 11
Explanation:
The concentration of OH⁻ in pure water is 10⁻⁷ M. If a substance increases OH⁻ concentration by 10⁴, the new concentration will be:
[OH⁻] = 10⁴ x 10⁻⁷ M = 10⁻³M
We can calculate pOH using it's definition:
pOH = -log[OH⁻] = -log (10⁻³) = 3
Then, we can find out pH using the following relation:
pH + pOH = 14
pH = 14 - pOH = 14 -3 = 11
Since pH = 11 is higher than 7, we can confirm that the substance is a base.
ObIel WiLll unt COl.. USSMS A certain chemical reaction releases 31.2 kJ/g of heat for each gram of reactant consumed. How can you calculate what mass of reactant will produce 1080. J of heat? Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols. mass X ? Explanation Check ducabon ARghtsserved T of Ue Phacy The I'm esc
Answer:
The expression to calculate the mass of the reactant is [tex]m = \frac{1.080kJ}{31.2kJ/g}[/tex]
Explanation:
The amount of heat released is equal to the amount of heat released per gram of reactant times the mass of the reactant. To keep to coherence between units we need to transform 1,080 J to kJ. We do so with proportions:
[tex]1,080J.\frac{1kJ}{10^{3}J } =1.080kJ[/tex]
Then,
[tex]1.080kJ=31.2\frac{kJ}{g} .m\\m = \frac{1.080kJ}{31.2kJ/g}[/tex]
Answer:
1. The expression is: [tex]m=\frac{E}{\Delta _rH}[/tex]
2. The computed mass is: [tex]m=0.0346g[/tex]
Explanation:
Hello,
In this case, we know the so called enthalpy of reaction whose symbol and value is shown below:
[tex]\Delta _rH=31.2\frac{kJ}{g}[/tex]
In addition, we know that the energy released by the involved reactant is:
[tex]E=1080 J[/tex]
Therefore, the expression to compute the required mass, based on the given units is:
[tex]m=\frac{E}{\Delta _rH}[/tex]
Finally, the computed mass turns out:
[tex]m=\frac{1080J*\frac{1kJ}{1000J} }{31.2\frac{kJ}{g}} \\m=0.0346g[/tex]
Best regards.
What is the chemical formula for Iron(III) hydroxide? Hydroxide is OH O a. Fe(OH)2 O b. FeOH Fe2OH O c. O d. Fe(OH)3 O e. Fe2(OH)3
The correct chemical formula for Iron(III) hydroxide is Fe(OH)3, which is formed by combining one iron ion with a +3 charge (Fe3+) with three hydroxide ions (OH-).
Explanation:The chemical formula for Iron(III) hydroxide is Fe(OH)3. This compound consists of an iron ion with a +3 charge (Fe3+) and three hydroxide ions (OH-). The number in parentheses after iron indicates the oxidation state of iron, which in this case is +3. The formula reflects the stoichiometry required to balance the charges, resulting in one Fe3+ for every three OH- ions to form the neutral compound. Therefore, the correct answer is Fe(OH)3.
The correct answer is d. Fe(OH)₃.
Air enters a diffuser witha velocity of 400 m/s, a pressure of 1 bar and temperature of 25 C. It exits with a temperature of 100 C. What is the exit velocity of the air? Assume there are no heat losses or change in potential energy Data:= 0.718 kJ/kg.°C. MW = 28.9 g/mol
Answer:
Exit velocity of air is 96.43 m/s.
Explanation:
Given that
[tex]V_1=400\ m/s[/tex]
[tex]T_1=25C[/tex]
[tex]T_2=100C[/tex]
For air
[tex]C_p=1.005\ KJ/kg.K[/tex]
Now from first law of thermodynamics for open system at steady state
[tex]h_1+\dfrac{V_1^2}{2000}+Q=h_2+\dfrac{V_2^2}{2000}+w[/tex]
For diffuser
[tex]h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}[/tex]
We know that
[tex]h=C_pT[/tex]
[tex]h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2000}[/tex]
[tex]1.005\times 25+\dfrac{400^2}{2000}=1.005\times 100+\dfrac{V_2^2}{2}[/tex]
[tex]V_2=96.43\ m/s[/tex]
So the exit velocity of air is 96.43 m/s.
When 0.491 grams of a protein were dissolved in 44 mL of benzene at 24.4 degrees C, the osmotic pressure was found to be 50.9 torr. Calculate the molar mass of the protein.
Answer:
4057.85 g/mol
Explanation:
Hello, the numerical procedure is shown in the attached file.
- In this case, since we don't have the density of the protein, we must assume that the volume of the solution is solely given by the benzene's volume, in order to obtain the moles of the solute (protein).
-Van't Hoff factor is assumed to be one.
Best regards.
To calculate the molar mass of the protein, use the formula M = (RT) / (V * P), where M is the molar mass, R is the ideal gas constant, T is the temperature, V is the volume, and P is the osmotic pressure.
