give one use of zinc
Which technique is best suited to each application?
a. In the second week of a four week biochemistry experiement, you have 50 fractions collected from a gel filtration column to determine which fractions contain lactate dehydrogenase. You are given only 400 uL of 0.100 mg/mL lactate dehydrogenase to prepare your calibration curve. 96-well microplate
b. Your environmental lab has 2000 samples to be analyzed for trace ammonia by next week. discrete analyzer.
c. Twenty water samples must be analyzed for Cl-, NH3, PO3-, and So during each work shift. flow injection analysis colorimeter.
d. Your professor heard you will be hiking the Appalachian Trail next summer. She asks you to collect 100-mL water samples from the ten streams with the highest concentration of phosphate.
Answer:
a. discrete analyzer
b. 96 well microplate
c. flow injection analysis
d. colorimeter
Explanation:
96 well microplates are instruments designed for sample collection and throughput screening. If an environment lab has collected 2000 samples then 96 well microplate is best suited application. Discrete analyzer is automated chemical analyzer which performs test on samples kept in discrete cells. Flow injection analysis is approach used for chemical analysis. It injects a plug of sample into a flowing carrier stream. Colorimeter is a device which measures absorbance of wavelength of light by a specific solution.
Calculate the specific heat of a piece of wood if 2000 g of wood absorbs 75,250 J of heat, and its temperature changes from 30°C to 50°C.
A 37.63
B) 0.53
C) 1.88
D
752.50
Answer:
C
Explanation:
The specific heat capacity=quantity of heat in joule/mass×change in temperature
from this question the quantity of heat is 75250,the mass is 2000 and the change in temperature is 50-30
which is 20
therefore
c=75250/2000×20
c=75250/40000
c=1.88
I hope this helps
What is the molecular geometry of CIO3F as predicted by the VSEPR model?
Multiple Choice
trigonal pyramidal
square planar
square pyramidal
tetrahedral
octahedral
Explanation:
since there are no lone pairs on the central atom, the shape will be tetrahedral
You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution?
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
Calculate the boiling point of a 3.5 % solution (by weight) of sodium chloride in water.
Kb of H2O = 0.512 oC/M
Answer: The boiling point of the solution is [tex]101.02^oC[/tex]
Explanation:
We are given:
3.5 % (by weight) NaCl
This means that 3.5 g of NaCl is present in 100 g of solution
Mass of solvent = Mass of solution - Mass of solute
Mass of solvent (water) = (100 - 3.5) g = 96.5 g
Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.
The expression for the calculation of elevation in boiling point is:
[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]
OR
[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)
where,
Boiling point of pure solvent (water) = [tex]100^oC[/tex]
Boiling point of solution = ?
i = Vant Hoff factor = 2 (for NaCl)
[tex]K_b[/tex] = Boiling point elevation constant = [tex]0.512^oC/m[/tex]
[tex]m_{solute}[/tex] = Given mass of solute (NaCl) = 3.5 g
[tex]M_{solute}[/tex] = Molar mass of solute (NaCl) = 36.5 g/mol
[tex]w_{solvent}[/tex] = Mass of solvent (water) = 96.5 g
Putting values in equation 1, we get:
[tex]\text{Boiling point of solution}-(100)=2\times 0.512\times \frac{3.5\times 1000}{36.5\times 96.5}\\\\\text{Boiling point of solution}=(1.02+100)^oC\\\\\text{Boiling point of solution}=101.02^oC[/tex]
Hence, the boiling point of the solution is [tex]101.02^oC[/tex]
Which acid or base (along with its corresponding salt) should be used to generate a buffer solution with pH around 3.5
Answer:
Formic acid
Sodium formiate
Explanation:
To determine acid or base that can generate a buffer solution with pH around 3.5, we have to think in the acid whose pKa = pH.
Although we have to also think in buffer capacity, a measure which can cause a change of 1 pH unit in 1 L of solution.
