If the shattered glassware is not noticed, there could be potential injury to the janitorial crew. So, clean broken glassware, melting point capillaries, pipets or glass test tubes should always be discarded in the broken glass box. So, option C is correct.
This lab safety advice aims to prevent and/or reduce accidents from broken glass when handling lab glassware.
It is acceptable to collect broken glass that is not contaminated with chemicals, infectious elements, or other dangerous items in certified broken glass bins or 5-gallon buckets with covers.
To reduce the chance of getting hurt, always pick up shattered glass with a brush and dustpan, tongs, or forceps and put it into a receptacle meant for throwing away broken glass. Broken glass must never be handled with care and must never be disposed of in regular trash. There are numerous vendors who sell glass disposal boxes. These containers include a plastic liner for disposing of wet glasses and a lid with a built-in cardboard flap to close the container when it is full.
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The complete question is:
Choose the one best response for why you discard clean broken glassware, melting point capillaries, pipets or glass test tubes in the broken glass box and not in the regular trash can.
a) Glassware could still have chemical residues that need to be properly disposed of.
b) If shattered glass is thrown away with regular trash, the stockroom cannot charge for it.
c) If the shattered glassware is not noticed, there could be potential injury to the janitorial crew.
d) Glasses could still contain chemical traces that need to be properly disposed of.
define each of the following as atomic element, molecular element, ionic compound, or molecular compound. h2s [ select ] xe [ select ] br2 [ select ] cacl2 [ select ] i2 [ select ] al2s3 [ select ] pb [ select ] no2
Xe and Pb are atomic elements. I₂ and Br₂ are molecular elements. CaCl₂ and Al₂S₃ are ionic compounds and NO₂ and H₂S are molecular compounds.
Atomic elements are the elements which cannot be broken down further and combine together to form molecules. Among the substances given to us, Xenon (Xe) and Lead (Pb) can be categorised as atomic elements. Molecular elements are formed when the same elements combine with each other. I₂ and Br₂ are molecular elements.
Ionic compounds are consist of ions bound together. CaCl₂ and Al₂S₃ are ionic compounds. Molecular compounds are formed up of two or more elements and NO₂ and H₂S are molecular compounds.
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you want to determine the heat of reaction when a sodium chloride solution is formed using a 10.0 g sample of nacl(s) and 50.0 ml of water in a coffee cup calorimeter. which of the following equations would you use? assume the specific heat of the solution is the same as the specific heat of water.
The heat of reaction when a sodium chloride solution is formed using a 10.0 g sample of NaCl(s) and 50.0 ml of water in a coffee cup calorimeter. The equation is :
E. q = ((60.0 g)(4.18 J/g K)(T f - T i))/(10.0 g/(58.44 g/mol))
Given that :
The mass of the NaCl = 10 g
The volume of sample = 50 mL
The molar heat of the reaction for NaCl ( sodium chloride ) = + 3.9 kJ/mol
The equation for the heat is given as :
q = m c ΔT
Where,
q = heat energy
m = mass
c = specific heat capacity
ΔT = change in temperature
Therefore the equation is given as :
q = m c ΔT
q = ( 50 g + 10 g) (4.184 J/g °C) (T f - T i)
q / mol = q water / mol NaCl
q = ((60.0 g)(4.18 J/g K)(T f - T i)) / (10.0 g/(58.44 g/mol))
The given question is incomplete, the complete question is :
You want to determine the molar heat of solution of sodium chloride using a 10.0 g sample of NaCl(aq) and 50.0 mL of water in a coffee cup calorimeter. Which of the following equations would you use? Assume the specific heat of the solution is the same as the specific heat of water.
