Final answer:
The equilibrium concentrations for CO and Cl₂ at 600 K are 0.540 M, and for COCl₂, it is 0.320 M, calculated using the equilibrium expression and the equilibrium constant (Kc).
Explanation:
To solve for the equilibrium concentrations of the substances involved in the reaction CO(g) + Cl₂(g) ⇌ COCl₂(g), we will use the provided equilibrium constant (Kc) and the initial concentrations of the reactants.
Let's let x be the amount of CO and Cl₂ that react to form COCl₂ at equilibrium. The initial concentration of CO and Cl₂ is 0.430 mol/0.500 L = 0.860 M.
At equilibrium, the concentrations will be:[CO] = [Cl₂] = 0.860 M - x[COCl₂] = xThe equilibrium expression is:At equilibrium, the concentrations are 0.239 M for both CO and Cl₂, and 0.621 M for COCl₂.
The equilibrium constant and initial concentrations were used to find these values. This calculation involves solving a quadratic equation.
To determine the concentrations of CO, Cl₂, and COCl₂ at equilibrium for the reaction CO(g) + Cl₂(g) ⇌ COCl₂(g), we start with the initial moles of each reactant and use the equilibrium constant, Kc.
Given:
Initial moles of CO and Cl₂: 0.430 mol eachVolume of flask: 0.500 LKc: 4.95 at 600 KFirst, calculate the initial concentrations: [CO]° = [Cl₂]° = (0.430 mol) / (0.500 L) = 0.860 M.
Let 'x' be the change in concentration of CO and Cl₂.
At equilibrium:
CO: 0.860 - xCl₂: 0.860 - xCOCl₂: xSet up the equilibrium expression:
Kc = [COCl]₂/ ([CO] [Cl₂]) = 4.95
4.95 = x / ((0.860 - x) (0.860 - x))
Solving for 'x' involves solving the quadratic equation: x = 0.621 M.
Therefore, at equilibrium:
[CO] = 0.860 - 0.621 = 0.239 M[Cl₂] = 0.860 - 0.621 = 0.239 M[COCl₂] = 0.621 MArrange each set of atoms in order of increasing IE₁:
(a) Sr, Ca, Ba (b) N, B, Ne (c) Br, Rb, Se (d) As, Sb, Sn
Answer:
(a) Ba < Sr < Ca
(b) B < N < Ne
(c) Rb < Se < Br
(d) Sn < Sb < As
acetone and ethanol Choose one or more: A. London dispersion B. dipole–dipole C. hydrogen bonding D. ion-induced dipole
Explanation:
The weak intermolecular forces which can arise either between nucleus and electrons or between electron-electron are known as dispersion forces. These forces are also known as London dispersion forces and these are temporary in nature.
Dipole-dipole interactions are defined as the interactions that occur when partial positive charge on an atom is attracted by partial negative charge on another atom.
When a polar molecules produces a dipole on a non-polar molecule through distribution of electrons then it is known as dipole-induced forces.
Hydrogen bonding is defined as a bonding which exists between a hydrogen atom and an electronegative atom like O, N and F.
Chemical formula of acetone is [tex]CH_{3}COCH_{3}[/tex]. Due to the presence of oxygen atom there will be partial positive charge on carbon and a partial negative charge on oxygen atom. Hence, dipole-dipole forces will exist in a molecule of acetone.
Whereas hydrogen bonding will exist in a molecule of ethanol ([tex]CH_{3}CH_{2}OH[/tex]). Since, hydrogen atom is attached with electronegative oxygen atom.
Whereas London dispersion forces will also exist in both acetone and ethanol molecule.
As an SN2 solvent for reactions like the alkylation of acetaminophen, water would be a polar selection. Water is more polar that 2-butanone or DMSO. Why are polar, aprotic solvents better that polar, protic solvents (like water) for SN2 reactions
Answer: polar protic solvents solvate the nucleophile necessary for attack on the substrate in SN2 substitution.
Explanation:
Aprotic solvents are solvents that lack protons such as dimethyl sulphoxide (DMSO). DMSO has no exposed positive end. The positive end is buried inside the molecular structure. As a result of this, the nucleophile is not solvated. If the nucleophile is solvated, the rate of SN2 reaction will reduce drastically because the nucleophile becomes unavailable to attack the substrate. This solvation normally occur in polar protic solvents such as water because of the exposed positive end of the molecule which interacts with the nucleophile thereby reducing the rate of SN2 reaction.
Polar, aprotic solvents are better than polar, protic solvents for SN2 reactions because the nucleophiles are relatively free to approach the electrophilic carbon of the substrate.
Explanation:Polar, aprotic solvents are better than polar, protic solvents for SN2 reactions because the nucleophiles are relatively free to approach the electrophilic carbon of the substrate.
In polar protic solvents, like water, the solvent forms a layer around the cation and anion through ion-dipole interactions. This solvent layer prevents nucleophiles from approaching the electrophilic carbon, making them unsuitable for SN2 reactions.
In polar, aprotic solvents, the anion (nucleophile) is relatively free because it is prevented from approaching the positive pole of the solvent due to steric hindrance.
Suggest an explanation for the observations that ethanol, C2H5OH, is completely miscible with water and that ethanethiol, C2H5SH, is soluble only to the extent of 1.5 g per 100 mL of water. ?
Explanation:
Ethanethiol is an organosulphur compound which had an alkyl group(CnH2n+1) and a -thiol group (-SH). A thiol group like an hydroxyl group has its Oxygen replaced with sulphur.
Ethanethiol shows little association of its -SH group with the water molecules by hydrogen bonding; both with water molecules and among themselves. Hence why they are less soluble in water or slightly soluble in water.
Ethanol is an organic compound which the general formula of CnH2n+1OH. Ethanol is soluble in water because of the reaction of the -OH group with the water molecules to form hydrogen bonds.
Note that as the alkyl chain increases, Solubility decreases. This is because the alkyl chain forms the hydrophobic surface of the molecule due to its non-interaction of the water molecules to form hydrogen bond.
Ethanol is completely miscible with water due to its polar nature and ability to form strong hydrogen bonds with water. However, ethanethiol is only partially soluble because it forms weaker hydrogen bonds with water due to its less polar sulfhydryl group.
Explanation:The miscibility and solubility of substances depend on the principle that 'like dissolves like'. This means that polar substances dissolve well in other polar substances, while non-polar substances dissolve well in other non-polar substances. Ethanol (C2H5OH) is a polar molecule because it contains a hydroxyl group (-OH), which forms hydrogen bonds with water, another polar substance. This makes ethanol completely miscible with water. On the other hand, ethanethiol (C2H5SH) contains a sulfhydryl group (-SH), which is less polar than the hydroxyl group and forms weaker hydrogen bonds with water. As a result, ethanethiol is only partially soluble in water.
