Answer: 13,68 ML
Explanation: In the problem density and mass are give & you have to figure out the volume of mercury. So use this formula:
Mass = Volume × Density
Volume = Mass ÷ Density
Volume = 186 g ÷ 3.6 g/ml
= 13.68 ml
Lewis structure of methyl metcaptain
Answer:
The lewis structure of the compounds can be drawn by making the skeleton of the molecule first. Then the different atoms are arranged and the electrons are arranged in their bonding pattern. The lone pair of the atoms, which are not involved in the bonding are represented by the dots.
So the lewis structures of both the compound methyl mercaptan has been in the attached image:
Spelling of methyl metcaptain is wrong, the correct spelling is methyl mercaptan.
Answer:
Methyl mercaptan is also known as Methanethiol with the chemical formula CH3SH and it is an organosulfur compound.
For lewis structure of methyl mercaptan (CH3SH), there are total 14 valence electrons. Four hydrogen atoms has one valence electron each, carbon has four valence electrons and sulfur has six valence electrons. Carbon form one bond with three hydrogen atoms by sharing one electron with each, carbon form one single bond with sulfur atom by sharing one electron with it and sulfur form one single bond with hydrogen. Sulfur left with four unpair electrons.
The speed of a car is 32.0 miles/hr . What is it's speed in m/s ? ( given 1 mile = 1609m )
Answer:14.3 m/s
Explanation:
32 x 1609 = 51488
51488 / 60 / 60 = 14.3
(60 minutes and 60 seconds so you can get unit of 1 second)
oxidation number of Se in Se8
Answer:
0.
Explanation:
All elements existing on their own in their free states has an oxidation number of 0.
The oxidation number of Se in Se8 is 0. The element Se in its elemental state has an oxidation number of 0.
Chemistry uses the idea of oxidation number, also known as oxidation state, to describe the charge that an atom seems to have when it forms a compound or ion. It is a notion that aids in our comprehension of how electrons are distributed within a molecule or ion. Oxidation numbers can be used to balance chemical equations and identify the many kinds of chemical reactions that may take place. For a better understanding of the reactivity and bonding of various elements in compounds, they offer insights into the transfer or sharing of electrons between atoms. Se in Se8 has an oxidation number of 0. Se has an oxidation number of 0 when it is in its purest form.
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Identify the type of solid for AgCl. Identify the type of solid for AgCl. metallic atomic solid nonbonding atomic solid molecular solid ionic solid networking atomic solid
Silver chloride (AgCl) is a crystalline solid substance that is composed of silver and chloride ions. AgCl is an ionic solid. Thus, option D is correct.
What are ionic solids?Ionic solids are substances that show the properties of solid matter and have ionic, positive, and negative charges in them. They are linked together by the attraction of the opposite charges.
The silver metal in the molecule has a positive charge and the chloride ions are negative in charge making them establish an ionic bond.
The solid molecule is held together by ionic bonds and not the covalent or other metallic bonds. The cations and anions of AgCl are linked together by the electrostatic forces that make their structure appear strong and brittle.
Therefore, AgCl has been known as an ionic solid.
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A chemist dissolves 867. mg of pure barium hydroxide in enough water to make up 170. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 degree C.) Be sure your answer has the correct number of significant digits.
Answer: The pH of the solution is 11.2
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in ml
moles of [tex]Ba(OH)_2[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.867g}{171g/mol}=0.00507mol[/tex] (1g=1000mg)
Now put all the given values in the formula of molality, we get
[tex]Molarity=\frac{0.00507\times 1000}{170}[/tex]
[tex]Molarity=0.0298[/tex]
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
[tex]pOH=-\log [OH^-][/tex]
[tex]Ba(OH)_2\rightarrow Ba^{2+}+2OH^{-}[/tex]
According to stoichiometry,
1 mole of [tex]Ba(OH)_2[/tex] gives 2 mole of [tex]OH^-[/tex]
Thus 0.0298 moles of [tex]Ba(OH)_2[/tex] gives =[tex]\frac{2}{1}\times 0.0298=0.0596[/tex] moles of [tex]OH^-[/tex]
Putting in the values:
[tex]pOH=-\log[0.0596]=2.82[/tex]
[tex]pH+pOH=14[/tex]
[tex]pH=14-2.82[/tex]
[tex]pH=11.2[/tex]
Thus the pH of the solution is 11.2
You've just synthesized a new molecule and need to purify it by recrystallization. You find that it is poorly soluble in water and highly soluble in ethanol, even when cooled in ice. What solvent should work in this situation
Answer:
Water is used as a solvent.