A. q = (50.0 g)(4.18 J/g K)(T i - T f)
B. q = ((10.0 g)(4.18 J/g K)(T f - T i))/(10.0 g/(58.44 g/mol))
C. q = (60.0 g)(4.18 J/g K)(T f - T i)
D. q = (60.0 g)(4.18 J/g K)(T i - T f)
E. q = ((60.0 g)(4.18 J/g K)(T f - T i))/(10.0 g/(58.44 g/mol))
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Determine the empirical formula of each of the following compounds from the percent composition: a. 7.8% carbon and 92.2% chlorine b. 10.0% C c. 0.80% H, 89.1% Cl
The empirical formula of a compound with 7.8% carbon and 92.2% chlorine is CCl₄ and that of a compound with 10.0% C, 0.80% H and 89.1% Cl is CHCl₃.
In order to determine the empirical formula of the compounds, we have to divide the percentage by mass of each of the elements by their relative atomic masses and then divide them through the lowest ratio.
Carbon = 7.8%
7.8/12 = 0.65
Chlorine = 92.2%
92.2 / 35.5 = 2.59
Dividing by the lowest ratio,
Carbon = 0.65/0.65 = 1
Chlorine = 2.59/0.65 = 3.98 ≅ 4
So the empirical formula is CCl₄.
Hydrogen = 0.80%
0.80/1 = 0.80
Chlorine = 89.1%
89.1/35.5 = 2.50
Carbon = 10.0%
10/12 = 0.83
Dividing by the lowest ratio,
Hydrogen = 0.80/0.80 = 1
Carbon = 0.83/0.80 = 1.03 ≅ 1
Chlorine = 2.50/0.80 = 3.1 ≅ 3
So the empirical formula is CHCl₃
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the enthalpy of the overall reaction represented above can be determined by adding the enthalpies of reactions 1 and 2. which of the following could be reaction 2 ?
Most probably the reaction 2 could be A: "( 2NO(g)+O2(g)→2NO2(g) )".
Reaction 2 is the second step of the overall reaction, which is the formation of NO2 from NO and O2. The overall reaction is (2O3(g)+2NO(g)→2NO2(g) + 2O2(g) ).
From the given options, the best match for reaction 2 is (NO(g) + O2(g) → 2NO2(g) ) because it is a balanced equation that forms 2 moles of NO2 from 1 mole of O2 and 1 m0le of NO. This is consistent with the balanced equation of the overall reaction, which forms 2 moles of NO2.
Option A: 2NO(g) + O2(g) → 2NO2(g) forms 2 moles of NO2 but it consumes 2 moles of NO which is not consistent with the overall reaction.
Option B: 5NO(g) + O2(g) → 3NO2(g) forms 3 moles of NO2 but it consumes 5 moles of NO, again this is not consistent with the overall reaction.
Option C: NO(g)+O2(g)→2NO2(g), it is a balanced equation that forms 2 moles of NO2 from 1 mole of O2 and 1 m0le of NO.
Therefore, based on the information provided, the correct option for reaction 2 is C: NO(g) + O2(g) → 2NO2(g)
"
Complete question
Reachtion 1 = (2O2(g)→3O2(g) ).
Reachtion 2 = ?
Overall recation = (2O3(g)+2NO(g)→2NO2(g) + 2O2(g) ).
the enthalpy of the overall reaction represented above can be determined by adding the enthalpies of reactions 1 and 2. which of the following could be reaction 2 ?
A: 2NO(g)+O2(g)→2NO2(g)
B: 5NO(g)+O2(g)→3NO2(g)
C: NO(g)+O2(g)→2NO2(g)
D: None
"
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control of metal-support interactions in heterogeneous catalysts to enhance activity and selectivity
Control of metal-support interactions (MSIs) in heterogeneous catalysts can enhance both activity and selectivity.
MSIs refer to the interactions between the metal catalyst and the support material it is placed on. By fine-tuning the MSIs, the activity of the catalyst can be increased by optimizing the distribution and dispersion of the metal particles. In addition, by controlling the MSIs, the selectivity of the catalyst can be improved, allowing for the production of specific products in a reaction. This is done by manipulating the electronic and geometric properties of the metal-support interface, which affect the adsorption and activation of reactant molecules.
In general, controlling MSIs is a powerful way to make heterogeneous catalysts work better.