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Given the reaction has a percent yield of 86.8 how many grams of aluminum iodide would be required to yield an actual amount of 73.75 grams of aluminum?
Answer:
Approximately [tex]1.29 \times 10^3[/tex] grams.
Explanation:
Let [tex]x[/tex] represent the number of grams of aluminum iodide required to yield that 73.75 grams of aluminum.
In most cases, the charge on each aluminum ion would be +3 while the charge on each iodide ion would be -1. For the charges to balance, there needs to be three iodide ions for every aluminum ion. Hence, the empirical formula for aluminum iodide would be [tex]\rm AlI_3[/tex].
How many moles of formula units in that [tex]x[/tex] grams of [tex]\rm AlI_3[/tex]? Start by calculating its formula mass [tex]M(\mathrm{AlI_3})[/tex]. Look up the relative atomic mass of aluminum and iodine on a modern periodic table:
Al: 26.982.I: 126.904.[tex]M(\mathrm{AlI_3}) = 1\times 26.982 + 3\times 126.904 = 410.694\; \rm g \cdot mol^{-1}[/tex].
[tex]n(\mathrm{AlI_3}) = \displaystyle \frac{m}{M} = \frac{x}{410.694}\;\rm mol[/tex].
Since there's one aluminum ion in every formula unit,
[tex]n(\mathrm{Al}) = n(\mathrm{AlI_3}) = \displaystyle \frac{x}{410.694}\; \rm mol[/tex].
How many grams of aluminum would that be?
[tex]m(\mathrm{Al}) = n \cdot M = \displaystyle \frac{x}{410.694}\; \times 26.982 = \frac{26.982}{410.694}\, x\; \rm g[/tex].
However, since according to the question, the percentage yield (of aluminum) is only [tex]86.8\%[/tex]. Hence, the actual yield of aluminum would be:
[tex]\begin{aligned}&\text{Actual Yield} \\ &= \text{Percentage Yield} \times \text{Theoretical Yield} \\ &= 86.8\% \times \frac{26.982}{410.694}\, x \\ &= 0.868 \times \frac{26.982}{410.694}\, x \\ &\approx 0.0570263\, x\; \rm g\end{aligned}[/tex].
Given that the actual yield is 73.75 grams,
[tex]0.0570263\, x = 73.75[/tex].
[tex]\displaystyle x = \frac{73.75}{0.0570263} \approx 1.29 \times 10^3\; \rm g[/tex].
To yield 73.75 grams of aluminum with a percent yield of 86.8%, approximately 1113.33 grams of aluminum iodide would be required.
Explanation:To determine the amount of aluminum iodide required to yield 73.75 grams of aluminum with a percent yield of 86.8%, we need to use stoichiometry and the concept of percent yield.
First, we need to determine the theoretical yield of aluminum iodide using the balanced chemical equation: 2 Al + 3 I2 → 2 AlI3
From the equation, we can see that 2 moles of Al react with 3 moles of I2 to produce 2 moles of AlI3. Therefore, the molar ratio between Al and AlI3 is 2:2.
To calculate the theoretical yield of AlI3, we need to convert the mass of Al to moles using the molar mass of Al. Then, we use the mole ratio to find the moles of AlI3, and finally, convert it back to grams using the molar mass of AlI3.
Using the molar mass of Al (26.98 g/mol), we find that the moles of Al is 73.75 g / 26.98 g/mol = 2.73 moles.
Since the molar ratio between Al and AlI3 is 2:2, the moles of AlI3 produced is also 2.73 moles.
Finally, we can calculate the mass of AlI3 using the molar mass of AlI3 (407.7 g/mol): 2.73 moles * 407.7 g/mol = 1113.33 g.
Therefore, approximately 1113.33 grams of aluminum iodide would be required to yield an actual amount of 73.75 grams of aluminum with a percent yield of 86.8%.
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If 39.0 g of C6H6 reacts with excess chlorine and produces 30.0 g of C6H5Cl in the reaction C6H6 + Cl2 → C6H5Cl + HCl , what is the percent yield of C6H5Cl?
Answer:
53.4 % is the percent yield
Explanation:
This is the reaction:
C₆H₆ + Cl₂ → C₆H₅Cl + HCl
First of all we need to know the moles of benzene we used
39 g . 1 mol / 78 g = 0.5 moles
Ratio is 1:1 so 1 mol of benzene produces 1 mol of chloride
0.5 moles of chloride were produced by 0.5 moles of benzene
We must calculate the mass of chloride we produced
0.5 mol . 112.45 g / 1 mol = 56.2g
Let's calculate the percent yield
(Yield produced / Theoretical yield ) . 100
(30 g / 56.2 g) . 100 = 53.4 %
Answer:
53.4%
ut quest
Does the reaction of a main-group nonmetal oxide in water produce an acidic or a basic solution? Write a balanced equation for the reaction of a Group 6A(16) nonmetal oxide with water.
Answer:
They form acidic solutions
SO2(g) + H2O(l) --> H2SO3(aq)
SO3(g) + H2O(l) --> H2SO4(aq)
Explanation:
Group 6 elements are elements on the periodic table with 6 atoms on their valence shell. Metallic properties increases down the group from oxygen through polonium. Examples are oxygen, polonium, sulphur etc.
Group 6 nonmetal oxides react with water to form acidic solutions, this is because the oxides of nonmetallic elements are basic in nature.
Sulphur exhibits a wide range of oxidation states ranging from +2 to +6. So its oxides are SO2 and SO3.