Explanation:To calculate the molar mass of the protein, we can use the formula:
M = (RT) / (V * P)
Where M is the molar mass, R is the ideal gas constant (0.0821 L.atm/mol.K), T is the temperature in Kelvin (24.4 + 273 = 297.4 K), V is the volume in liters (44 mL / 1000 = 0.044 L), and P is the osmotic pressure in atm (50.9 torr / 760 = 0.067 atm).
Substituting the values into the formula:
M = (0.0821 * 297.4) / (0.044 * 0.067) = 25567.14 g/mol
Therefore, the molar mass of the protein is approximately 25567.14 g/mol.
A weak acid. What is the pH of a 0.1 M solution of acetic acid (pKa = 4.75)?
(Hint: Let x be the concentration of H+ ions released from acetic acid when it dissociates. The solutions to a quadratic equation of the form ax^2 + bx + c = 0 are x = (-b +- squareroot (b^2- 4ac)/2a.
Answer:
pH of acetic acid solution is 2.88
Explanation:
[tex]pK_{a}=4.75[/tex]
or, [tex]-log(K_{a})=4.75[/tex]
or, [tex]K_{a}=10^{-4.75}=1.78\times 10^{-5}[/tex]
We have to construct an ICE table to determine concentration of [tex]H^{+}[/tex] and corresponding pH. Initial concentration of acetic acid is 0.1 M.
[tex]CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+H^{+}[/tex]
I(M): 0.1 0 0
C(M): -x +x +x
E(M): 0.1-x x x
So, [tex]\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}=K_{a}[/tex]
or, [tex]\frac{x^{2}}{0.1-x}=1.78\times 10^{-5}[/tex]
or, [tex]x^{2}+(1.78\times 10^{-5}\times x)-(1.78\times 10^{-6})=0[/tex]
So, [tex]x=\frac{-(1.78\times 10^{-5})+\sqrt{(1.78\times 10^{-5})^{2}+(4\times 1\times 1.78\times 10^{-6})}}{2\times 1}[/tex](M)
so, [tex]x=1.33\times 10^{-3}M[/tex]
Hence [tex]pH=-log[H^{+}]=-log(1.33\times 10^{-3})=2.88[/tex]
A liquid mixture contains 1 kg of water, 1.9 lb of ethanol, and 4.6 lb of methyl acetate. What is the weight fraction of ethanol in this mixture? Report your answer to the nearest hundredths place.
Answer: The weight fraction of ethanol in the mixture is
Explanation:
We are given:
Mass of water = 1 kg = 2.205 lb (Conversion factor: 1 kg = 2.205 lb)
Mass of ethanol = 1.9 lb
Mass of ethyl acetate = 4.6 lb
Mass of mixture = [2.205 + 1.9 + 4.6] = 8.705 lb
To calculate the percentage composition of ethanol in mixture, we use the equation:
[tex]\%\text{ composition of ethanol}=\frac{\text{Mass of ethanol}}{\text{Mass of mixture}}\times 100[/tex]
Mass of mixture = 8.705 lb
Mass of ethanol = 1.9 lb
Putting values in above equation, we get:
[tex]\%\text{ composition of ethanol}=\frac{1.9lb}{8.705lb}\times 100=21.8\%[/tex]
Hence, the weight fraction of ethanol in the mixture is 21.8 %
Define "Dispersed System", "Dispersion Medium", and "Dispersed Particles
Answer:
Dispersion system is a system in which certain particles are scattered in a continuous liquid or solid medium. The two phases present in this system are the dispersed particles and dispersion medium. These phases may or may not be present in the same state.
In a dispersion system, the particles that are dispersed are known as the dispersed particles and the medium in which the particles are dispersed is known as the dispersion medium.
We know that one mole of Oxygen gas contains two moles of O atoms (remember that O is a diatomic element in nature and is written O2). If we have a sample of O2 that contains 1.0000 x 10-20 mol, we will have (blank)
atoms of O in the sample.
Answer:
12.4 × 10∧3 atoms
Explanation:
Given data:
moles of oxygen molecule= 1.0000 x 10-20 mol
atoms =?
Solution:
32 g O2 = 1 mol = 6.02 × 10∧23
1.0000 x 10∧-20 mol × 6.02 × 10∧23 × 2 = 12.4 × 10∧3 atoms
How many atoms are found in 4.20 g of Magnesium?
Answer:
There are 1.041×10²³ atoms in 4.20g of Magnesium.
Explanation:
To find the amount of atoms in 4.20 g of Magnesium we need de molar mass of Mg: 24.305 g/mol
According to Avogadro number there are 6.022×10²³ particles in 1 mol, so the number of atoms of Mg is:
[tex]4.20 g Mg*\frac{1 molMg}{24.305gMg} *\frac{6.022*10^{23}atoms }{1 mol Mg} = 1.041*10^{23}atoms Mg[/tex]
Answer: The number of atoms found in given amount of magnesium is [tex]1.042\times 10^{23}[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of magnesium = 4.20 g
Molar mass of magnesium = 24.31 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of magnesium}=\frac{4.20g}{24.31g/mol}=0.173mol[/tex]
According to mole concept:
1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms
So, 0.173 moles of an element contains = [tex]0.173\times 6.022\times 10^{23}=1.042\times 10^{23}[/tex] number of atoms
Hence, the number of atoms found in given amount of magnesium is [tex]1.042\times 10^{23}[/tex]
2.4 Bromium has two naturally occurring isotopes: 79Br, with an atomic weight of 78.918 amu, and 81Br, with an atomic weight of 80.916 amu. If the average atomic weight for Br is 79.903 amu, calculate the fraction-of-occurrences of these two isotopes.