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which of the following would decrease the rate of a chemical reaction? adding more reactants stirring the reaction more vigorously squeezing the reactants into a smaller volume increasing the temperature raising the activation energy
Raising the activation energy would decrease the rate of a chemical reaction.
The rate of a chemical reaction refers to the speed at which the reactants are converted into products. There are several factors that can affect the rate of a chemical reaction, including the concentration of reactants, temperature, and the presence of catalysts or inhibitors.
Adding more reactants would generally increase the rate of a chemical reaction, as there would be more particles present to collide and react. Stirring the reaction more vigorously would also increase the rate of the reaction, as it would increase the chances of successful collisions between reactant particles. Squeezing the reactants into a smaller volume would also increase the rate of the reaction, as it would increase the concentration of reactants and the chances of successful collisions.
On the other hand, increasing the temperature would generally increase the rate of a chemical reaction, as it would increase the kinetic energy of the reactant particles and increase the chances of successful collisions.
However, raising the activation energy would decrease the rate of a chemical reaction, as it would require more energy for the reactant particles to overcome the activation energy barrier and undergo a successful reaction. Activation energy is the energy required to start a chemical reaction.
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From the formula for calcium acetate,Ca(C2H3O2)2, calculatethe mass of carbon that can be obtained from 65.3 g of thecompound.
The mass of carbon that can be obtained from 65.3 g of the compound is 19.83 grams of carbon,
First, you should calculate the % of carbon in calcium acetate
% of carbon - (12.01 x 4)/(40.08+12.01 x 4+ 1.008 x 6+16x 4)
% of carbon-48.04/158.168 x 100%
% of carbon 30.37% In 65.3 g of compound,
there are 65.3 x 30.37% -19.83 grams of carbon.
What do you mean by compound?
A object created by joining several different elements or parts. Particularly: a separate substance created by the chemical fusion of two or more substances in a certain weight-to-volume ratio. a fresh antibiotic substance.
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3. For the carbon in the structure shown, determine the followi
H
(a) Hybridization:
(b) Molecular shape_
(c) Number of electron domains:
(d) Bond Angle:
The carbon that is SP2 hybridized in the compound is carbon 4 (C4).
What is the hybridization?When we talk about the hybridization of the carbon atom, we mean the fact that the orbitals that are in the carbon atom are mixed in such a way that the orbitals that have the appropriate energy to be able to participate in the chemical bonds are formed.
As such looking at the compound that we have been shown in the image, we can see that not all the carbon atoms would be at the same state of hybridization.
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would you expect the following compound to have a dipole moment? if the molecule has a dipole moment, specify its direction. select the single best answer.
The compound to have a dipole moment, the correct statement about the dipole moment is 4) The molecule has no dipole moment.
The molecule is CF₄, carbon tetrafluoride. The bond between the C - F is the polar covalent bond because of the greater electronegativity difference between the C and the F atoms. Even though it has no dipole moment because of the geometry that is tetrahedral geometry. The all the C - F bonds are arrange in such a manner that the dipole moment will cancel out each other.
Thus, the CF₄ molecule has no dipole moment. The molecule is the non - polar molecule.
This question is in complete , the complete question is :
Would you expect the following compound to have a dipole moment? if the molecule has a dipole moment, specify its direction. select the single best answer. CF₄
1) The dipole moment is oriented from the fluorine atom towards the carbon atom.
2) The dipole moment is oriented from the carbon atom towards the fluorine atom.
3) The orientation cannot be determined.
4) The molecule has no dipole moment.
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The density of air at ordinary atmospheric pressure and 25 ∘C is 1.19 g/L. What is the mass of the air in a room that measures 14.5×16.5×6.0 ft?
Answer:
40,827.34 g
Explanation:
First find the volume of the room and then multiply it by the density of air at ordinary atmospheric pressure and 25 ∘C.