Explanation:
In order to purify the new molecule, recrystallization occurs in which the new molecule should be added in water and heated the water in order to increase the solubility of the solution. The new molecule dissolve in water while the impurity remains undissolved which can be removed from the solution and then remain the solution to be cold down and the new molecule will again undissolved and the molecule can be extracted without any impurities. We use water instead of ethanol due to lower solubility.
A student was tasked with creating 450.0 mL of a 1.75 M HCl solution from a 4.50 M HCl stock solution.
How much of the stock solution should be transferred in order to conduct the dilution?
A. 1,160 mL
B. 1.16 L
C. 175L
D. 0.175L
Answer:
Option D is correct.
The volume of the 4.50 M stock solution required to be diluted to the specifications given is 0.175 L.
Explanation:
The law of dilution gives that the number of moles in both solutions stay the same. Mathematically, the law is given as
C₁V₁ = C₂V₂
The student wants to make 450 mL of 1.75 M of HCl.
We can find the number of moles of HCl in this required solution.
Number of moles = (Concentration in mol/L) × (Volume in L)
Concentration in mol/L = 1.75 M
Volume in L = (450/1000) = 0.45 L
Number of moles = 1.75 × 0.45 = 0.7875 mole
The stock solution has a concentration of 4.50 M
Volume of Stock solution needed
= (Number of moles)/(Concentration of stock solution in mol/L)
= (0.7875/4.5) = 0.175 L
Hence, the volume of the 4.50 M stock solution required to be diluted to the specifications given is 0.175 L.
Hope this Helps!!!
How many ng(nanogram) are in 3.3 mg(milligram)?
Answer:
There are [tex]3.3(10^6)[/tex] nanograms in 3.3 milligrams
Explanation:
The conversion is 1 milligram is equal to [tex]1(10^6)[/tex] nanograms. Use the base 10 decimal system to help you.
Calculate the theoretical percentage of water for the following hydrates.
(a) manganese(II) monohydrate, MnSO4 H2O
(b) manganese(II) tetrahydrate, MnSO4 4H2O
Answer:
(a) [tex]\% H_2O=10.65\%[/tex]
(b) [tex]\% H_2O=32.2[/tex]
Explanation:
Hello.
For this questions we must consider the ratio of the molar mass of water to hydrated compound molar mass as shown below:
(a) In this case, we can consider that inside the manganese (II) sulfate monohydrate, whose molar mass is 169.02 g/mol, there is one water molecule that has a molar mass of 18 g/mol, for which the theoretical percentage of water is:
[tex]\% H_2O=\frac{18g/mol}{169.0g/mol} *100\%\\\\\% H_2O=10.65\%[/tex]
(b) In this case, we can consider that inside the manganese (II) sulfate tetrahydrate, whose molar mass is 223.1 g/mol, there are four water molecules that have a molar mass of 4*18 g/mol, for which the theoretical percentage of water is:
[tex]\% H_2O=\frac{4*18g/mol}{223.1g/mol} *100\%\\\\\% H_2O=32.27\%[/tex]
Best regards.