Buffer capacity does not only depend on the concentration of its components but also of the relationship between that concentrations.
When pH = pKa, buffer capacity is maximum which means that the concentration of conjugated species is the same and the ability to oppose pH changes is maximum.
One example with pH = pKa or nearly if:
COOH⁻ + Na⁺ → NaCOOH
HCOOH + H₂O → COOH⁻ + H₃O⁺ Ka: 1.8×10⁻⁴
pKa = 3.74
Help me, please
Help me, please
Answer:.......
Explanation:
3. At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume
be if the temperature were changed to 22 C and the pressure to 1.25 atmospheres
Answer:
The volume will be 185.83 mL.
Explanation:
Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T} =k[/tex]
Where P = pressure, T = temperature, k = Constant
Boyle's law says that the volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure. Boyle's law is expressed mathematically as:
P*V=k
Where P = pressure, V = volume, k = Constant
Finally, Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, which is kept at a constant pressure. For a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement of the gas molecules. .
In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:
[tex]\frac{V}{T} =k[/tex]
Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:
[tex]\frac{P*V}{T} =k[/tex]
Studying two different states, an initial state 1 and a final state 2, it is satisfied:
[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]
In this case:
P1= 720 torr V1= 256 mLT1= 35 C= 308 K (being 0 C= 273 K)P2= 1.25 atm= 950 torr (being 1 atm= 760 torr)V2= ?T2= 22 C= 295 KReplacing:
[tex]\frac{720 torr*256 mL}{308 K} =\frac{950 torr*V2}{295 K}[/tex]
Solving:
[tex]V2= \frac{295K}{950 torr} *\frac{720 torr*256 mL}{308 K}[/tex]
V2= 185.83 mL
The volume will be 185.83 mL.
The reversible reaction: 2SO2(g) O2(g) darrw-tn.gif 2SO3(g) has come to equilibrium in a vessel of specific volume at a given temperature. Before the reaction began, the concentrations of the reactants were 0.060 mol/L of SO2 and 0.050 mol/L of O2. After equilibrium is reached, the concentration of SO3 is 0.040 mol/L. What is the equilibrium concentration of O2
Answer:
0.030 M
Explanation:
Step 1: Make an ICE chart
2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)
I 0.060 0.050 0
C -2x -x +2x
E 0.060-2x 0.050-x 2x
Step 2: Find the value of x
The concentration of SO₃ at equilibrium is 0.040 mol/L. Then,
2x = 0.040
x = 0.020
Step 3: Calculate the concentration at equilibrium of O₂
[O₂] = 0.050 - x = 0.050 - 0.020 = 0.030 M
The equilibrium concentration of oxygen is 0.030 M.
A reversible reaction is a reaction that can move either in the forward or in the reverse direction. We have the reaction; 2SO2(g) + O2(g) ⇄ 2SO3(g). We can now set up the ICE table as shown below;
2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)
I 0.060 0.050 0
C -2x -x +2x
E 0.060-2x 0.050-x 2x
At equilibrium;
2x = 0.040 M
x = 0.040 M/2 = 0.020 M
For oxygen;
0.050-x
0.050 M - 0.020 M = 0.030 M
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How many atoms are in protons
Answer:
the number of protons in a atom is unique to each element
Explanation:
protons are about 99.86% as massive neutrons. The number of protons in a atom is unique to each element .For example carbon atoms have six protons in an atom is referred to as the atomic number of that element
If a 520 mg sample of technetium-99 is used for diagnostic procedure, how much of Tc-99 remains after 30.0h? Half life of Tc-99 is 6.0 hours.
0.28 M Ca(NO3)2
Express your answer using two significant figures.
Answer:
Mass=Moles × RFM
Mass= 0.28M× 164
Mass= 45.92 grammes
What kind of element is Phosphorus is
Answer:
NON-METAL
Explanation:
Phosphorus is a non-metal that sits just below nitrogen in group 15 of the periodic table. This element exists in several forms, of which white and red are the best known.