In this case, the room has a length of 14.5 ft, a width of 16.5 ft, and a height of 6.0 ft.
volume = 14.5 ft x 16.5 ft x 6.0 ft = 1209.25 ft^3
To convert ft^3 to L, we can use the conversion factor 1ft^3 = 28.316846592 L
1209.25 ft^3 = 1209.25 * 28.316846592 = 34,416.99 L
Calculate the mass of the air in the room by multiplying the volume of the room by the density of air:
mass = 34,416.99 L x 1.19 g/L = 40,827.34 g
So, the mass of the air in the room is 40,827.34 g
What is the correct reading of the volume in the pictured buret? Make sure to report your reading with the appropriate significant figures. Select one: a. 39.9 mL b. 39.5 mL c. 40.0 mL d. 39.950 mL e. 39.95 mL f. 40.05 mL
The correct reading of the volume in the pictured burette is e) 39.95 mL. The picture is attached below.
The purpose of the burette reading is to show that how much the solution has been dispensed, not the how much the burette contains. As compared to the volumetric glassware, the zero scale on the burette is written on the top. The Measurement scale on the burette is at the 1.0 intervals.
The intervals can be calculated as :
Intervals = (larger value - small value) / 2
Intervals = 1
The reading will be 39.95 mL.
Thus, according to the picture of the burette, the reading of the burette is the 39.95 mL.
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Deduce the structure of a compound of molecular formula C4H9Br which exhibits the following 1H NMR spectrum.
The structure C₄H₉Br that is 2-bromo-2-methylpropane , exhibits the ¹H NMR is as follows :
CH₃
|
CH₃ - CH - CH₃
|
Br
In the 2-bromo-2-methylpropnae molecule, all the 9 protons has the same chemical environment and there no splitting of the ¹H NMR resonance. All the protons are chemically equivalent, and they will have the same resonance frequency in the NMR.
In the molecule , 9 protons of the 2-bromo-2-methylpropane, as all the hydrogen atoms are having the same chemical environment. In the the Chemical shift (a) of the 1.80 ppm on the ¹H NMR spectrum for the 2-bromo-2-methylpropane is present.
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Consider the composition. information from the SDS. What is a possible name of this compound?
A. Glycine
B. Glycol
C. Glycogen
D. Glycerol
Answer: B
Explanation:
x consists of a flexible container with eight particles of a gas as shown. what happens to the pressure of the system when x is converted to the representations in (a), (b), and (c)?
When a gas is heated in a flexible container with constant pressure, the statement "The gas particles travel quicker" is accurate.
option (C)
The container's walls are struck more forcefully by the gas particles inside the container.
(a) Gas molecules lack a fixed form and a fixed volume. They grow till they fill their container. Gas molecules move readily both inside and outside of the container because they are fluid. Unless compressed, gas molecules are not very dense.
(b) Gas molecules mix and disperse as they diffuse and effuse (travel through small holes).
(c) The container's walls are struck more forcefully by the gas particles inside the container. The gas container's capacity expands as a result of the impact.
The gas container's capacity expands as a result of the impact. Gas molecule characteristics
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The complete question follows
x consists of a flexible container with eight particles of a gas as shown. what happens to the pressure of the system when x is converted to the representations in (a), (b), and (c)?
(a) pressure increased
(b) pressure decreased
(c) pressure is constant
a gas is compressed from a volume of 5.42 l to a volume of 2.75 l by an external pressure of 748 torr. which of the following correctly reflect the calculations required to determine the work done on/by the system (the gas)? select all that apply.
The one which correctly reflect the calculation required to find the work done is a) W = - P ΔV = 266 J
The initial volume = 5.42 L
The final volume = 2.75 L
The change in volume , ΔV = final volume - initial volume
= 2.75 - 5.42
The change in volume , ΔV = - 2.67 L
The work done is expressed as :
Work done, W = - P ΔV
where,
ΔV = - 2.67 L
P = 748 torr = 0.984 atm
W = - P ΔV
W = - ( 0.98 × - 2.67 ) × 101.3 J ( 1 atm ⋅L = 101.3 J )
W = 266 J
The work done is 266 J.
