Answer:
even I have the same dought
What is the pressure in millimeters of mercury of 0.0150 mol of helium gas with a volume of 213 mL at 50. C? (Hint: You must convert each quantity into the correct units (L, atm, mol and K) before substituting into the ideal gas law.)
Explanation:
0.08206 L atm mol-1K-1
pv=nRT
Px213 x10^-³ = 0.0150 x 0.08206 x 323
px213 x10^-³ = 0.398
p = 0.398/213 x10^-³
p = 1.87 x 10^-6atm
p = 0.0014mmHg
please brainliest
what are structure for each of the compounds listed. Explain why the name given is incorrect. And give a correct name in each case
(1) 1-methylbutane
(2) 1,1,3-trimethylhexane
(3) 5octyne
(4)2-ethyl-1-propanol
(5)2,2-dimethyl1-3-butanol
Answer:
Explanation:
(1) 1-methylbutane
(2) 1,1,3-trimethylhexane
(3) 5octyne
(4)2-ethyl-1-propanol
(5)2,2-dimethyl1-3-butanol
A buffer solution is all of the following EXCEPT: a solution that contains both a weak acid and its conjugate base. a solution that resists a change in pH when a base is added. a solution that regulates pH because it is such a strong acid or base. a solution that resists a change in pH when an acid is added. All of the above are true.
Answer:
A SOLUTION THAT REGULATES pH BECAUSE IT IS SUCH A STRONG ACID OR BASE
Explanation:
A buffer solution is an aqueous solution consisting of a weak acid and its conjugate base. It is an aqueous solution used to keep the pH of solution at a nearly constant value in various chemical processes. It resists change in pH when either a strong acid or a strong base is added. So it is very essential in various chemical applications and even in the human body as the blood pH is kept in nearly constant value by the bicarbonate buffer system in conjunction with the kidneys. The buffer solution is able to keep this nearly constant range of values because of the equilibrium between the weak acid and its conjugate base. So therefore, the incorrect statement in the options is that buffer solution is a solution that regulates pH because it is such a strong acid or base. The other options are correct.
Which correctly lists three characteristics of minerals?
solid, crystal structure, definite chemical composition
organic, crystal structure, definite chemical composition
human-made, solid, organic
crystal structure, definite chemical composition, human-made
Answer:a
Explanation:
The three characteristics of minerals are that they are solid, have definite crystal structure and definite chemical composition.
What are minerals?Minerals are defined as a chemical compound which has a well -defined composition and possesses a specific crystal structure.It occurs naturally in the pure form.
If a compound occurs naturally in different crystal structure then each structure is considered as a different mineral.The chemical composition of a mineral varies depending on the presence of small impurities which are present in small quantities.
Some minerals can have variable proportions of two or more chemical elements which occupy equivalent position in the crystal structure.It may also have variable composition which is split into separate species.
Physical properties of minerals include color,streak, luster,specific gravity and cleavage.
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How much heat will be absorbed by a 63.1 g piece of aluminum (specific heat = 0.930 J/g･°C) as it changes temperature from 23.0°C to 67.0°C?
Answer:
[tex]Q=2582J=2.58kJ[/tex]
Explanation:
Hello,
In this case, for us to compute the absorbed heat, we apply the following equation:
[tex]Q=m_{Al}Cp_{Al}(T_2-T_1)[/tex]
Whereas we use the mass, specific heat and temperature change for the piece of aluminium, thus, we obtain:
[tex]Q=63.1g*0.930\frac{J}{g*\°C}*(67.0-23.0)\°C\\ \\Q=2582J=2.58kJ[/tex]
It is positive as the heat is entering, therefore the temperature raises.
Best regards.
What is the electron configuration for N (nitrogen)?