what is the modification of the Dalton's atomic theory
Answer:
Dalton's theory had to be modified after mass spectrometry experiments demonstrated that atoms of the same element can have different masses because the number of neutrons can vary for different isotopes of the same element
Question 4
2 pts
A careless chemistry student performed a chemical reaction where his theoretical yield of
Magnesium oxide was 57.82 grams, but he actually produced 12.89 grams. What is his percent yield
for this experiment? (include the number with 4 significant figures but no units)
Answer:
22.29%
Explanation:
Percent yield = experimental yield / theoretical yield * 100
= 12.89 / 57.82 * 100 = 22.29%
A 1.0 L buffer solution is 0.300 M HC2H3O2 and 0.045 M LiC2H3O2. Which of the following actions will destroy the buffer?
a) Adding 0.050 moles of HC2H3O2
b) Adding 0.075 moles of HCl
c) Adding 0.0500 moles of LiC2H3O2
d) Adding 0.050 moles of NaOH
e) None of the above will destroy the buffer.
Answer:
b) Adding 0.075 moles of HCl
Explanation:
A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa (Weak base with its conjugate acid).
The buffer of the problem is the acetic acid / lithium acetate.
The addition of any moles of the acid and the conjugate base will not destroy the buffer, just would change the pH of the buffer. Thus, a and c will not destroy the buffer.
The addition of an acid (HCl) or a base (NaOH), produce the following reactions:
HCl + LiC₂H₃O₂ → HC₂H₃O₂ + LiCl
The acid reacts with the conjugate base to produce the weak acid.
And:
NaOH + HC₂H₃O₂ →NaC₂H₃O₂ + H₂O
The base reacts with the weak acid to produce conjugate base.
As the buffer is 1.0L, the moles of the species of the buffer are:
HC₂H₃O₂ = 0.300 moles
LiC₂H₃O₂ = 0.045 moles
The reaction of HCl with LiC₂H₃O₂ consume all LiC₂H₃O₂ -because there are an excess of moles of HCl that react with all LiC₂H₃O₂-
As you will have just HC₂H₃O₂ after the reaction, the addition of b destroy the buffer.
In the other way, 0.0500 moles of NaOH react with the HC₂H₃O₂ but not consuming all HC₂H₃O₂, thus d doesn't destroy the buffer.
If 25 mL of a HCl solution of unknown concentration was neutralized with 10 mL of a 0.30 M NaOH solution, what was the original concentration of the HCl solution
Answer:
0.12 M
Explanation:
Step 1: Write the balanced equation
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the reacting moles of NaOH
10 mL of a 0.30 M NaOH solution react.
[tex]0.010L \times \frac{0.30mol}{L} = 3.0 \times 10^{-3} mol[/tex]
Step 3: Calculate the reacting moles of HCl
The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 1/1 × 3.0 × 10⁻³ mol = 3.0 × 10⁻³ mol.
Step 4: Calculate the concentration of HCl
3.0 × 10⁻³ mol of HCl are in 25 mL of solution.
[tex]M = \frac{3.0 \times 10^{-3} mol}{0.025L} = 0.12 M[/tex]
What is science?
O A. Attempting to solve ethical problems through research
O B. Learning about the physical world through observation
O C. Determining a body of knowledge that never changes
O D. Using rules and patterns to predict what humans will do
SUBMIT
Answer:
The answer is B
Explanation:
Answer:
B. Learning about the physical world through observation
Explanation:
A p e x 2021 :) Trust me!
How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solution using 100 mL of ethyl acetate per extraction if the partition coefficient is 8
Answer:
4.94g of material
Explanation:
Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:
Kp = 8 = Concentration in Ethyl acetate / Concentration in water
100mL of a 5% solution contains 5g of material in 100mL of water. Thus:
8 = X / 100mL / (5g-X) / 100mL
Where X is the amount of material in grams that comes to the organic phase.
8 = X / 100mL / (5g-X) / 100mL
8 = 100X / (500-100X)
4000 - 800X = 100X
4000 = 900X
4.44g = X
Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.
And will remain 5g-4.44g = 0.56g.
In the second extraction:
8 = X / 100mL / (0.56g-X) / 100mL
8 = 100X / (56-100X)
448 - 800X = 100X
448 = 900X
0.50g = X
In the second extraction, you will extract 0.50g of material
Thus, after the two extraction you will lost:
4.44g + 0.50g = 4.94g of material
Which substance is a base? a. HCOOH b. RbOH c. H2CO3 d. NaNO3
Answer:
rboh
Explanation:copied from previous
Among the following , RbOH is a base as it has hydroxyl ions which are generated on dissociation.
