Answer:
FalseExplanation:
Target organ toxins are chemicals that can cause adverse effects or disease states manifested in specific organs of the body. Toxins do not affect all organs in the body to the same extent due to their different cell structures.
Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol
Answer:
Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol
Explanation:
According to IUPAC rules, the name of a compound is:
Prefix+root word+suffix
1) Select the longest carbon chain and it gives the root word.
2) The substituents give the prefix.
3) The functional group gives the secondary suffix and the type of carbon chain gives the primary suffix.
The structure of the given compounds are shown below:
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Which statement describes an electron?
EEEE
It has a positive charge and is located in the nucleus.
O It has a positive charge and is located in orbitals around the nucleus.
It has a negative charge and is located in the nucleus.
O It has a negative charge and is located in orbitals around the nucleus.
Answer:
It has a negative charge and is located in orbitals around the nucleus
Explanation:
The statement describes an electron is " It has a negative charge and is located in orbitals around the nucleus."
What is electron?The electron would be a subatomic particle with a negatively one elementary charge electric charge.
What is nucleus?Protons, that have a positive charge, as well as neutrons, which have no electrical charge, make up the nucleus. Quarks were subatomic particles that make up protons but also neutrons.
Electrons were present surrounding the atom's nucleus, in contrast to protons as well as neutrons, that are contained within the nucleus at its core. Negative electrons were drawn to the positive nucleus since the electric charges of opposite polarity attract one another.
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Identify the options below that are results of adding a catalyst to a chemical system.
The reaction rates are increased.
The reaction quotient is unaffected.
The reaction quotient decreases.
The equilibrium constant is unaffected.
Answer:
The correct options are a, b and d
Explanation:
A catalyst is a substance that increases the rate of a chemical reaction by reducing the activation energy. Le Catelier's principle explains how a substance or an "action" can affect a reaction in equilibrium.
The principle states that when a change is made to the conditions of a reacting system at equilibrium, the position of the equilibrium moves to counteract the change made. These changes are change in temperature, pressure, volume and/or concentration. These changes will either cause the equilibrium to shift forward or backward.
However, the presence of a catalyst DOES NOT affect a chemical equilibrium/equilibrium constant nor does it affect the reaction quotient because the same amount of reactants and products are available just as in uncatalyzed reaction except that the reaction proceeds faster (which does not affect equilibrium).
The rate of reaction is given as the time required by the reactant to convert into the product. The addition of catalyst increases the rate of reaction, while the reaction quotient and the equilibrium remain unaffected.
What is a catalyst?A catalyst is a chemical or compound that adds to the reaction and lowers the activation energy by providing an alternative path to the reaction.
The catalyst takes part in the reaction but did not consume in the chemical reaction.
The equilibrium and the reaction quotient are dependent on the conversion of the reactant to the product. The catalyst is not used in the reaction and thus did not affect the reaction quotient or the equilibrium.
Hence, options A, B, and D are correct for the use of catalysts in the chemical reaction.
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Which one of the following molecule is planer?
a. NF3 c. PH3
b. BH3 d. NCl3
Answer:
option a
hope helps you
have a great day
What are the uses of Sulphuric acid?
Answer:
The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.
The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.
Pls pls help me me pls
Answer:
Danger
Explanation:
Rank each of the following gases in order of increasing urms assuming equivalent amounts and all gases are at the same temperature and pressure where 1 has the lowest urms and 4 has the highest urms.
a. Gas 1 : H2S
b. Gas: He
c. Gas 3: NF3
d. Gas 4: H2O
The Urms refers to the root mean square speed of the gas. The order of increasing Urms for the gases shown in the question; NF3 < H2S < H2O < He.
What is the Urms?The Urms refers to the root mean square speed of the gas. This is ultimately dependent on the relative molecular mass of the gases when they are maintained at the same temperature.
Now, let us look at the order of increasing Urms for the gases shown in the question; NF3 < H2S < H2O < He.
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what is the machine used to check melting point called?
Answer:
Melting-point apparatus
Which equation obeys the law of conservation of
mass?
Answer:2C4H10+2C12+12O2 4CO2+CC14+H20
In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced if 5.00 g of vanillin are added to 1 L of water at 25 C?
Answer:
The full amount (5.00 g) will be dissolved in 1 L of water at 25°C.
Explanation:
The molecular weight (MW) of Vanillin (C₈H₈O₃) is calculated from the chemical formula as follows:
MW(C₈H₈O₃) = (12 g/mol x 8) + (1 g/mol x 8) + (16 g/mol x 3) = 152 g/mol
If 0.070 mol of C₈H₈O₃ are soluble per liter of water at 25°C, the maximum mass that can be dissolved in 1 L is:
0.070 mol x 152 g/mol = 10.64 g
Since 5.00 g is lesser than the maximum amount that can be dissolved (10.64 g), the added amount will be completely dissolved in 1 L of water at 25°C.