A. 1521p5
B. 1s22s23s 3p
O cly 1s22s22p3
O D. 1s 2s22p 3s
SUBMIT
Answer:
[tex]1s^22s^22p^3[/tex]
Explanation:
Nitrogen has the atomic number = 7
So, No. of electrons = 7
Electronic Configuration:
[tex]1s^22s^22p^3[/tex]
Remember that:
s sub shell holds upto 2 electrons while p sub shell upto 6
Consider the equilibrium reaction. 4A+B↽−−⇀3C After multiplying the reaction by a factor of 2, what is the new equilibrium equation?
How many liters of a 1 M NaOH stock solution would you need to make 785 mL of a 215 mM NaOH dilution? (m.w. = 40.00 g/mol) Do not include units in your answer; report your answer in the requested units.
Answer:
0.1688L of the 1M NaOH stock solution
Explanation:
A 215mM = 0.215M solution of NaOH contains 0.215 moles per liter. As you want to prepare 785mL = 0.785L of the 0.215M you will need:
0.785L × ( 0.215mol / L) = 0.1688 moles of NaOH.
These moles of NaOH comes from the 1M stock solution, that means the volume of 1M NaOH solution you need is:
0.1688 moles NaOH × (1L / 1mol) =
0.1688L of the 1M NaOH stock solutionThe volume of the stock solution of 1 M NaOH stock solution needed to make 785 mL of a 215 mM NaOH dilution is 0.169 L
From the question given above, the following data were obtained:
Molarity of stock (M₁) = 1 M
Volume of diluted solution (V₂) = 785 mL = 785 / 1000 = 0.785 L
Molarity of diluted solution (M₂) = 215 mM = 215 / 1000 = 0.215 M
Volume of stock solution needed (V₁) =?
The volume of the stock solution needed can be obtained as follow:
M₁V₁ = M₂V₂1 × V₁ = 0.215 × 0.785
V₁ = 0.169 LTherefore, the volume of the stock solution needed is 0.169 L
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The nitration of aromatic compounds is a highly exothermic reaction that generally uses catalysts that tend to be corrosive (e.g., HNO3 /H2SO4 ). A less corrosive employs N2 O5 as the nitrating agent as illustrated below: If this reaction is conducted in an adiabatic CSTR, what is the reactor volume and space time necessary to achieve 35 percent conversion of N2O5? The reaction rate is first order in A and second order in B. Data: HRX = -370.1 kJ/mol CpA = 84.5 J/(mol-K) CpB = 137 J/(mol-K) Cpc = 170 J/(mol-K) CpD = 75 J/(mol-K) To = 303 K FAo = 10 mol/min FBo = 30 mol/min v = 1000 L/min CAo =0.01 mol/L
Answer:
-370.1 kJ/mol CpA = 84.5 J/(mol-K) CpB = 137 J/(mol-K) Cpc = 170 J/(mol-K) CpD = 75 J/(mol-K) To = 303 K FAo = 10 mol/min FBo = 30 mol/min v = 1000 L/min CAo =0.01 mol/L
Explanation:
The nitration of aromatic compounds is a highly exothetion that generally uses catalysts that tend to be corrosive (e.g., HNO3 /H2SO4 ). A less corrosive employs N2 O5 as the nitrating agent as illustrated below: If this reaction is conducted in an adiabatic CSTR, what is the reactor volume and space time necessary to achieve 35 percent conversion of N2O5? The reaction rate is first order in A and second order in B. Data: HRX = -370.1 kJ/mol CpA = 84.5 J/(mol-K) CpB = 137 J/(mol-K) Cpc = 170 J/(mol-K) CpD = 75 J/(mol-K) To = 303 K FAo = 10 mol/min FBo = 30 mol/min v = 1000 L/min CAo =0.01 mol/L
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rmic
why we used petrol for vehicles not water?
Answer:
Water isn't combustible. There is nothing you can add to it other than gasoline that will make it even remotely combustible. Now, through electrolysis, it can be broken down into hydrogen and oxygen, which could be burned in an internal combustion engine.
because the water in itself does not produce energy
warming oceans temperatures directly lead to all of the following except A sea level rising B coral bleaching C ocean deoxigenation D pollution
Answer:
D. pollution
Explanation:
Increase in ocean temperature is one of the major consequence of global warming which directly leads to rise in sea level, coral bleaching and ocean deoxygenation.