What is a base?
According to the Arrhenius concept, base is defined as a substance which yields hydroxyl ions on dissociation.These ions react with the hydrogen ions of acids to produce salt in an acid-base reaction.
Bases have a pH higher than seven as they yield hydroxyl ions on dissociation.They are soapy in touch and have a bitter taste.According to the Lowry-Bronsted concept, base is defined as a substance which accepts protons .Base react violently with acids to produce salts .Aqueous solutions of bases can be used to conduct electricity .They can also be used as indicators in acid-base titrations.
They are used in the manufacture of soaps,paper, bleaching powder.Calcium hydroxide ,a base is used to clean sulfur dioxide gas while magnesium hydroxide can be used as an antacid to cure acidity.
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What would have happened to your % Transmittance reading and to your calculations of Keq if the spectrophotometer had been set at 520 nm rather than 447nm
Answer:
On the off chance that the wavelength(λ) maximum worth has been changed to 520 nm from 470 nm on the spectrophotometer, less light would be absorbed and in this way %T would be higher than the one found at 470 nm. The wavelength utilized at 520 nm isn't adequate for the excitation and consequently lesser light is absorbed by the arrangement.
Explanation:
A spectrophotometer is an analytical equipment used to quantitatively gauge the transmission(passage) or impression of visible light, UV light or infrared light through a medium.
Transmittance (τ) is the ratio of the brilliant or luminous flux at a given wavelength that is transmitted to that of the incident radiation.
where, Keq is the equilibrium constant.
On the off chance that the wavelength(λ) has been changed to 520 nm from 470 nm on the spectrophotometer, less light would be absorbed and in this way %T would be higher than the one found at 470 nm.
What happens to Transmittance?A spectrophotometer is an analytical equipment used to quantitatively gauge the transmission(passage) or impression of visible light, UV light or infrared light through a medium. Transmittance (τ) is the ratio of the brilliant or luminous flux at a given wavelength that is transmitted to that of the incident radiation. The wavelength utilized at 520 nm isn't adequate for the excitation and consequently lesser light is absorbed by the arrangement. As the concentration goes up, more radiation is absorbed and the absorbance goes up. Therefore, the absorbance is directly proportional to the concentration.
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There are five constitutional isomers with the molecular formula C6H14. When treated with chlorine at 300°C, isomer A gives a mixture of two monochlorination products. Under the same conditions, isomer B gives a mixture of five monochlorination products, isomer C gives four monochlorination products, and isomer D gives a mixture of three monochlorination products. From this information, draw the structural formula of isomer D.
Answer:
Pentane or 2,2-dimethylbutane
Explanation:
I've numbered the isomeric hexanes from 1 to 5 and labelled the sets of equivalent hydrogens.
The results are
Isomer 1— three sets of equivalent hydrogens
Isomer 2— five sets of equivalent hydrogens
Isomer 3— four sets of equivalent hydrogens
Isomer 4— two sets of equivalent hydrogens
Isomer 5— three sets of equivalent hydrogens
Each set will give one monochloro substitution product.
4 = A. Two monochloro isomers.
2 = B. Five monochloro isomers.
3 = C. Four monochloro isomers.
Isomers 1 and 5 each give three monochloro isomers.
Thus, we cannot assign Structure D definitively.
D is either pentane or 2,2-dimethylbutane.
How many grams of LiNO3 must be added to 25.0 g of water to prepare a 5.00% (m/m) solution of LiNO3?
Answer:
1.25 g.
Explanation:
5% or 25 g
= 0.05 * 25
= 1.25 g (answer).
What is the rate constant of a reaction if rate = 1.5 (mol/L)/s, [A] is 1 M, [B] is
3 M, m = 2, and n = 1?
k=
rate
[A]" [B]"
A. 0.17
B. 13.5
C. 0.5
D. 4.5
[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Rate Constant.