A gas is compressed from a volume of 5.42 l to a volume of 2.75 l by an external pressure of 748 torr. which of the following correctly reflect the calculations required to determine the work done on/by the system (the gas)? select all that apply.
a) W = - P ΔV = 266 J
b) W = + P ΔV = - 266 J
c) none of the above
d) both a and b
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which of the following has the correct curved arrows placement to show resonance for the given allylic carbocation>
An allylic lone pair is the appropriate positioning for the curved arrows to indicate resonance for the specified allylic carbocation.
What is a carbocation and how does one create one?The carbocation is indeed an organic molecule that develops when two valence electrons, which are typically shared by two atoms of carbon, are lost from such a carbonyl group that already possesses four bonds. As a result, a carbon atom with three rather than four bonds and a positive charge is created.
What distinguishes a carbocation?All carbon atoms in carbocations, also referred to as carbonium ions, are positively charged. A ion is indeed a positive charge, and the "carbo" part of the term refers to an atom of carbon.
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A gas sample at a certain temperature has an initial volume of 5.345 L. What is the initial temperature of the sample if the volume and the temperature are changed to 4.345 L and 898.0 K, respectively? Assume that the pressure and the amount of the gas remain constant.
Answer:
1104 K
Explanation:
We can use Charles' Law: [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
Plugging in the numbers gives us the initial temperature to be 1104 K
write a balanced equation for the synthesis reaction that occurred in experiment 1. include physical states.
The equation NaCl + AgNO3 => AgCl + NaNO3 incorporates the physical states for the synthesis reaction that took place in experiment 1. Factors are denoted by symbols in chemical equations.
When two or more chemical equations combine in a synthesis reaction, the result is A + B AB. This form makes it easy to spot synthesis reactions because there are more reactants than products. When two or more reactants react, a larger compound is produced depending on the direction of the reaction and the physical characteristics of the reactants. In 1615, chemist Jean Beguin of France formulated the first chemical equation. Chemical processes are present everywhere, beginning with food metabolism.
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chromium and cadmium can be removed from wastewater by treatment with sodium hydroxide. show the net ionic reactions of cr3 (aq) with naoh(aq) and cd2 (aq) with naoh(aq).
The following are the net ionic processes for removing chromium (III) and cadmium from wastewater using sodium hydroxide: Sodium hydroxide with chromium (III): Cr3+(aq) + 3NaOH(aq) -> Na3Cr(OH)3.
Cadmium in the presence of sodium hydroxide: Cd2+(aq) + 2NaOH(aq) -> Na2Cd(OH)2. Metal ions react with hydroxide ions in both processes to generate insoluble metal hydroxide precipitates that can be readily removed from the solution. Precipitation is a treatment method that is excellent for removing heavy metal ions from wastewater.chromium and cadmium can be removed from wastewater by treatment with sodium hydroxide. show the net ionic reactions of cr3 (aq) with naoh(aq) and cd2 (aq) with naoh(aq).The following are the net ionic processes for removing chromium (III) and cadmium from wastewater using sodium hydroxide: Sodium hydroxide with chromium (III): Cr3+(aq) + 3NaOH(aq) -> Na3Cr(OH)3.
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in part 1, sort the orbitals in order of energy, with the lowest-energy orbital at the bottom. then in part 2, consider how many of these orbitals are filled with electrons.
The lowest energy orbital will be closest to the nucleus of an atom. Each and every orbital is filled with electrons in the order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, and so on.
The lowest energy sublevel always be the 1s sublevel, which consists of one orbital. The single electron of an hydrogen atom will be occupy the 1s orbital when the atoms are in its ground state. As we proceed with atoms with multiple electrons, those electrons are added to the next lowest sublevel: 2s, 2p, 3s, and so on. Orbitals will be ranked in the increasing order of orbital energy as follows: 1s < 2s = 2p < 3s = 3p = 3d <4s = 4p = 4d= 4f.
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T/F rf values should be kept in the range of 0.2 to 0.8 for the most effective separation when more than one compound is present in the sample.