Warming ocean temperatures do not leads to pollution directly whereas pollution leads to warming ocean temperatures. So, in the case of pollution, the effect is opposite.
Hence, the correct option is D.
in an endothermic reaction the ____ have more energy than the ____?
Answer: products; reactants
Explanation: as the endothermic reactions are tye one which absorbs energy
Given Ba, Li, Na, Cs, and Be, arrange the group 1 and 2 elements in order of increasing reactivity to water (H2O).
1. Be∠Na∠ Li∠ Ba∠ Cs
2. Be∠ Li∠ Ba∠ Na∠ Cs
3. Cs∠ Na∠ Be∠ Li∠ Ba
4. Li∠ Cs∠ Na∠ Be∠ Ba
I think it's the third option, but the Beryllium and Barium are messing me up! And a brief explanation too. Thanks!
Answer:
Be∠ Li∠ Ba∠ Na∠ Cs
Explanation:
Beryllium does not react with water. It is the only alkaline earth metal that does not react with water because of its small size and high ionization energy. Beryllium differs considerably from other members of group two, its compounds when anhydrous show a considerable degree of covalent character.
As the atomic number of the group two elements increases, their ionization energies decreases and their electrode potentials become more negative hence their reactivity increases down the group. This implies that barium will have a very negative electrode potential comparable to that of the alkali metals, hence it reacts considerably with water.
The reactivity of alkali metals with water increases down the group. Lithium reacts quietly with water, sodium and potassium react with water with increasing vigour while rubidium and cesium react with water with exceptional violence.
This little explanation, is the reason behind the option chosen as the answer.
30. A. An organic compound - A (C4H80) forms phenyl
hydrazone with phenyl hydrazine and reduces Fehling's
mpt any two questions:
solution. It has negative iodoform test. Identify the
organic compound A.
Answer:
Methyl ethyl ketone
Explanation:
Compound 'A' forms phenyl hydrazone, so it must be a carbonyl compound. Since it also gives a negative iodoform test, so it can't be an aldehyde.
'A' on reduction gives propane. So, it must be butanone. Ketone reacts with phenyl hydrazine to form phenyl hydrazone but gives a negative iodoform test.
Thus, the correct answer is - Methyl ethyl ketone
half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution
Answer:
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
Explanation:
In reduction-oxidation reaction two reactions take place, one is oxidation and the other is reduction reaction. In an oxidation reaction, there is the loss of an electron whereas in the reduction reaction there is gain of electron occus.
Reduction reaction occurs on the cathode, in a reduction of water there is gain of 2 electrons to gaseous hydrogen in basic aqueous solution. half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution-
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
Which of the following best identifies where long-range order would be found?
ОООО
in amorphous solids
in crystalline solids
in thermal plasmas
in nonthermal plasmas
Answer:
in crystalline solids
Hope this answers your question, good luck
The crystalline solids represent the best identification where the long-range order should be found.
What are crystalline solids?Crystalline solids refer to the solid where the atoms, molecules should be make the arrangement. The smallest & repeated pattern of this solid should be called as the unit cell. The unit cell should be treated as the bricks in the wall which means it should be homogenous in the nature and repeated
Therefore, The crystalline solids represent the best identification where the long-range order should be found.
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. A compound X, containing C, H, and O was found to have a relative molar mass of 6o
whilst 20.0g of X contained 8.0g of C and 1.33g of H. Calculate the empirical formula
of compound X, and hence determine its molecule formula. show calculation
Answer:
empirical formula = CH2O
molecular formula = C2H4O2
Explain why phosphorus has a low melting point.
Answer:
Phosphorus has a low melting point because the intramolecular forces holding it together is London Dispersion Forces.
Explanation:
London Dispersion Forces (LDF) are the weakest intramolecular forces. You don't need to break the covalent bonds, but rather the Van Der Waals' Forces. If LDF are the weakest forces, then the melting point is low.