A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipitate. (a) Write the molecular equation for the reaction. (b) What is the limiting reactant in the reaction? (c) How many grams of precipitate potentially form?
Answer:
0.21 g
Explanation:
The equation of the reaction is;
NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)
Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles
Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles
Since the reaction is 1:1, NaCl is the limiting reactant.
1 mole of NaCl yields 1 mole of AgCl
0.00147 moles of NaCl yields 0.00147 moles of AgCl
Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol
= 0.21 g
Carbonic anhydrase is strongly inhibited by the drug acetazolamide, which is used as a diuretic (i.e., to increase the production of urine) and to lower excessively high pressure in the eye (due to accumulation of intraocular fluid) in glaucoma.
a. True
b. False
Answer:
a. True
Explanation:
There is strong inhibition of Carbon Anhydrase by Aceta-zolamide Carbonic Anhydrase. The drug acetazolamide is used as diuretic which increase the urine production in human body. It lowers pressure in eye in glaucoma.
2) If a brick has a length of 13.77 cm, a width of 8.50 cm, and a height of 5.12 cm:
a) What is the volume of the brick?
b) If the brick has a mass of 895.3 g, what is its density?
Answer:
a. 599 cm³
b. 1.49 g/cm³
Explanation:
A. VolumeVolume is the amount of space an object occupies. Since this is a brick, the object is a rectangular prism. The formula for the volume of a rectangular prism is the product of length, width, and height.
[tex]V= l *w*h[/tex]
The brick's length (l) is 13.77 centimeters, the width (w) is 8.50 centimeters, and the height (h) is 5.12 centimeters. Substitute these values into the formula.
[tex]V= 13.77 \ cm * 8.50 \ cm * 5.12 \ cm[/tex]
Multiply the numbers together.
[tex]V= 117.045 \ cm^ 2* 5.12 \ cm[/tex]
[tex]V= 599.2704 \ cm^3[/tex]
The original measurements have at least 3 significant figures, so our answer must have 3. For the number we calculated, that is the ones place. The 2 in the tenths place tells us to leave the 9 in the ones place.
[tex]V \approx 599 \ cm^3[/tex]
[tex]\bold {The \ volume \ of \ the \ brick \ is \ approximately \ 599 \ cubic \ centimeters}}[/tex]
2. DensityDensity is the amount of matter in a specified space. The formula for density is mass over volume.
[tex]d= \frac{m}{v}[/tex]
The mass of the brick is 895.3 grams and we just found the volume to be 599.2704 cubic centimeters. Substitute the values into the formula.
[tex]d= \frac{895.3 \ g}{599 \ cm^3}[/tex]
Divide.
[tex]d= 1.494657763 \ g/cm^3[/tex]
Round to three significant figures. For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 9 in the hundredth place.
[tex]d \approx 1.49 \ g/cm^3[/tex]
[tex]\bold {The \ density\ of \ the \ brick \ is \ approximately \ 1.49 \ grams /cubic \ centimeters}}[/tex]
Sobre ações relacionadas ao aquecimento global, assinale somente as alternativas corretas:
a) ( x) As ações humanas não influenciam no aumento da temperatura do planeta.
b) ( ) As mudanças climáticas são intensificadas pela emissão de gases das atividades humanas.
c) ( ) A queima de combustíveis fósseis e de florestas são as principais ações humanas que liberam gases que intensificam o efeito estufa.
d) ( ) O efeito estufa é um fenômeno natural.
e) ( ) Se as águas dos oceanos ficarem mais quentes, os furacões não terão tanta força.
Answer:
123456788012346778901234567890
What is the molarity of a solution containing 150 g of zinc sulfate (ZnSO4) per liter?