Here, the "K" is the Rate Constant.
so the ANSWER IS C.) 0.5
The rate of constant is 0.5.
The answer is option C.
How do find the rate of constant?To determine the fee regulation from a desk, you have to mathematically calculate how differences in molar concentrations of reactants affect the response charge to parent out the order of every reactant. Then, plug in values of the response charge and reactant concentrations to discover the particular rate constant.
What's the rate of constant?The particular rate constant is the proportionality steady touching on the rate of the reaction to the concentrations of reactants. The fee regulation and the unique charge constant for any chemical response need to be decided experimentally. The price of the fee regular is temperature-established.
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MnBr2 molecular or ionic:
name:
molecular or ionic
Answer: Manganese(II)bromide
Explanation:
Because it is a chemical compound composed of manganese and bromine
What is T2, if T1= 500 k, v1=10L, V2=8L,P1=600 torr,P2=200 torr?
Answer:
T2 = 133.333°K
Explanation:
Using Combined Gas Laws:
(600 torr)(10L)/500°K = (200 torr)(8L)/x°K
[tex]\frac{600 torr(10L)}{500K} =\frac{200 torr(8L)}{xK}[/tex]
Cross multiply:
x°K (600 torr)(10L) = 500°K(200 torr)(8L)
Divide:
x°K = (500°K(200 torr)(8L))/(600 torr)(10L)
[tex]xK = \frac{500K(200 torr)(8L)}{600 torr(10L)}[/tex]
x = 400/3°K or 133.333°K
When a 21.5-g sample of LiCl was added to 195 g of water at a temperature of 20.00°C in a calorimeter, the temperature increased to 39.26°C. How much heat is involved in the dissolution of the LiCl?
Answer:
[tex]63.09KJ/mol[/tex]
Explanation:
In this case, to calculate the heat of solution (KJ/mol) we have to take into account the mass of water, the specific heat of the water and the temperature change, so:
[tex]m=195~g[/tex]
[tex]c=4.18~J/g^{\circ}C[/tex]
Δ[tex]T=39.26^{\circ}C[/tex]
With this in mind, we can use the equation:
[tex]Q=m*c*T[/tex]
If we plug the values into the equation we will have:
[tex]Q=195~g*4.18~J/g^{\circ}C*39.26^{\circ}C[/tex]
[tex]Q=32000.83~J[/tex]
Now, with the mass value (21.5 g) and the molar mass of LiCl (42.39g/mol) we can calculate the moles of LiCl:
[tex]21.5~g~LiCl\frac{1~mol~LiCl}{42.39~g~LiCl}=0.507~mol~LiCl[/tex]
Now, in the heat of solution, we have KJ/mol units. Therefore, we have to convert from J to KJ:
[tex]32000.83~J\frac{1~KJ}{1000~J}=32~KJ[/tex]
Finally, we can divide by the moles of LiCl:
[tex]\frac{32~KJ}{0.507~mol}=63.09KJ/mol[/tex]
So, for each mole of LiCl, we have 63.09 KJ involved in the dissolution process.
I hope it helps!
The two reactions above, show routes for conversion of an alkene into an oxirane. If the starting alkene is cis-3-hexene the configurations of the oxirane products, A and B are Product A: _______ Product B: _______ Will either of these two oxirane products rotate the plane of polarization of plane polarized light? _____
Answer:
Product A: cis; no
Product B: cis: no
Explanation:
Two common methods of forming oxiranes from alkenes are:
Reaction with peroxyacids Formation of a halohydrin followed by reaction with base
1. Reaction with peroxyacids
(a) Stereochemistry
The reaction with a peroxyacid is a syn addition, so the product has the same stereochemistry as the alkene.
The starting alkene is cis, so the product is cis-2,3-diethyloxirane.
(b) Configuration
The product is optically inactive because it has an internal plane of symmetry.
It will not rotate the plane of polarized light.