Answer:
because the intramolecular forces holding it together is London Dispersion Forces.
Explanation:
Suppose you have a solid that looks like gold but you believe it to be fool’s gold. The mass of the solid is 23.5 grams. When the solid is lowered into a graduated cylinder the water level rises from 47.5 to 52.2 mL. Is the simple fool’s gold
Answer:
The sample is fool's gold
Explanation:
Density is defined as the ratio between mass in grams and volume in mililiters.
A sample of pure gold has a density of 19.3g/mL.
Using Archimedes' principle, the volume of the sample is:
52.2mL - 47.5mL = 4.7mL
As the mass of the sample is 23.5g, the density is:
23.5g / 4.7mL = 5g/mL
The denisty of the sample is very different to density of pure gold, that means:
the sample is fool's gold
A piece of bismuth with a mass of 22.5 g is submerged in 46.3 cm3 of water in a graduated cylinder. The water level increases to 48.6 cm3. What is the density of indium to the proper number of significant figures
Answer:
[tex]\rho = 9.78\frac{g}{cm^3}[/tex]
Explanation:
Hello,
In this case, in order to compute the density of bismuth we need to apply the formula:
[tex]\rho =\frac{m_{Bi}}{V_{Bi}}[/tex]
Nonetheless, the volume is computed by the difference:
[tex]V_{Bi}=48.6-46.3=2.30cm^3[/tex]
Therefore:
[tex]\rho = \frac{22.5g}{2.30cm^300}\\\\\rho = 9.78\frac{g}{cm^3}[/tex]
Regards.
The following skeletal oxidation-reduction reaction occurs under basic conditions. Write the balanced OXIDATION half reaction.
Bi(OH)3 + NO2 → Bi + NO3-
Answer:
[tex]N^{4+}O_2+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+H_2O[/tex]
Explanation:
Hello,
In this case, for the given reaction, we first start by the writing of the oxidation states of all the involved elements:
[tex]Bi^{3+}(OH)^-+N^{4+}O^{2-}_2\rightarrow Bi^0+(N^{5+}O^{2-}_3)^-[/tex]
In such a way, we are noticing nitrogen is undergoing an increase in its oxidation state, therefore it is being the oxidized species, for which the oxidation half reaction, should be (considering basic conditions):
[tex]N^{4+}O_2+H_2O+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+2H^++2OH^-\\\\N^{4+}O_2+H_2O+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+2H_2O\\\\N^{4+}O_2+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+H_2O[/tex]
Best regards.
Arrange the following set of atoms in order of decreasing atomic size: Sn, I, Sr
Which atom has the largest atomic size?
a) Sn
b) I
c) Sr
Which atom has the smallest atomic size?
a) Sn
b) I
c) Sr
Need help with chemistry question
Answer:
See explanation
Explanation:
In this case, we have to check two variables:
1) The leaving group
2) The carbon bonded to the leaving group.
Let's check one by one:
2-chloro-3-methylbutane
In this molecule, the leaving group is "Cl", the carbon bonded to the leaving group has two neighbors. Therefore, we have a secondary substrate.
1-phenylpropan-1-ol
In this molecule, the leaving group is "OH", the carbon bonded to the hydroxyl group has two neighbors also. So, we have a secondary substrate.
(E)-pent-3-en-2-yl 4-methylbenzenesulfonate
In this case, the leaving group is "OTs" (Tosylate), the carbon bonded to the tosylate group has as a neighbor a double bond. Therefore, we have an allylic substrate.
3a-bromooctahydro-1H-indene
In this molecule, the leaving group is "Br", the carbon bonded to the bromine has three neighbors. So, we have a tertiary substrate.
1-iodo-3-methylbutane
In this molecule, the leaving group is "I", the carbon bonded to the iodide has only one neighbor. So, we have a primary substrate.
See figure 1
I hope it helps!