Example:
SO2(g) + H2O(l) --> H2SO3(aq)
SO3(g) + H2O(l) --> H2SO4(aq)
How many grams of product can be produced by reacting 5.0 gram of aluminum and 22 grams of bromine? 2 Al + 3 Br2 > 2 AlBr3
Answer:
The mass of AlBr3 is 24.5 grams
Explanation:
Step 1: Data given
Mass of aluminium = 5.0 grams
Mass of bromine = 22.0 grams
Molar mass of aluminium = 26.98 g/mol
Molar mass of br2= 159.8 g/mol
Step 2: The balancced equation
2 Al + 3 Br2 → 2 AlBr3
Step 3: Calculate moles Al
Moles Al = mass Al / molar mass Al
Moles Al = 5.0 grams / 26.98 g/mol
Moles Al = 0.185 moles
Step 4: Calculate moles Br
Moles Br = 22.0 grams / 159.8 g/mol
Moles Br = 0.138 moles
Step 5: Calculate limiting reactant
For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3
Br2 is the limiting reactant. It will completely be consumed (0.0313 moles)
Al is in excess. There will react 2/3*0.138 = 0.092 moles
There will remain 0.185- 0.092 = 0.093 moles Al
Step 6: Calculate moles AlBr3
For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3
For 0.0313 moles Br2 we'll have 2/3*0.138 = 0.092 moles AlBr3
Step 7: Calculate mass AlBr3
Mass AlBr3 = moles * molar mass
Mass AlBr3 = 0.092 moles * 266.69 g/mol
Mass AlBr3 = 24.5 grams
The mass of AlBr3 is 24.5 grams
Final answer:
To find out the mass of aluminum bromide produced, we calculate the moles of aluminum and bromine, identify the limiting reactant, and then calculate the mass of the product from the moles of the limiting reactant using the stoichiometry of the reaction.
Explanation:
The question involves a stoichiometry calculation to determine the amount of product formed from a chemical reaction between aluminum and bromine (2 Al + 3 Br2 → 2 AlBr3). To solve this, we need to identify the limiting reactant and use it to calculate the mass of the product formed. First, the moles of each reactant is determined using their molar masses. Next, the stoichiometry of the balanced chemical equation is used to find the moles of product that can be formed from the limiting reactant.
Then, the moles of product are converted to grams using the molar mass of the product. This step-by-step process will lead us to find the mass of aluminum bromide (AlBr3) that can be produced.
An x-ray has a wavelength of 1.3 Å. Calculate the energy (in J) of one photon of this radiation.
The energy of a photon with a wavelength of 1.3 Å is calculated using Planck's equation. The wavelength is first converted to frequency, and then substituted into Planck's equation along with Planck's constant. The result, 1.53 x 10^-15 J, is the energy of the photon in Joules.
Explanation:The energy of a photon can be calculated using the Planck's equation E = hv, where 'v' is the frequency of radiation and 'h' is the Planck's constant (6.63 x 10^-34 J.s). In this case, we need to convert wavelength to frequency using the equation v = c/λ, where 'c' is the speed of light and 'λ' is the wavelength.
The given wavelength of 1.3 Å needs to be converted to meters: 1.3 Å = 1.3 x 10^-10 m. Then, the frequency can be calculated as: v = (3 x 10^8 m/s) / (1.3 x 10^-10 m) = 2.31 x 10^18 Hz.
Substituting these values into Planck's equation we get: E = (6.63 x 10^-34 J.s) x (2.31 x 10^18 Hz) = 1.53 x 10^-15 J. So, the energy of a photon of this radiation in Joules is 1.53 x 10^-15 J.
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The energy of one photon of X-ray radiation can be calculated using Planck's equation. In this case, the energy of one photon with a wavelength of 1.3 Å is approximately 1.53 × 10^-15 J.
Explanation:The energy of one photon of X-ray radiation can be calculated using Planck's equation. The formula E = hv, where E is energy, h is Planck's constant, and v is frequency. In this case, we need to convert the wavelength of the X-ray to frequency using the relationship c = λv, where c is the speed of light.
First, convert the wavelength from angstroms to meters: 1.3 Å = 1.3 × 10^(-10) m.
Then, use the equation c = λv to solve for v: (3.00 × 10^8 m/s) = (1.3 × 10^(-10) m)v.
Solve for v: v = (3.00 × 10^8 m/s) / (1.3 × 10^(-10) m) = 2.31 × 10^18 Hz.
Finally, calculate the energy using E = hv: E = (6.63 × 10^(-34) J·s)(2.31 × 10^18 Hz) = 1.53 × 10^(-15) J.
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A balloon filled with helium has a volume of 6.9 L. What is the mass, in grams, of helium in the balloon?
Answer : The mass of helium gas in the balloon is, 1.23 grams.
Explanation : Given,
Volume of helium gas = 6.9 L
First we have to calculate moles of helium gas at STP.
As we know that, 1 mole of substance occupy 22.4 L volume of gas.
As, 22.4 L volume of helium gas present in 1 mole of helium
So, 6.9 L volume of helium gas present in [tex]\frac{6.9L}{22.4L}\times 1mole=0.308mole[/tex] of helium
Now we have to calculate the mass of helium gas.
[tex]\text{Mass of He gas}=\text{Moles of He gas}\times \text{Molar mass of He gas}[/tex]
Molar mass of He gas = 4 g/mol
[tex]\text{Mass of He gas}=0.308mol\times 4g/mol=1.23g[/tex]
Thus, the mass of helium gas in the balloon is, 1.23 grams.
To find the mass of helium in the balloon, we can use the ideal gas law equation to calculate the number of moles of helium, and then convert it to grams using the molar mass. The mass of helium in the balloon is 27.14 grams.
To find the mass of helium in the balloon, we need to know the density of helium. The ideal gas law can be used to calculate the density of a gas. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation gives us n = PV/RT. We can then use the molar mass of helium (4 g/mol) to convert the number of moles to grams. Since the volume of the balloon and the pressure are given, we can plug in the values and solve for the mass of helium.
Given:
Volume (V) = 6.9 L
Molar mass of helium = 4 g/mol
R = 8.31 J/mol · K
After performing the calculations, the mass of helium in the balloon is 27.14 grams.
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While exploring a coal mine, scientists found plant fossils in the ceiling of the mine which had been preserved by an earthquake. Samples taken from one of the fossils have carbon‑14 activities of 40.0 counts/min. A reference sample of the same size, from a plant alive today, has a carbon‑14 activity of 160.0 counts/min.
If carbon‑14 has a half‑life of 5730 years, what is the age of the plant fossil in years?
Answer:
11552.45 years
Explanation:
Given that:
Half life = 5730 years
[tex]t_{1/2}=\frac{\ln2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac{\ln2}{t_{1/2}}[/tex]
[tex]k=\frac{\ln2}{5730}\ years^{-1}[/tex]
The rate constant, k = 0.00012 years⁻¹
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given that:
The rate constant, k = 0.00012 years⁻¹
Initial concentration [tex][A_0][/tex] = 160.0 counts/min
Final concentration [tex][A_t][/tex] = 40.0 counts/min
Time = ?
Applying in the above equation, we get that:-
[tex]40.0=160.0e^{-0.00012\times t}[/tex]
[tex]e^{-0.00012t}=\frac{1}{4}[/tex]
[tex]-0.00012t=\ln \left(\frac{1}{4}\right)[/tex]
[tex]t=11552.45\ years[/tex]
An electric current transports 93.0 C of charge in 601 milliseconds. Calculate the size of the electric current. Be sure your answer has the correct unit symbol and the correct number of significant digits.