2. Halohydrin formation
(a) Stereochemistry
The halogenation of the alkene proceeds via a cyclic halonium ion.
The backside displacement of halide ion by alkoxide is also stereospecific, so a cis alkene gives a cis epoxide.
The product is cis-2,3-diethyloxirane.
(b) Configuration
The cyclic halonium ion has an internal plane of symmetry, as does the product (meso).
The oxirane will not rotate the plane of polarized light.
(a) How many stereoisomers are possible for 4-methyl-1,2-cyclohexanediol? ___ (b) Name the stereoisomers formed by oxidation of (S)-4-methylcyclohexene with osmium tetroxide. If there is only one stereoisomer formed, leave the second space blank. Isomer #1: Isomer #2: (c) Is the product formed in step (b) optically active? _____
Answer:
See explanation
Explanation:
For the first part of the question, we have to check the chiral carbons in 4-methyl-1,2-cyclohexanediol. In this case carbons, 1 and 2 are chiral, if we have 2 chiral carbons we will have 4 isomers. We have to remember that formula 2^n in which "n" is the number of chiral carbons, so:
2^n = 2^2 = 4 isomers
And the isomers that we can have are:
1) (1R,2S)-4-methylcyclohexane-1,2-diol
2) (1S,2S)-4-methylcyclohexane-1,2-diol
3) (1S,2S)-4-methylcyclohexane-1,2-diol
4) (1S,2R)-4-methylcyclohexane-1,2-diol
See figure 1
For the second part of the question, we have to remember that the oxidation with [tex]OsO_4[/tex] is a syn addition. In other words, the "OHs" are added in the same plane. In this case, we have the methyl group with a wedge bond, so the "OH" groups will have a dashed bond due to the steric hindrance. Due to this we only can have 1 isomer ((1S,2R,4S)-4-methylcyclohexane-1,2-diol). Finally, on this molecule, we dont have any symmetry planes (this characteristic will cancel out the optical activity), so the product of this reaction has optical activity.
See figure 2
I hope it helps!
Need help asap!!! Can someone please help me?
Answer:
The answer is option C.
voltage (V) = Current( I ) × Resistance (R)
V = IR
I = V/ R
V = 10V R = 20Ω
I = 10/20 = 1/2
I = 0.50A
Hope this helps
How many carbon atoms are represented by the model below
A) 0
B) 3
C) 2
D) 1
Answer:
Option (B) 3.
Explanation:
In the model represented above, the two extreme represent carbon atoms since no other group are attached to it. The joint at the middle also represent carbon atom.
Thus, we can write a more simplify illustration for the model above as
C—C—C
From the above illustration, we can see that the model contains 3 carbon atom.
Helium (He) is the lightest noble gas component of air, and xenon (Xe) is the heaviest. Perform the following calculations, using R = 8.314 J/(mol·K) and ℳ in kg/mol. (a) Find the rms speed of He in winter (0.°C) and in summer (30.°C). Enter your answers in scientific notation. × 10 m/s (winter) × 10 m/s (summer) (b) Find the ratio of the rms speed of He to that of Xe at 30.°C. (rate He)/(rate Xe) (c) Find the average kinetic energy per mole of He and of Xe at 30.°C. Enter your answers in scientific notation. × 10 J/mol for He × 10 J/mol for Xe (d) Find the average kinetic energy per molecule of He at 30.°C. Enter your answer in scientific notation. × 10 J/He atom
Answer:
Explanation:
Hello,
Data;
R = 8.314J/(mol.K)
Temp (winter) = 0°C = (0 + 273.15)K = 274.15K
Temp.(summer) = 30°C = (30 + 273.15)K = 303.15K
Molar mass of He = 4g/mol = 0.004kg/mol
Molar mass of Xe = 131.29g/mol = 0.131kg/mol
a) rms speed in winter and summer
Vrms = √(3RT/M)
R = gas constant
T = temperature of the gas
M = molar mass of the gas
In winter,
Vrms = √(3×8.314×273.15) / 0.004
Vrms = 1.30×10³m/s
In summer
Vrms = √(3×8.314×303.15) / 0.004
Vrms = 1.37×10³m/s
b) Vrms of Xe at 30°C
Vrms = √(3 × 8.314 × 303.15) / 0.131
Vrms of Xe = 240.24m/s = 2.40×10²m/s
Vrms of He = 1.37×10³m/s
Rate of He / rate of Xe = 1.37×10³ / 2.40×10²
Rate of He / rate of Xe = 5.7
c) K.E per mole
At 30°C
K.E of He = (3/2) × 8.314 × 303.15
K.E of He = 3.78×10³J/mol
K.E of Xe = (3/2) × 8.314 × 303.15
K.E of Xe = 3.78×10³J/mol
d) K.E per molecule = ½mv²
K.E of He molecule = ½ × 0.004 × 1.37×10³
K.E = 2.74J
The rms speed of He in winter and in summer are 1.30×10³m/s & 1.37×10³m/s respectively, ratio of the rms speed of He to that of Xe at 30 degree celsius is 5.7, average kinetic energy per mole of He and of Xe and average kinetic energy per molecule of He are discussed below.