How many moles of gaseous boron trifluoride, BF3, are contained in a 4.3410 L bulb at 788.0 K if the pressure is 1.220 atm What is the complete ground state electron configuration for the neon atom
Answer:
n= 0.08186
{He}2s^2 2p^6
Explanation:
PV=nRT
n=PV/RT
n= (1.220 atm)(4.3410 L) / (0.0821 atm*L/mol*K)(788.0 K)
n=0.08186
As for the electron configuration:
Ne:
{He} 2s^2 2p^6
or long hang:
1s^2 2s^2 2p^6
[tex][Ne]=1s^22s^22p^{10}[/tex]
Given:
A gaseous boron trifluoride in a 4.3410 L bulb at 788.0 K, if the pressure is 1.220 atm.Neon atom.To find:
The moles of gaseous boron trifluoride in a container.The electronic configuration of neon in the ground state.1.
The pressure of the gaseous boron trifluoride = P = 1.220 atm
The volume of the gas in bulb = V = 4.3410 L
The moles of the gaseous boron trifluoride = n
The temperature of gaseous boron trifluoride = T = 788.0 K
Using an ideal gas equation:
[tex]PV = nRT\\\\1.220 atm\times 4.3410 L=n\times 0.0821 atm L/mol K\times 788.0 K\\\\n=\frac{1.220 atm\times 4.3410 L}{0.0821 atm L/mol K\times 788.0 K}\\\\n=0.08186 mol[/tex]
The moles of gaseous boron trifluoride is 0.08186 moles.
2.
The atomic number of neon atom = 10
The electronic configuration in the ground state is the most stable arrangement of the electrons in the lowest energy levels.
The ground state electronic configuration of neon is:
[tex][Ne]=1s^22s^22p^{10}[/tex]
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A compound has a mass percentage of 53.46% C, 6.98% H, and 39.56% O. What is the empirical formula for this compound
Answer:
The empirical formula is: C₂H₃O
Explanation:
The empirical formula, also known as the “minimum formula”, is the simplest expression to represent a chemical compound and indicates the elements that are present and the minimum integer ratio between its atoms.
The percentage composition is the percentage by mass of each of the elements present in a compound.
Having 100 g of the compound as a base, it is possible to express the percentages in grams. That is, assuming you have 100 g of the compound, you have 53.46 g of C , 6.98 g of H , and 39.56 g of O.
Taking into account the molecular mass of each substance, the number of relative atoms of each chemical element is calculated:
C: [tex]53.46 g *\frac{1 mol}{12.01 g } = 4.45 moles[/tex]
H:[tex]6.98 g *\frac{1 mol}{1.01 g } = 6.91 moles[/tex]
O:[tex]39.56 g *\frac{1 mol}{16g } = 2.47 moles[/tex]
Now you divide each value obtained by the least of them:
C: [tex]\frac{4.45 moles}{2.47 moles}= 1.80[/tex]
H:[tex]\frac{6.91 moles}{2.47 moles}= 2.8[/tex]
O:[tex]\frac{2.47 moles}{2.47 moles}=1[/tex]
Decimals approach the nearest integer, then:
C: 2
H: 3
O: 1
Therefore the empirical formula is: C₂H₃O
Answer: The other guy is wrong. The correct answer is C9H14O5
Explanation:
et the mass percentages from the percent composition represent grams in a total mass of 100g. Use the grams to calculate the number of moles of each atom present.
%C:53.46%molC=53.46gC=53.46gC×1molC12.011gC=4.451molC
%H:6.98%molH=6.98gH=6.98gH×1molH1.008gH=6.924molH
%O:39.56%molO=39.56gO=39.56gO×1molO15.999gO=2.473molO
Divide by the smallest number of moles.
subscriptC=4.451molC2.473molO≈1.800≈95
subscriptH=6.924molH2.473molO≈2.800≈145
subscriptO=2.473molO2.473molO=1
Now, multiply each subscript by 5 to achieve whole number subscripts. Therefore, the empirical formula is C9H14O5.
Perform the following
mathematical operation, and
report the answer to the correct
number of significant figures.