Answer:
155 A
Explanation:
Given data
Transported charge (Q): 93.0 CoulombElapsed time (t): 601 ms = 0.601 secondsWe can determine the magnitude of the electric current (I) using the following expression.
I = Q / t
I = 93.0 C / 0.601 s
I = 155 C/s
I = 155 A
The magnitude of the electric current is 155 Ampere.
What numbers should be dialed into the p-10 display if a volume of 3.7 l is to be measured?
Following numbers should be dialed:
Tens: 3 (30 µL)Ones: 7 (7 µL)Tenths: 0 (0.1 µ)The display of the P-10 pipette consists of three digits: tens, ones, and tenths. These digits represent different places in the measurement scale. Here's how you would set it for a volume of 3.7 µL:
Tens Section (Leftmost Digit - Tens Place):
Dial in the digit 3. This represents 3 tens of microliters (µL). Essentially, it sets the pipette to 30 µL.
Ones Section (Middle Digit - Ones Place):
Dial in the digit 7. This represents 7 ones of microliters (µL).
Tenths Section (Rightmost Digit - Tenths Place):
Dial in the digit 0. This represents 0.1 µL. However, for your measurement of 3.7 µL, you don't need to adjust this digit, as you don't have any tenths to add.
Putting it all together:
Tens: 3 (30 µL)
Ones: 7 (7 µL)
Tenths: 0 (0.1 µL, but not needed in this case)
So, with the tens set to 3 and the ones set to 7, you have dialed in a total volume of 37 µL.
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Final answer:
For measuring a volume of 3.7 L with a micropipette, it’s not feasible using a P-10 pipettor due to its limited range. If the intended volume is 3.7 µL, dial 037 into the P-10 display. It's essential to use the pipettor within the correct volume range to avoid damage.
Explanation:
To measure a volume of 3.7 L (3700 mL) using a P-10 micropipette, which has a maximum volume of 10 µL (0.01 mL), would not be possible because the micropipette's volume range is much smaller than the volume you need to measure. However, if you meant to measure 3.7 µL, the numbers to dial on the P-10 display would be 037, indicating that you need to measure 3.7 µL. You should rotate the volume adjustment knob until the display shows these three digits. It is crucial to operate within the pipettor's specified volume range for accuracy and to prevent damage to the equipment.
Caution: Never turn the indicator dial beyond the micropipette's volume limits. To measure volumes accurately with a micropipette, ensure you are using the correct model for the desired volume range and adjust the dial accordingly. Using the P10 model, you can measure volumes as small as 1 µL up to 10 µL, which is its upper volume limit. Always consult the pipettor's manual or your lab guidelines when unsure about how to correctly set the volume on the micropipette.
Paraffin oil has a boiling point greater than 370°C. What was the purpose of adding paraffin to the reaction flask when cracking dicyclopentadiene? (Select all that apply.)
1) Paraffin serves as the hydrogen-donor in the reduction of dicyclopentadiene to cyclopentadiene.
2) Paraffin prevents the formation of free radicals that interfere with the cracking process.
3) Paraffin as a solvent results in more homogeneous heat distribution in the reaction vessel.
4) Paraffin is inflammable, which helps control the risk of fire during the cracking process.
5) Paraffin induces the formation of free radicals that are needed for the cracking process to proceed.
6) Paraffin prevents the reaction vessel from running dry.
Answer:
3) Paraffin as a solvent results in more homogeneous heat distribution in the reaction vessel.
Explanation:
First off, cyclopentadiene is obtained by “cracking” dicyclopentadiene. The cracking is carried out at 300 °C.
How do we Know if the reaction mixture have reached 300°C for cracking to occur? You probably thinking we use a thermometer..
The temperature of the reaction should be about 300 degrees and a normal thermometer would explode because it would reach past the heating limit.
So what do we then use?
We use a solvent whose boiling point is above the temperature needed for the reaction to occur and that's where paraffin comes into play. The moment the solvent begins to boil, we are certain that the temperature needed for the reaction has been reached.
The correct answer is option 3.
Final answer:
Paraffin is added to the reaction flask when cracking dicyclopentadiene mainly as a solvent to ensure homogeneous heat distribution and to prevent the reaction vessel from running dry due to its high boiling point.
Explanation:
The purpose of adding paraffin to the reaction flask when cracking dicyclopentadiene can be explained as follows:
Paraffin acts as a solvent providing a more homogeneous heat distribution in the reaction vessel (Option 3). This uniform heat distribution is crucial for the reaction to proceed efficiently and safely.
Given that paraffin oil has a high boiling point, it helps in preventing the reaction vessel from running dry by maintaining an adequate volume of liquid during the high-temperature process (Option 6).
While paraffin oil itself is inflammable, contributing to safer laboratory conditions, this is not directly related to its role in cracking dicyclopentadiene. Therefore, the fourth claim is factually correct about paraffin but irrelevant in this context.
What is the partial pressure of carbon dioxide, in atm, in a container that contains 3.63 mol of oxygen, 1.49 mol of nitrogen, and 6.16 mol of carbon dioxide when the total pressure is 871 mmHg? (enter your answer with 3 significant figures and no units)
Answer: The partial pressure of carbon dioxide is 476.
Explanation:
We are given:
Moles of oxygen gas = 3.63 moles
Moles of nitrogen gas = 1.49 moles
Moles of carbon dioxide gas = 6.16 moles
Total number of moles = [3.63 + 1.49 + 6.16] = 11.28 moles
Mole fraction of a substance is given by:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]
Mole fraction of carbon dioxide, [tex]\chi_{CO_2}=\frac{n_{CO_2}}{n_T}[/tex]
So, [tex]\chi_{CO_2}=\frac{6.16}{11.28}=0.546[/tex]
To calculate the partial pressure of carbon dioxide, we use the equation given by Raoult's law, which is:
[tex]p_{CO_2}=p_T\times \chi_{CO_2}[/tex]
where,
[tex]p_A[/tex] = partial pressure of carbon dioxide = ?
[tex]p_T[/tex] = total pressure = 871 mmHg
[tex]\chi_{CO_2}[/tex] = mole fraction of carbon dioxide = 0.546
Putting values in above equation, we get:
[tex]p_{CO_2}=871mmHg\times 0.546\\\\\chi_{CO_2}=475.6mmHg=476.mmHg[/tex]
Hence, the partial pressure of carbon dioxide is 476.
(a) Calculate the Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.