How do we calculate the root mean square velocity?Root mean square velocity of the gases will be calculated as:
Vrms = √(3RT/M), where
R = universal gas constant = 8.314J/(mol.K)
M = molar mass of gas in kg/mol
T = temperature in K
Root mean square velocity of Helium at 0°C or 273.15K and 30°C or 303.15K will be calculated as:At 273.15K-
Vrms = √(3×8.314×273.15) / 0.004 = 1.30×10³m/s
At 303.15K-
Vrms = √(3×8.314×303.15) / 0.004 = 1.37×10³m/s
Vrms (winter×summer) = 1.30×10³m/s × 1.37×10³m/s = 1.78×10⁶m/s
Vrms of Xe at 303.15KVrms = √(3 × 8.314 × 303.15) / 0.131
Vrms of Xe = 240.24m/s = 2.40×10²m/s
Vrms of He = 1.37×10³m/s
Rate of He / rate of Xe = 1.37×10³ / 2.40×10²
Rate of He / rate of Xe = 5.7
Average kinetic energy will be calculated as:K.E = 3RT / 2N, where
N = avgadros number = 6.022×10²³ atoms/mole
Average kinetic energy per mole of He & Xe at 303.15K is same as:
K.E = 3/2(1.38 × 10⁻²³J/K)(303.15K) = 6.2 × 10⁻²⁵J/mol
Average kinetic energy per molecule of He will be calculated as:K.E of He molecule = ½ × 0.004 × 1.37×10³
K.E = 2.74J
Hence, calculations for the given points are described above.
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Una cierta cantidad de gas se encuentra a la presión de 790 mm Hg cuando la temperatura es de 25ºC. Calcula la presión que alcanzará si la temperatura sube hasta los 200ºC.
Respuesta:
1.25 × 10³ mmHg
Explicación:
Presión inicial (P₁): 790 mmHgTemperatura inicial (T₁): 25 °CPresión final (P₂): ?Temperatura final (T₂): 200 °CPaso 1: Convertir las temperaturas a Kelvin
Cuando trabajamos con gases ideales debemos convertir las temperaturas a Kelvin. Usaremos la siguiente ecuación:
K = °C + 273.15
K = 25°C + 273.15 = 298 K
K = 200°C + 273.15 = 473 K
Paso 2: Calcular la presión final
Usaremos la ley de Gay-Lussac.
[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}\\P_2 = \frac{P_1 \times T_2}{T_1} = \frac{790mmHg \times 473K}{298K} = 1.25 \times 10^{3} mmHg[/tex]
Aqueous calcium chloride reacts with aqueous potassium carbonate in a double-displacement reaction. Write a balanced equation to describe this reaction. Include states of matter in your answer. Click in the answer box to open the symbol palett
Answer: [tex]CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)[/tex]
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
The balanced reaction between aqueous calcium chloride reacts with aqueous potassium carbonate is shown as:
[tex]CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)[/tex]