5.446 x 0.14156
Define dew point in complete sentences
g A laboratory analysis of an unknown compound found the following composition: C 75.68% ; H 8.80% ; O 15.52%. What is the empirical formula of the compound?
Answer:
THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O
Explanation:
The empirical formula for the unknown compound can be obtained by following the processes below:
1 . Write out the percentage composition of the individual elements in the compound
C = 75.68 %
H = 8.80 %
O = 15.52 %
2. Divide the percentage composition by the atomic masses of the elements
C = 75 .68 / 12 = 6.3066
H = 8.80 / 1 = 8.8000
O = 15.52 / 16 = 0.9700
3. Divide the individual results by the lowest values
C = 6.3066 / 0.9700 = 6.5016
H = 8.8000 / 0.9700 = 9.0722
O = 0.9700 / 0.9700 = 1
4. Round up the values to the whole number
C = 7
H = 9
O = 1
5 Write out the empirical formula for the compound
C7H90
In conclusion, the empirical formula for the unknown compound is therefore C7H9O
Steam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a tank with of methane gas and of water vapor, and when the mixture has come to equilibrium measures the amount of carbon monoxide gas to be .Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.
The given question is incomplete, the complete question is:
Calculating an equilibrium constant from a partial equilibrium... Steam reforming of methane (CH) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 25.0L tank with 8.0 mol of methane gas and 1.9 mol of water vapor, and when the mixture has come to equilibrium measures the amount of carbon monoxide gas to be 1.5 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to 2 significant digits.
Answer:
The correct answer is 2.47.
Explanation:
Based on the given information, the equation for the synthesis gas is,
CH₄ (g) + H₂O (g) ⇔ CO (g) + 3H₂ (g)
Based on the given information, 25.0 L is the volume of the tank, the concentration of CH₄ is 8.0 mol, the concentration of water vapor is 1.9 mol, and the concentration of CO gas is 1.5 mol.
Therefore, 25 L of the solution comprise 8.0 mole of CH₄. So, 1 L of the solution will comprise 8.0 / 25 mole CH₄,
= 0.32 mole of CH₄
Thus, the concentration of CH₄ or [CH₄] will be 0.32 mole/L or 0.32 M.
Similarly, the concentration of H₂O or [H₂O] will be 1.9/25 = 0.076 M
and [CO] is 1.5/25 = 0.06 M
The concentration equilibrium constant for the steam will be,
Kc = [CO] pH₂ / [CH₄] [H₂O] (Here pH₂ is the partial pressure of H₂)
Now lets us assume that the reaction has taken place in a constant atmospheric pressure, therefore, pH₂ will be equal to 1.
= 0.06 M/0.32 M × 0.076 M
= 2.47
A sample of gold weighs 1.2 oz. The sample is pounded into a thin rectangular sheet with an area of 400. sq. ft. The density of gold is 19.3 grams per cm3. What is the thickness of the foil in centimeters?Given: 28.35 g = 1 oz, 1 ft = 12 inches, 1 inch = 2.54 cm
Answer:
thickness of the gold sheet = 4.74 * 10⁻⁶ cm
Explanation:
mass of gold sample = 1.2 oz,; area of rectangular gold sheet = 400 sq. ft
Converting mass of gold from oz. to g
1 oz. = 28.35 g
mass of gold sample in grams = 1.2 * 28.35 g = 34.02 g
Converting from feet to cm;
1 feet = 12 * 2.54 cm = 30.48 cm
1 sq. ft = (30.48)² = 929.0304 cm²
area of gold in cm² = 400 * 929.0304 cm² = 371612.16 cm²
Since the density of a solid is constant
Density = mass/volume
Volume = mass/density
where volume =area * thickness
therefore, area * thickness =mass/density
thickness = mass/(density * area)
substituting the value; thickness = 34.02 g/(19.3 gcm⁻³ *371612.16 cm²)
thickness of the gold sheet = 4.74 * 10⁻⁶ cm