(b) What is the energy (in J) of the atom in part (a)?
(c) What is the energy of an Li²⁺ ion when its electron is in the n = 3 orbit?
(d) Why are the answers to parts (b) and (c) different?
Explanation:
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
Formula used for the radius of the [tex]n^{th}[/tex] orbit will be,
[tex]r_n=\frac{n^2\times 52.9}{Z}pm[/tex] (in pm)
where,
[tex]E_n[/tex] = energy of [tex]n^{th}[/tex] orbit
[tex]r_n[/tex] = radius of [tex]n^{th}[/tex] orbit
n = number of orbit
Z = atomic number
a) The Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.
Z = 1
[tex]r_3=\frac{3^2\times 52.9}{1} pm[/tex]
[tex]r_3=476.1 pm[/tex]
476.1 pm is the Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.
b) The energy (in J) of the atom in part (a)
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
[tex]E_3=-13.6\times \frac{1^2}{3^2}eV=1.51 eV[/tex]
[tex]1 eV=1.60218\times 10^{-19} Joules[/tex]
[tex]1.51 eV=1.51\times 1.60218\times 10^{-19} Joules=2.4210\times 10^{-19} Joules[/tex]
[tex]2.4210\times 10^{-19} Joules[/tex] is the energy of n = 3 orbit of a hydrogen atom.
c) The energy of an Li²⁺ ion when its electron is in the n = 3 orbit.
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
n = 3, Z = 3
[tex]E_3=-13.6\times \frac{3^2}{3^2}eV = -13.6 eV[/tex]
[tex]=-13.6eV = -13.6\times 1.60218\times 10^{-19} Joules=2.179\times 10^{-18} Joules[/tex]
[tex]2.179\times 10^{-18} Joules[/tex] is the energy of an Li²⁺ ion when its electron is in the n = 3 orbit.
d) The difference in answers is due to change value of Z in the formula which is am atomic number of the element..
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
A metal crystallizes in a body-centered cubic unit cell. The radius of one atom = 2.30 x 10 -8 cm. The density of the metal is 0.867 g/cm^3 . What is molar mass of metal?
Answer:
25.41 g/mol is molar mass of metal.
Explanation:
Number of atom in BCC unit cell = Z = 2
Density of metal = [tex]0.867 g/cm^3[/tex]
Edge length of cubic unit cell= a = ?
Radius of the atom of metal = r = [tex]2.30\times 10^{-8} cm[/tex]
[tex]a=2\times r=2\times 2.30\times 10^{-8} cm=4.60\times 10^{-8} cm[/tex]
Atomic mass of metal =M
Formula used :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density
Z = number of atom in unit cell
M = atomic mass
[tex](N_{A})[/tex] = Avogadro's number
a = edge length of unit cell
On substituting all the given values , we will get the value of 'a'.
[tex]0.867 g/cm^3=\frac{2\times M}{6.022\times 10^{23} mol^{-1}\times (4.60\times 10^{-8} cm)^{3}}[/tex]
[tex]M = 25.41 g/mol[/tex]
25.41 g/mol is molar mass of metal.
To calculate the molar mass of a metal with a body-centered cubic unit cell, you need to calculate the edge length of the unit cell. Once you have the edge length and the density of the metal, you can determine the molar mass.
Explanation:The molar mass of a metal can be calculated using the formula:
Molar mass = (density × Avogadro's number) / (volume of one atom)
To calculate the volume of one atom, we need to know the type of unit cell and the edge length of the unit cell. Given that the metal crystallizes in a body-centered cubic unit cell, the edge length can be calculated using the formula:
Edge length = 4R / sqrt(3),
where R = radius of the atom.
Using the given radius of the atom (2.30 x 10-8 cm), we can calculate the edge length of the unit cell. With the edge length and the density of the metal (0.867 g/cm3), we can then calculate the molar mass of the metal.
In basic enzyme kinetics, a large Km indicates that the substrate binds A. Permanently B. Transiently C. Covalently D. with low specificity E. Weakly
Answer: D
Explanation:
Km value is a signature of the enzyme. It is the characteristic feature of a particular enzymes for a specific substrate. Km denotes the affinity of the enzyme for substrate. The lesser the numerical value of Km, the affinity of the enzyme for the substrate is more.
In the velocity x substrate graph in a fixed quantity if enzyme. As substrate concentration is increase, the velocity is also increasing at the initial phase but the curve fatten afterwards. This is because as more substrate is added, all enzymes molecules become saturated. Further increase in substrate cannot make any effect in the reaction velocity.
The maximum velocity is called Vmax. Km is the concentration of substrate that Vmax is half.
The larger the numerical value of Km, the lesser the enzyme binds the substrate
Final answer:
In basic enzyme kinetics, a large Km value signifies that the substrate binds weakly to the enzyme, indicating a lower affinity. A high Km necessitates a higher substrate concentration for effective enzyme activity, making it a critical parameter in enzymology and biotechnology.
Explanation:
In basic enzyme kinetics, a large Km indicates that the substrate binds weakly. The Km value, or the Michaelis constant, provides a measure of the affinity of the enzyme for its substrate. A high Km value suggests that a higher concentration of the substrate is required to achieve half of the maximum rate of reaction (Vmax), indicating a lower affinity of the enzyme for the substrate. Conversely, a low Km value signifies a high affinity, meaning the enzyme can easily bind the substrate even at low concentrations.
Understanding the implications of Km values is essential in enzymology and for designing better enzymes through biotechnology. A larger Km value generally indicates weaker enzyme-substrate interactions, suggesting that the enzyme requires a higher concentration of substrate to be effective. This parameter is crucial in the study of enzyme kinetics, drug design, and understanding metabolic pathways.
Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:
(a) Cl (b) Na (c) Ca
Explanation:
Chlorine atom has an atomic number of 17
Electronic configuration
Cl = 1s2 2s2 2p6 3s2 3p5
Cl- = 1s2 2s2 2p6 3s2 3p6
Soduim atom has an atomic number of 11
Electronic configuration
Na = 1s2 2s2 2p6 3s1
Na+ = 1s2 2s2 2p6
Calcium atom has an atomic number of 20
Electronic configuration
Ca = 1s2 2s2 2p6 3s2 3p6 4s2
Ca2+ = 1s2 2s2 2p6 3s2 3p6
The monatomic ions most likely formed by Cl, Na, and Ca are Cl-, Na+, and Ca2+, respectively.
Explanation:(a) The most likely monatomic ion formed by Chlorine (Cl) is Cl-. The electronic configuration of Cl is 1s2 2s2 2p6 3s2 3p5. By gaining one electron, Cl achieves a stable octet and becomes a chloride ion (Cl-).