Answer:
49.3% 81Br and 50.7% 79Br or [tex]\frac{493}{1000}[/tex] 81Br and [tex]\frac{507}{1000}[/tex] 79BR
Step-by-step explanation:
To get the fraction-of-occurrences of the isotopes we must write the following equation. x is the isotopic abundance of 81Br, we can use 1 - x to get the isotopic abundance of 79Br.
(78.918)(1 - x) + (80.916)(x) = 79.903
78.918 - 78.918x + 80.916x = 79.903
1.998x = 0.985
x = 0.493
0.493 × 100 = 49.3% 81Br
100 - 49.3 = 50.7% 79Br
49.3% 81Br and 50.7% 79Br or [tex]\frac{493}{1000}[/tex] 81Br and [tex]\frac{507}{1000}[/tex] 79BR
Calculate the mass of manganese in potassium permanganate.
Answer:
The percentage mass of Manganese is 34.76%
Explanation:
The formula of potassium permanganate is KMnO4
Its molar mass is 158.034 g/mol.
The molar mass of Mn is 54.938 g/mol.
Percentage mass of Mn is:
[tex]Mn=\frac{54.938}{158.034}100[/tex]
Mn=34.76%
A consistent explanation of known observations is called A) an experiment. B) a prediction. C) a hypothesis. theory
A theory is a well-substantiated explanation of observed phenomena that has undergone rigorous testing and validation through the scientific method.
Explanation:A consistent explanation of known observations is called a theory. In the realm of science, a theory is more than just a simple guess; it is a well-substantiated explanation that is grounded in a significant body of evidence. This evidence consists of a series of facts and observations that, when considered collectively, give a reliable account of a part of the natural world. To arrive at a theory, the scientific method is typically employed, involving systematic observations, forming hypotheses, conducting experiments, and refining those hypotheses based on experimental results. When a hypothesis withstands rigorous testing and is capable of explaining a large and diverse range of phenomena, it can evolve into a theory. Scientific theories are valued because they offer comprehensive explanations that are testable and can be modified in the face of new evidence.
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Conversion of flow quantities by the continuity equation Water flows through a 3-inch, schedule 40 steel pipe (inside diameter- 3.068 inches) with an average velocity of 4 ft/s. Compute: (a) mass flow rate in lb/h, (b) volumetric flow rate in ft /h, and (c) mass velocity (mass flux) in lb/h-ft.
Answer:
a) m = 2398256.64 Ib/hr
b) Q = 38433.6 ft^3/hr
c) J = = 898560 lb/h-ft2
Explanation:
a) To get the mass flow rate in lb/h we are going to use this formula:
m = ρνA
note: we have to change some units to reach to the final units
when m is the mass flow rate which we need to calculate
ρ is the density of water in lb/ft3 = 62.4 lbs/ft3 it is a constant numbers you can get it from anywhere
ν is the velocity of the water = 4 ft/s we need to change it to ft/h unit so
when 1 ft/s = 3600 ft/hr (note : 3600 is a result of converting hr to sec 60min*60sec)
so the velocity of the water = 4 ft/s * 3600 ft/hr= 14400 ft/hr
now , we need to get the area of the pipe in feet also when:
A = π*diameter * length
diameter = 3.068 inches
when 1 inch = 0.0833333333 feet so,
diameter in feet = 3.068 * 0.0833333333 =.300 0.2557ft
and length = 40 * 0.0833333333 = 3.333 ft
so the area of the pipe in feet^2 = 3.14*0.255*3.33 =2.669 ft^2
by substitution in the mass flow rate:
m = 62.4 lbs/ft3 * 14400 ft/hr *2.669 ft^2 = 2398256.64 Ib/hr
b) now to calculate volumetric flow rate in ft /h we are going to use this formula:
Q = AV
when Q is the volume flow rate
A is the cross-sectional area filled by water =2.669 ft^2
V is the average velocity of the water = 14400 ft/hr
so , by substitution:
Q = 2.669 ft^2* 14400 ft/hr= 38433.6 ft^3/hr
C) To calculate mass velocity (mass flux) in lb/h-ft we are going to use this formula:
J = m / A
when J is the mass flux
and m is the mass flow rate which we calculated above = 2398256.64 Ib/hr
and A is the cross section area of the pipe which we calculated above and = 2.669 ft^2
so, by substitution:
J = 2398256.64 Ib/hr / 2.669 ft^2
= 898560 lb/h-ft2