You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all 0.100 M, are available to you: HCOOH, CH3COOH, H3PO4 , NaCHOO, NaCH3COO, and NaH2PO4. What would be the best combination to make the required buffer solution? Select one:
a. NaH2PO4 and NaCHOO
b. H3PO4 and NaH2PO4
c. NaH2PO4 and HCOOH
d. CH3COOH and NaCH3COO e. HCOOH and NaCHOO
can someone helo me with this
Answer:
e. HCOOH and NaCHOO
Explanation:
For a buffer solution, both an acid and its conjugate base are required.
With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.
Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:
H₃PO₄ pKa = 2.12CH₃COOH pKa = 2.8HCOOH pKa = 3.74The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO
En la fermentación del alcohol, la levadura convierte la glucosa en etanol y dióxido de carbono:
C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
Si reaccionan 5.97 g de glucosa y se recolectan 1.44 L de CO2 gaseoso, a 293 K y 0.984 atm, ¿cuál
es el rendimiento porcentual de la reacción
Answer:
88.9%
Explanation:
Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:
5.97 g ÷ 180 g/mol = 0.0332 molDespués calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:
0.0332 mol C₆H₁₂O₆ * [tex]\frac{2molCO_2}{1molC_6H_{12}O_6}[/tex] = 0.0664 mol CO₂Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:
0.984 atm * 1.44 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 Kn = 0.0590 molFinalmente calculamos el rendimiento porcentual:
0.0590 mol / 0.0664 mol * 100% = 88.9%the force of attraction between non polar molecules are what (a)electrovalent bond (b)covalent bond (c)Hydrogen bond (d)Van der waals forces
Answer:
d. van der waals force
Explanation:
Van der Waals force :
the weakest intermolecular forceand consist of dipole-dipole force and dispersion force.
If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?
Answer:
0.74 M
Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 5.90 M
Volume of stock solution (V₁) = 0.250 L
Volume of diluted solution (V₂) = 2 L
Molarity of diluted solution (M₂) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5.90 × 0.250 = M₂ × 2
1.475 = M₂ × 2
Divide both side by 2
M₂ = 1.475 / 2
M₂ = 0.74 M
Thus, the molarity of the diluted solution is 0.74 M
4.005 X 74 X 0.007 = 2.10049
Answer:
2.07459
Explanation:
this is the correct answer.
If a hydrogen atom and a helium atom have the same kinetic energy:________
a. the wavelength of the hydrogen atom will be about 4 times longer than the wavelength of the helium atom.
b. the wavelength of the hydrogen atom will be about 2 times longer than the wavelength of the helium.
c. the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.
d. the wavelength of the helium atom will be about 2 times longer than the wavelength of the hydrogen atom.
e. the wavelength of the helium atom will be about 4 times longer than the wavelength of the hydrogen atom.
Answer: If a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.
Explanation:
The relation between energy and wavelength is as follows.
[tex]E = \frac{hc}{\lambda}\\[/tex]
This means that energy is inversely proportional to wavelength.
As it is given that energy of a hydrogen atom and a helium atom is same.
Let us assume that [tex]E_{hydrogen} = E_{helium} = E'[/tex]. Hence, relation between their wavelengths will be calculated as follows.
[tex]E_{hydrogen} = \frac{hc}{\lambda_{hydrogen}}[/tex] ... (1)
[tex]E_{helium} = \frac{hc}{\lambda_{helium}}[/tex] ... (2)
Equating the equations (1) and (2) as follows.
[tex]E_{hydrogen} = E_{helium} = E'\\\frac{hc}{\lambda_{hydrogen}} = \frac{hc}{\lambda_{helium}} = E'\\\lambda_{helium} = \lambda_{hydrogen} = E'[/tex]
Thus, we can conclude that if a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.
Di- n- pentyl ether can be converted to 1- bromopentane by treatment with HBr through essentially a(n) ________ mechanism.
Answer:
SN1 mechanism
Explanation:
The mechanism of this reaction is shown in the image attached.
The Di- n- pentyl ether is first protonated. The CH3(CH2)4OH is now a good leaving group as shown.
The attack of the bromide ion on the cation formed completes the mechanism to yield 1- bromopentane as shown in the mechanism.