(b) The most likely monatomic ion formed by Sodium (Na) is Na+. The electronic configuration of Na is 1s2 2s2 2p6 3s1. By losing one electron, Na achieves a stable octet and becomes a sodium ion (Na+).
(c) The most likely monatomic ion formed by Calcium (Ca) is Ca2+. The electronic configuration of Ca is 1s2 2s2 2p6 3s2 3p6 4s2. By losing two electrons, Ca achieves a stable octet and becomes a calcium ion (Ca2+).
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How many grams of sulfur dioxide (SO2) are there in a 3.50 L container at 100°C and a pressure of 2.00 atm?
Answer: The mass of sulfur dioxide is 14.6 grams
Explanation:
To calculate the number of moles, we use the equation given by ideal gas, which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the sulfur dioxide = 2.00 atm
V = Volume of the sulfur dioxide = 3.50 L
T = Temperature of the mixture = [tex]100^oC=[100+273]K=373K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of sulfur dioxide = ?
Putting values in above equation, we get:
[tex]2.00atm\times 3.50L=n_{SO_2}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 373K\\\\n_{SO_2}=\frac{2.00\times 3.50}{0.0821\times 373}=0.228mol[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of sulfur dioxide = 0.228 moles
Molar mass of sulfur dioxide = 64 g/mol
Putting values in above equation, we get:
[tex]0.228mol=\frac{\text{Mass of sulfur dioxide}}{64g/mol}\\\\\text{Mass of sulfur dioxide}=(0.228mol\times 64g/mol)=14.6g[/tex]
Hence, the mass of sulfur dioxide is 14.6 grams
Reduced molecules become _____ upon donating an electron in a redox reaction. Choose one: A. saturated B. oxidized C. bound to oxygen D. reduced
Answer: Option B. oxidized
Explanation:
Define each of the following wave phenomena, and give an example of where each occurs: (a) refraction; (b) diffraction; (c) dispersion; (d) interference.
Explanation:
Refraction is the phenomenon of change in speed of light when the light when it passes from the medium of one optical density to a medium of another optical density. The light rays also show a bending when passing through the interface of such medium.Diffraction is the process of spreading of the light or any system of wave after passing through a narrow gap of the order of wavelength of the wave.Dispersion is the phenomenon of acquiring the velocity corresponding to its wavelength by a wave.Interference of waves is the phenomenon of interaction of of two waves when they meet and superimpose on one another to give a wave of resultant properties of both the source waves.On a balance, you have beakers of AgNO3 A g N O 3 solution and NaCl N a C l solution. When mixed, they will form AgCl(s) A g C l ( s ) . What will happen to the mass of the new mixture?
Answer:
It remains the same.
Explanation:
From the law of conservation of mass which states that mass in an isolated system can neither be created nor destroyed in a chemical reaction.
According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants, that is total mass of the system is unchanged after the chemical reaction.
So therefore, the reaction of
AgNO3(aq) + HCl(aq) --> AgCl(aq) + HNO3(aq)
the mass of the solution before and after the reaction is unchanged.
A gas contained in a piston cylinder assembly undergoes a process from state 1 to state 2 defined by the following relationship and given properties. Determine the final volume (V2) of the gas.P*V
Answer:
V2 = 1/4
Explanation: P1 = 1; P2 = 2; V1 = 1/2; V2 = ?
P1*V1 = P2*V2
∴ V2 = P1*V1/P2 = {1 X 1/2}/2 = 1/4
Write the electron configuration for the monatomic ions formed from the following elements (which form the greatest concentration of monatomic ions in seawater): (a) Cl (b) Na (c) Mg (d) Ca (e) K (f) Br (g) Sr (h) F
Answer : The electron configuration for the monatomic ions are shown below.
Explanation :
For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.
When an unequal number of electrons and protons then it leads to the formation of ionic species.
Ion : An ion is formed when an atom looses or gains electron.
When an atom looses electrons, it will form a positive ion known as cation.
When an atom gains electrons, it will form a negative ion known as anion.
(a) The given element is, Cl (Chlorine)
As we know that the rubidium element belongs to group 17 and the atomic number is, 17
The ground-state electron configuration of Cl is:
[tex]1s^22s^22p^63s^23p^5[/tex]
This element will easily gain 1 electron and form [tex]Cl^-[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Cl ion is:
[tex]1s^22s^22p^63s^23p^6[/tex]
(b) The given element is, Na (Sodium)
As we know that the sodium element belongs to group 1 and the atomic number is, 11
The ground-state electron configuration of Na is:
[tex]1s^22s^22p^63s^1[/tex]
This element will easily lose 1 electron and form [tex]Na^{+}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Na ion is:
[tex]1s^22s^22p^6[/tex]
(c) The given element is, Mg (Magnesium)
As we know that the magnesium element belongs to group 2 and the atomic number is, 12
The ground-state electron configuration of Mg is:
[tex]1s^22s^22p^63s^2[/tex]
This element will easily lose 2 electrons and form [tex]Mg^{2+}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Mg ion is:
[tex]1s^22s^22p^6[/tex]
(d) The given element is, Ca (Calcium)
As we know that the calcium element belongs to group 2 and the atomic number is, 20
The ground-state electron configuration of Ca is:
[tex]1s^22s^22p^63s^23p^64s^2[/tex]
This element will easily lose 2 electrons and form [tex]Ca^{2+}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Ca ion is:
[tex]1s^22s^22p^63s^23p^6[/tex]
(e) The given element is, K (Potassium)
As we know that the potassium element belongs to group 1 and the atomic number is, 19
The ground-state electron configuration of K is:
[tex]1s^22s^22p^63s^23p^64s^1[/tex]
This element will easily lose 1 electron and form [tex]K^{+}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of K ion is:
[tex]1s^22s^22p^63s^23p^6[/tex]
(f) The given element is, Br (Bromine)
As we know that the bromine element belongs to group 17 and the atomic number is, 35
The ground-state electron configuration of Rb is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^5[/tex]
This element will easily gain 1 electron and form [tex]Br^-[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Br ion is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
(g) The given element is, Sr (Strontium)
As we know that the strontium element belongs to group 2 and the atomic number is, 38
The ground-state electron configuration of Rb is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^2[/tex]
This element will easily lose 2 electrons and form [tex]Sr^{2+}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of Sr ion is:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]
(h) The given element is, F (Fluorine)
As we know that the fluorine element belongs to group 17 and the atomic number is, 9
The ground-state electron configuration of F is:
[tex]1s^22s^22p^5[/tex]
This element will easily gain 1 electron and form [tex]F^{-}[/tex] ion which attain stable noble gas electronic configuration.