The statement is true that rf values should be kept in the range of 0.2 to 0.8
Polar compoundsPolar will have higher Rf values than nonpolar compounds under the same developing conditions. Rf values should be kept in the range of 0.2 to 0.8 for the most effective separation when more than one compound is present in the sample. The less polar the solvent, the higher the Rf value.The fastest moving spot has the highest Rf value. polar (fastest moving), and the spot with the lowest polargo through the given link below to know more about rf values:-
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How many atoms are in 30.1 L of Water vapor at STP?
First ⬆️. Second ⬆️
I hope this helps ✌
Given the following balanced equation, determine the rate of reaction with respect to [O2]: 2O3 (g).....yields.....3O2 (g)
a. Rate = -2deltaO2 over delta t
b. Rate = -2/3delta O2 over delta t
c. Rate = +1/3delta O2 over delta t
d. Rate = +3delta O2 over delta t
e. it is not possible to determine without more information
The balanced equation is :
2O₃(g) -----> 3O₂(g)
The rate of the reaction with [O₂] :The correct option c) Rate = + (1/3) Δ[O₂] /Δt
The balanced chemical equation is as follows :
2O₃(g) -----> 3O₂(g)
The rate of the reaction is the speed of the reaction at which the product is formed from the reaction in the chemical reactions. The rate for the chemical reaction is expressed by the decrease in the concentration of the reactants or the increase in the concentration of the products:
The rate of reaction with respect to the [O₂] is expressed as :
The rate = + (1/3) Δ[O₂] /Δt
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Calculate the number of grams of sulfuric acid in 1 gallon of battery acid if the solution has a density of 1.31 g/ml and is 38.7 % sulfuric acid by mass.
Answer:
4.956 g/gallon * 0.387 = 1.91 g of sulfuric acid in 1 gallon of battery acid.
Explanation:
To calculate the number of grams of sulfuric acid in 1 gallon of battery acid, we can use the density of the solution and the percentage of sulfuric acid by mass.
First, we need to convert the density from g/ml to g/gallon. Since 1 gallon is equal to 3.785 L, we can multiply the density of 1.31 g/ml by 3.785 to get 4.956 g/gallon.
Next, we can use the percentage of sulfuric acid by mass to find the mass of sulfuric acid in the solution. To do this, we can multiply the mass of the solution by the percentage of sulfuric acid by mass, which is 0.387.
The hottest temperature yet recorded in Phoenix, Arizona, was 122 O F on June 26, 1990. IVhat is that temperature in 0 C? In K?
The temperature in Celsius and kelvin is 50°C and 323.15 K.
The three most used temperature scales for usage in business, science, and daily life are Kelvin, Celsius, and Fahrenheit.
A scale in Kelvin is absolute. There are no degree symbols after its values, which begin at absolute zero.
Both Celsius and Fahrenheit are relative scales. The degree symbol is used to indicate Fahrenheit and Celsius temperatures.
Given the temperature is 122°F
Temperature in Celsius is -
⇒ C = (F - 32)/1.80
⇒ C = (122 - 32) / 1.80
⇒ C = 90 / 1.80
⇒ C = 50°C
Temperature in Kelvin is -
K = 5/9(F - 32) + 273.15
⇒ K = 5/9 (122 - 32) + 273.15
⇒ K = 323.15 K
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the reaction of 2,2-dimethyl-1-propanol, also known by the common name neopentyl alcohol with hbr gives 2-bromo-2-methylbutane as the major product. giving the structures of the reactant and the product of the reaction, show mechanistic explanation for the product of the reaction.
The reaction of 2,2-dimethyl-1-propanol, also known as the neopentyl alcohol with HBr gives 2-bromo-2-methylbutane as the major product. It is a SN₁-mechanism.
Neopentyl alcohol is a primary alcohol, but the bulky group has hindered it. The backside attack is not possible, so instead of the SN₂ mechanism SN₁ followed to get the desired rearranged product. The reaction will forms a rearranged stable carbocation from the primary carbocation to the tertiary carbocation.