Given the following list of densities, which materials would float in a molten vat of lead provided that they do not themselves melt? Densities (g/mL): lead = 11.4, glass = 2.6, gold = 19.3, charcoal = 0.57, platinum = 21.4.
a. gold and platinum
b. glass and charcoal
c. gold, platinum, glass and coal
d. gold and charcoal
e. None of these
Answer:
b. glass and charcoal
Explanation:
Step 1: Given data
Density of Pb: 11.4 g/mLDensity of Glass: 2.6 g/mLDensity of Au: 19.3 g/mLDensity of charcoal: 0.57 g/mLDensity of platinum: 21.4 g/mLStep 2: Determine which material will float in molten lead
Density is an intrinsic property of matter. Less dense materials float in more dense materials. The materials whose density is lower than that of lead and will therefore float on it are glass and charcoal.
write any two things that should be remembered while writing chemical equation
Answer:
the product and the reactant must be balanced
if u are required to give the mechanism if the reaction it must be written
If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?
Answer:
1.88 A
Explanation:
Let's consider the reduction of copper in an electrolytic cell.
Cu²⁺ + 2 e⁻ ⇒ Cu
We can calculate the charge used to deposit 12.3 g of Cu using the following relations.
The molar mass of Cu is 63.55 g/mol.1 mole of Cu is deposited when 2 moles of electrons circulate.1 mole of electrons has a charge of 96486 C (Faraday's constant).The charge used is:
[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]
We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.
5.50 h × 3600 s/1 h = 1.98 × 10⁴ s
The current used is:
I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A
If I have 25g of Sodium, how much Sodium Chloride will I theoretically create?
O 50g NaCl
0 58.3g NaCl
O 63.7g Naci
0 35.4g NaCl
Answer:
64 g
Explanation:
Step 1: Write the balanced equation
2 Na + Cl₂ ⇒ 2 NaCl
Step 2: Calculate the moles corresponding to 25 g of Na
The molar mass of Na is 22.98 g/mol.
25 g × 1 mol/22.98 g = 1.1 mol
Step 3: Calculate the moles of NaCl formed from 1.1 moles of Na
The molar ratio of Na to NaCl is 2:2. The moles of NaCl formed are 2/2 × 1.1 mol = 1.1 mol.
Step 4: Calculate the mass corresponding to 1.1 moles of NaCl
The molar mass of NaCl is 58.44 g/mol.
1.1 mol × 58.44 g/mol = 64 g
A scientific hypothesis is
ANSWER:
predictive.
testable.
explanatory.
all of the above.
Answer:
All of the above.
Explanation:
For a scientific hypothesis to be considered a hypothesis, it has to be testable. When conducting a lab experiment, it also allows the tester to predict what might occur during and after the experimentation. They are also explanatory. For example, theories are hypotheses that have been verified and can explain why something in nature takes place.
To what volume (in mL) would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN?
Answer:
To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN
Explanation:
Dilution is the reduction of the concentration of a chemical in a solution and consists simply of adding more solvent.
In a dilution the amount of solute does not vary. But as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.
In a solution it is fulfilled:
Ci* Vi = Cf* Vf
where:
Ci: initial concentration Vi: initial volume Cf: final concentration Vf: final volumeIn this case:
Ci= 1.40 MVi= 20 mLCf= 0.088 MVf= ?Replacing:
1.40 M* 20 mL= 0.088 M* Vf
Solving:
[tex]Vf=\frac{1.40 M* 20 mL}{0.088 M}[/tex]
Vf= 318.18 mL
To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN
By how many times would you expect Al2(SO4)3 to depress the F.P of water compared to sucrose C12H22011 ?
Answer:
By how many times would you expect Al2(SO4)3 to depress the F.P of water compared to sucrose C12H22011 ?.
Explanation:
The freezing point of a pure solvent decreases further by adding a nonvolatile solute.
This is called depression in freezing point.
When an ionic solute is dissolved then the depression in the freezing point is proportional to the number of ions present in the solution.
In aluminum sulfate, there are five ions formed as shown below:
[tex]Al_2(SO_4)_3(aq)->2Al^3^+(aq)+3SO_4^2^-(aq)[/tex]
But sucrose is a covalent compound and it does not undergo dissociation.
Hence, aluminum sulfate decreases the freezing point of water by five times compared to sucrose.
Explanation:
g Calculate the number of grams of aluminum that is produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0A.
Answer:
3.36 grams Al°(s)
Explanation:
Given AlCl₃(s), determine the mass (grams) of Al°(s) produced from electrolysis of Aluminum Chloride at 10.0 amps for 1.00 hour.