The full ground-state electron configuration of F ion is:
[tex]1s^22s^22p^6[/tex]
An ion having only one atom is called a mono-atomic ion and is represented as the symbol of the element with mass and charge in subscript and superscript.
The ion that loses an electron is called a cation while the ion that acquires an electron is called an anion.
The electronic configuration of the following ions are:
a. Cl (Chlorine):
Chlorine belongs to group 17 in the periodic table and has an atomic number of 17.The ground state electronic distribution of chlorine ion:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}[/tex]
To attain noble gas configuration Cl will gain one electron to form [tex]Cl^{-}[/tex].The full ground-state electron distribution of Cl ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}[/tex]
b. Na (Sodium):
The sodium element has the atomic number 11 and belongs to group 1.Ground-state electron arrangement of Na is:[tex]1s^{2}2s^{2}2p^{6}3s^{1}[/tex]
It loses an electron to attain noble gas configuration and forms [tex]Na^{+}[/tex].The full ground-state electron arrangement of Na ion is:[tex]1s^{2}2s^{2}2p^{6}[/tex]
c. Mg (Magnesium):
It has atomic number 12 and belongs in the 2nd group.Ground-state electron distribution of Mg is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}[/tex]
To attain noble gas configuration it loses two electrons and forms[tex]Mg^{2+}[/tex].
The full ground-state electron distribution of Mg ion is:[tex]1s^{2}2s^{2}2p^{6}[/tex]
d. Ca (Calcium):
The atomic number is 20 and belongs to group 2.The ground-state electron arrangement of Ca is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}[/tex]
It loses two electrons to attain a noble gas configuration and forms [tex]Ca^{2+}[/tex].The full ground-state electron arrangement of Ca ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}[/tex]
e. K (Potassium):
The atomic number is 19 and belongs to group 1.The ground-state electron distribution of K is:
[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}[/tex]
To attain noble gas configuration it loses one electron and forms [tex]K^{+}[/tex].The full ground-state electron distribution of K ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}[/tex]
f. Br (Bromine):
The bromine element has the atomic number 35 and belongs to group 17.The ground-state electron arrangement of Br ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}[/tex]
It gains one electron to attain noble gas configuration and forms [tex]Br^{-}[/tex].The full ground-state electron arrangement of Br ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}[/tex]
g. Sr (Strontium):
It belongs to the 2nd group and has atomic number 38.The ground-state electron arrangement of Sr ion is:
[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}[/tex]
It loses two-electron to attain noble gas configuration and forms [tex]Sr^{2+}[/tex].The full ground-state electron arrangement of Sr ion is:[tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}[/tex]
h. F (Fluorine):
It belongs to group 17 and has the atomic number 9.The ground-state electron distribution of F ion is:[tex]1s^{2}2s^{2}2p^{5}[/tex]
It gains one electron to form [tex]F^{-}[/tex] to attain the noble gas configuration.The full ground-state electron distribution of F ion is:[tex]1s^{2}2s^{2}2p^{6}[/tex]
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1. 0.16 m Pb(CH3COO)2 A. Highest boiling point 2. 0.17 m NiBr2 B. Second highest boiling point 3. 8.8×10-2 m Al2(SO4)3 C. Third highest boiling point 4. 0.53 m Urea(nonelectrolyte) D. Lowest boiling point
Explanation:
[tex]\Delta T_b=T_b-T[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]Delta T_b=iK_b\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] =Elevation in boiling point
[tex]K_b[/tex] = boiling point constant of solvent
i = van't Hoff factor
m = molality
As we can see that molality is directly proportional ti elevation in boiling point, so higher the molality of the solution at more high temperature it will boil.
1) 0.16 m [tex]Pb(CH_3COO)_2[/tex]
i = 3
[tex]\Delta T_b=3K_b\times 0.16 m=K_b\times 0.48 m[/tex]
Third highest boiling point
2) 0.17 m [tex]NiBr_2[/tex]
i = 3
[tex]\Delta T_b=3K_b\times 0.17 m=K_b\times 0.51 m[/tex]
Second highest boiling point
3) [tex]8.8\times 10^{-2} m[/tex] of [tex]Al_2(SO_4)_3[/tex]
i = 5
[tex]\Delta T_b=5\times K_b\times 8.8\times 10^{-2} m=K_b\times 0.44 m[/tex]
Lowest boiling point
4) 0.53 m Urea
i = 1
[tex]\Delta T_b=1\times K_b\times 0.53 m =K_b\times 0.53 m[/tex]
Highest boiling point
The boiling point of a solution depends on the concentration and nature of the solute. Electrolytes increase the boiling point, while nonelectrolytes do not. In this case, the solutions with Pb(CH3COO)2, NiBr2, and Al2(SO4)3 have higher boiling points than the solution with urea.
Explanation:The boiling point of a solution is influenced by the concentration and nature of the solute. In this case, we are given four solutions with different concentrations and solutes. To determine the boiling point, we need to consider the number of particles the solute breaks into when it dissolves. Electrolytes like Pb(CH3COO)2, NiBr2, and Al2(SO4)3 break into ions and increase the boiling point while urea, a nonelectrolyte, does not break into ions and has a lower boiling point.
0.16 m Pb(CH3COO)2 has the highest boiling point because it is an electrolyte that breaks into one Pb2+ ion and two CH3COO- ions.0.17 m NiBr2 has the second highest boiling point because it is an electrolyte that breaks into two Ni2+ ions and four Br- ions.8.8×10-2 m Al2(SO4)3 has the third highest boiling point because it is an electrolyte that breaks into two Al3+ ions and three SO42- ions.0.53 m Urea (nonelectrolyte) has the lowest boiling point because it does not break into ions and does not increase the boiling point.Learn more about Boiling points of solutions here:https://brainly.com/question/35155444
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Determine the speed of sound at 20 oC in (a) air, (b) helium, and (c) natural gas (methane). Express your answer in m/s.
The speed of sound in different mediums varies based on their properties. In air at 20 oC, the speed of sound is approximately 343 m/s. In helium, it is approximately 972 m/s, and in natural gas (methane), it is approximately 454 m/s.
Explanation:The speed of sound in a medium depends on the properties of that medium. In general, sound travels faster in materials with higher densities and elastic properties. The equation to calculate the speed of sound in a medium is given by v = sqrt(E/p), where v is the speed of sound, E is the elastic modulus, and p is the density of the medium.
(a) In air at 20 oC, the speed of sound is approximately 343 m/s.