Protonation of the Neopentyl alcohol will converts the hydroxy group into a good leaving group. Water will leaves form a primary carbocation rearranged through the methyl shift, which is thus forming a stable tertiary carbocation. Bromide ion will also attacks the carbocation to form a product 2-bromo-2-methylbutane. Mechanism will showing the formation of 2-bromo-2-methylbutane.
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the linear order of amino acids (blank structure), with the varying properties of their side chains (or blank groups), determines what blank and blank structures will form to produce a protein. the resulting unique blank shapes of proteins are key to their specific and diverse functions.
The linear order of amino acids (primary structure), with the varying properties of their side chains (R groups), determines what secondary and tertiary structures will form to produce a protein. The resulting unique three-dimensional shapes of proteins are key to their specific and diverse functions.
Amino acids are the building blocks of proteins, and the order of their assembly is known as the primary structure.
Each amino acid has a side chain, or R group, that is composed of different atoms, giving each amino acid a unique property that affects the structure and function of the proteins they create.
Depending on the primary structure and the chemical characteristics of the side chains, proteins can assume a variety of secondary and tertiary structures. It is these unique three-dimensional shapes that enable proteins to carry out their specific and varied functions.
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question mode multiple choice question which one of the following irreversibilities describes the force that opposes the relative motion between two bodies in contact with one another? multiple choice question. friction heat transfer chemical reactions mixing of two fluids
The force that resists the relative motion of two bodies is referred to as irreversibility. Examples of this include chemical reactions, mixing two fluids, and electrical resistance.
Why do we use the term "chemical"?
Any material with a known composition is a chemical. To put it another way, a chemical always consists of the same "substance." There are some substances in nature, like water. Chlorine and other chemicals are made at factories.
What instances from chemistry?
All of these things are made of chemical, some natural and others synthetic, including air, water, iron, fuel, clothes, stone, equipment, plastics, plants, and food. Chemicals can be made up of several different chemicals or they can be specific chemical molecules like water. Ammonia, oxygen, water, carbon dioxide, and other substances are all present in air.
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Given the density of Au is 19.3 g/cm^{ 3} , determine the mass of gold in an ingot with the dimensions of 10. in x 4.00 in x 3.00 in. Give your answer as a number with correct sig figs only.
The mass of the gold with the density of 19.3 g/cm³ with dimensions of 10.0 in x 4.00 in x 3.00 in is 38,011.48 grams.
How to calculate mass?The mass of a substance can be calculated from the density by using the following expression;
Density = mass ÷ volume
According to this question, the dimensions of gold is given as 10.0 in x 4.00 in x 3.00 in. The density of gold is 19.3 g/cm³.
Volume of gold = 10 × 4 × 3 = 120in³
120in³ is equivalent to 1966.45cm³
19.3g/cm³ = mass ÷ 1966.45cm³
mass = 19.3 × 1966.45
mass = 38,011.48 grams.
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what is the formula of manganese which contains 50.48% oxygen? Ar (Mn) = 64.94
(Hint: Oxide is the result of combining Mn and O.) Mn (Mass of the element, in g) (assuming 100 g of compound) elements in moles The smallest moles or moles of an element Multiplier mathematical model O 1. 49.5 2. _50,5 3.
What is the empirical equation for a manganese oxide with a 49.5 manganese content?(Hint: Oxide is the result of combining Mn and O.) Mn (Mass of the element, in g) (assuming 100 g of compound) elements in moles The smallest moles or moles of an element Multiplier mathematical model O 1. 49.5 2. _50,5 3.The chemical having the formula MnO2 is known as manganese dioxide.By elemental manganese oxidation, When oxygen in the environment interacts with elemental manganese, MnO2 is the result. In nature, manganese is typically found as manganese dioxide since elemental manganese cannot exist due to this reaction.(Hint: Oxide is the result of combining Mn and O.) Mn (Mass of the element, in g) (assuming 100 g of compound) elements in moles The smallest moles or moles of an element Multiplier mathematical model O 1. 49.5 2. _50,5 3.To learn more about manganese dioxide refer to:
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