AlCl₃(s) + 378.3°F (=192.4°C) => Al⁺³(l) + 3Cl⁻(l)
formula wt. Al° = 27g/mol
Faraday Constant (F°) = 96,500 amp·sec
? grams Al°(s) = 10.0amps x (1 mole e⁻/96,500amp-sec) x (1 mole Al°(s)/3 mole e⁻) x (27g Al°(s)/1 mole Al°(s)) x 3,600 sec = 3.36 grams Al°(s)
The 3.36 grams of aluminum are produced in 1 hour by the electrolysis of molten AlCl₃ when 10A current is passed.
What is electrolysis?Electrolysis is a process that uses an electrical current to break chemical compounds. The electric current is passed through the substance to bring the chemical change by gain or loss of electrons.
The electrolysis of the aluminum chloride in the molten state is represented as:
AlCl₃ → Al³⁺ + 3Cl⁻
At cathode: Al³⁺ + 3e⁻ → Al (s)
Given, the current. I = 10 A and t = 1 hr = 3600 s
We know that the current is calculated from the equation: I = q/t
q = I× t
q = (10A) × (3600s)
q = 36 × 10³ C
We know, 96500 C of the charge has electrons = 1 mol
36 × 10³ C of the charge has electrons = 0.373 mol
3 moles of electrons required to produce aluminum = 1 mol
0.373 mol of electrons will produce aluminum = 0.373/3 = 0.124 mol
We know that, the mass of one mole of Al = 27g
The mass of 0.124 mol of Al = 27 × 0.124 = 3.36 g
Therefore, the aluminum produced in 1 hour by the electrolysis of molten AlCl₃ is equal to 3.36 grams.
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How does temperature, agitation, and particle size affect solubility?
Answer:
At higher temperatures, particles move faster and collide more, increasing solubility rates.
Agitation increases solubility rates as well, by bringing fresh solvent into contact with the undissolved solute
The smaller the particle size, the higher (faster) solubility rate. Vice versa, the bigger the particle size, the lower (slower) solubility rate.
Explanation:
Identify the possible quantitative analysis you can do using only the 28.02 g/mol as a unit factor. Select one or more:
Answer:
Calculate the moles of N2 molecules in 3.94 grams of nitrogen.
Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules.
Explanation:
Calculate the moles of N2 molecules in 4.73 liters of nitrogen gas. FALSE. You can't make this conversion using only the conversion factor with units of g/mol. To convert liters to moles are necessaries pressure, temperature and volume of the gas to use PV = nRT
Calculate the grams of N2 in 10.58 liters of nitrogen gas. FALSE. As explained, you need, P,V and T to find the moles of the gas. With the moles you can find the mass using the conversion factor of 28.02g/mol
Calculate the moles of N2 molecules in 3.94 grams of nitrogen. TRUE. You can find the moles of N2 as follows:
3.94g N2 * (1mol/28.02g) = 0.14 moles of N2 molecules
Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules. TRUE. The mass in 5.03x10²⁰ moles of nitrogen molecules is:
5.03x10²⁰ moles * (28.02g/mol) = 1.4x10²²g of nitrogen.
Que es la actividad física y en qué mejora
The metal tantalum becomes superconducting at temperatures below 4.483 K. Calculate the temperature at which tantalum becomes superconducting in degrees Celsius.
Answer:
The correct answer is "-268.667°C".
Explanation:
Given:
Temperature,
= 4.483 K (below)
Now,
The formula of temperature conversion will be:
⇒ [tex]T(^{\circ} C)=T(K)-273.15[/tex]
By putting the values, we get
⇒ [tex]=4.483-273.15[/tex]
⇒ [tex]=-268.667^{\circ} C[/tex]
Thus the above is the correct answer.
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was , calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Answer:
molar heat of combustion = -5156 *10³ kJ/mol
Explanation:
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Step 1: Data given
Mass of naphthalene = 1.435 grams
Initial temperature of water = 20.28 °C
Final temperature of water = 25.95 °C
heat capacity of the bomb plus water was 10.17 kJ/°C
Molar mass naphtalene = 128.2 g/mol
Step 2:
Qcal = Ccal * ΔT
⇒with Qcal =the heat of combustion
⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C
⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C
Qcal = 10.17 kJ/°C * 5.67 °C
Qcal = 57.7 kJ
Step 3: Calculate moles
Moles naphthalene = 1.435 grams / 128.2 g/mol
Moles naphthalene = 0.01119 moles
Step 4: Calculate the molar heat of combustion
molar heat of combustion = Qcal/ moles
molar heat of combustion = -57.7 kJ/ 0.01119 moles
molar heat of combustion = -5156 *10³ kJ/mol