(b) In helium at 20 oC, the speed of sound is approximately 972 m/s.
(c) In natural gas (methane) at 20 oC, the speed of sound is approximately 454 m/s.
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Final answer:
The speed of sound in air at 20.0°C is 343 m/s. For other gases like helium and methane at the same temperature, their speeds of sound are higher due to their lower molecular masses. Helium has a much higher speed of sound, while methane's speed of sound is closer to but slightly higher than air.
Explanation:
The speed of sound in air at a given temperature can be calculated using the following formula, where °C represents degrees Celsius:
v = 331 m/s + (0.6 m/s/°C) × temperature
For air at 20.0°C, the speed of sound is:
v = 331 m/s + (0.6 m/s/°C) × 20.0°C
v = 331 m/s + 12 m/s
v = 343 m/s
For other gases like helium and methane at the same temperature, their speeds of sound are higher due to their lower molecular masses compared to air.
The speed of sound in helium and methane can be determined from tables or calculated using specific formulas accounting for the properties of these gases.
Helium: 972 m/sNatural Gas (Methane): 443 m/sRank the ions in each set in order of decreasing size, and explain your ranking:
(a) Se²⁻, S²⁻, O²⁻ (b) Te²⁻, Cs⁺, I⁻ (c) Sr²⁺, Ba²⁺, Cs⁺
The size of an ion is determined by the effective nuclear charge and the number of electrons in the ion. In isoelectronic ions, the size decreases with increasing effective nuclear charge. In ions with the same number of electrons, the size increases with increasing atomic number.
Explanation:(a) In this set, the ions are all isoelectronic, meaning they have the same number of electrons. Therefore, their sizes are determined by the effective nuclear charge. The effective nuclear charge increases from Se²⁻ to S²⁻ to O²⁻ because the number of protons in the nucleus increases. So, the sizes of the ions decrease from Se²⁻ > S²⁻ > O²⁻.
(b) In this set, the ions have different numbers of electrons. The size of an ion generally increases with the addition of electrons. So, the sizes of the ions decrease from I⁻ > Te²⁻ > Cs⁺.
(c) In this set, the ions have the same number of electrons but different atomic numbers. The size of an ion generally increases with increasing atomic number. So, the sizes of the ions decrease from Cs⁺ > Sr²⁺ > Ba²⁺.
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The size of an ion is determined by the number of electrons surrounding it. In each set, the size of the ions decreases as you move along the set. The order of decreasing size for each set is explained based on the position of the elements on the periodic table.
Explanation:(a) Se²⁻, S²⁻, O²⁻: The size of an ion is determined by the number of electrons surrounding it. As you move down a group on the periodic table, the size of the ions generally increases. Therefore, the order of decreasing size is O²⁻, S²⁻, Se²⁻.
(b) Te²⁻, Cs⁺, I⁻: In this set, the ionic size increases from Te²⁻ to I⁻. This is due to the decrease in the number of protons in the nucleus as you move down the periodic table. Cs⁺ is significantly smaller than Te²⁻ and I⁻ due to the removal of an electron.
(c) Sr²⁺, Ba²⁺, Cs⁺: Again, the size of the ions increases as you move down the periodic table. Therefore, the order of decreasing size is Cs⁺, Sr²⁺, Ba²⁺.
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How many grams of the molecule glucose C6H12O6 would be required to make 1 L of a 0.5 M solution of the molecule?
Answer:
90g
Explanation:
We need to find the molar mass of C6H12O6. This is done below:
MM of C6H12O6 = (12x6) + (12x1) + (6x16) = 72 + 12 + 96 = 180g/mol
From the question, we obtained:
Volume = 1L
Molarity = 0.5M
Number of mole = Molarity x Volume
Number of mole = 0.5 x 1 = 0.5mol
Mass conc of C6H12O6 = number of mole x MM = 0.5 x 180 = 90g
To make 1 L of a 0.5 M solution of glucose, one would need 90.078 grams of glucose, based on the molecular weight calculation of C6H12O6.
To calculate how many grams of glucose (C6H12O6) are needed to make 1 L of a 0.5 M solution, you first need to calculate the molar mass of glucose. The molar mass can be found by adding up the atomic masses of all the atoms in one molecule of glucose. The atomic masses are approximately carbon (C) 12.01 g/mol, hydrogen (H) 1.008 g/mol, and oxygen (O) 16.00 g/mol. So, the molar mass of C6H12O6 is:
(6 × 12.01) + (12 × 1.008) + (6 ×16.00) = 180.156 g/molNext, you use the molarity equation which is:
Molarity (M) = moles of solute / liters of solution
For a 0.5 M solution, you require 0.5 moles of glucose per liter of solution. Using the molar mass calculated earlier, the mass of glucose needed is:
(0.5 moles) ×(180.156 g/mol) = 90.078 g
Therefore, you would need 90.078 g of glucose to make 1 L of a 0.5 M solution of glucose.
A chemist measures the energy change during the following reaction: 1. This reaction is:______. a. endothermic. b. exothermic. 2. Suppose 70.1 g of NO2 react. Will any heat be released or absorbed? A. Yes, absorbed. B. Yes, released. C. No. 3. If you said heat will be released or absorbed, calculate how much heat will be released or absorbed?
The question is incomplete. the complete question is:
A chemist measures the energy change during the following reaction: [tex]2NO_2(g)\rightarrow N_2O_4[/tex] [tex]\Delta H=-55.3kJ[/tex] 1. This reaction is:______. a. endothermic. b. exothermic.
2. Suppose 70.1 g of NO2 react. Will any heat be released or absorbed? A. Yes, absorbed. B. Yes, released. C. No.
3. If you said heat will be released or absorbed, calculate how much heat will be released or absorbed?
Answer:1. b. exothermic
2. Yes , released
3. 42.0 kJ
Explanation:
1. Endothermic reactions are those in which heat is absorbed by the system and exothermic reactions are those in which heat is released by the system.
[tex]\Delta H[/tex] for Endothermic reaction is positive and [tex]\Delta H[/tex] for Exothermic reaction is negative.
2. [tex]2NO_2(g)\rightarrow N_2O_4[/tex] [tex]\Delta H=-55.3kJ[/tex]
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{70.1g}{46g/mol}=1.52moles[/tex]
2 moles of [tex]NO_2[/tex] reacts, energy released = 55.3 kJ
1.52 moles of [tex]NO_2[/tex] reacts, energy released =[tex]\frac{55.3 kJ}{2}\times 1.52=42.0kJ[/tex]
Thus 42.0 kJ heat will be released when 70.1 g of [tex]NO_2[/tex] react.