Answer:
0.93 M
Step-by-step Explanation:
First, we have to calculate the molar mass (MM) of ZnSO₄ by using the molar mass of each chemical element:
MM(ZnSO₄) = 65.4 g/mol Zn + 32 g/mol S + (16 g/mol x 4) = 161.4 g/mol
Then, we divide the mass of ZnSO₄ into its molar mass to obtain the number of moles:
moles ZnSO₄ = mass/MM = 150 g/(161.4 g/mol)= 0.93 mol
Since the molarity of a solution expresses the number of moles of solute per liter of solution, we calculate the molarity (M) as follows:
M = moles ZnSO₄/1 L = 0.93 mol/1 L = 0.93 M
Write a balanced half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution. Be sure to add physical state symbols where appropriate.
Answer:
2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)
Explanation:
Let's consider the unbalanced half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution.
H₂O(l) ⇒ H₂(g)
First, we will perform the mass balance. We will balance oxygen atoms by multiplying H₂O by 2 and adding 2 OH⁻ to the right side.
2 H₂O(l) ⇒ H₂(g) + 2 OH⁻(aq)
Then, we perform the charge balance by adding 2 electrons to the left side.
2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)
The following list of properties is most descriptive of a(n) ______________. Low melting point, non-conductor of electric current. Group of answer choices
Answer:
This question is incomplete
Explanation:
This question is incomplete but the completed question is below;
"The following list of properties is most descriptive of a(n) ______
1. High melting point, conductor of electricity when dissolved in water
2. Low melting point, non-conductor of electric current
3. Malleable, ductile, insoluble in water.
The choices for all 3 are: a) metallic solid b) molecular solid c) ionic solid d) all of these e) none of these f) more than one of these"
1. Ionic solid: Ionic solids are solids that have ionic/electrovalent bonds holding it's constituent molecules together. These bonds are strong bonds that involve the transfer of electrons from one constituent atom (the metal) to another constituent atom (the nonmetal). This strong bond causes the solid to have a high melting and boiling point. Also, when dissolved in water, the constituent atoms (involved in the electron transfer) dissociate to form ions (become charged) and thus easily carry electric charges (i.e conduct electricity).
Examples are Sodium Chloride and Potassium Iodide
2. Molecular solids: These are solids whose constituent molecules are held together by a weak bond/force known as Van der Waal forces. This forces are easily broken down when subjected to heat and thus the molecular solids have a low melting point. Also, these solids do not have carriers of heat or electric charges in them and are thus non-conductors of electric current.
Examples are Ice (frozen water) and sucrose
3. Metallic solids: These of solids made from constituent metal atoms only. The nuclei of these constituent metal atoms have the ability to move past one another without disrupting there metallic bonding; it is for this reason they are malleable and ductile. There constituent atoms however do not dissociate in water and are thus insoluble in water.
Examples are aluminium and copper crystal.
A gas sample containing a constant number of gas molecules has a volume of 2.70 L at a constant pressure and a temperature of 25.0o C. What would be the volume (in Liters) of this gas sample at 75.0o C? Round your answer to 3 sig fig
Answer:
[tex]\boxed {\boxed {\sf 8.10 \ L}}[/tex]
Explanation:
This question asks us find the volume of a gas sample given a change in temperature. Since the pressure remains constant, we only are concerned with the variables of temperature and volume.
We will use Charles's Law. This states the volume of a gas is directly proportional to the temperature of a gas. The formula is:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
The gas starts at a volume of 2.70 liters and a temperature of 25.0 degrees Celsius.
[tex]\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{T_2}[/tex]
The temperature is increased to 75.0 degrees Celsius, but the volume is unknown.
[tex]\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{75.0 \textdegree C}[/tex]
We are solving for the volume at 75 degrees Celsius, so we must isolate the variable V₂.
It is being divided by 75.0 °C. The inverse operation of division is multiplication, so we multiply both sides of the equation by 75.0 °C.
[tex]75.0 \textdegree C *\frac {2.70 \ L}{25.0 \textdegree C}=\frac{V_2}{75.0 \textdegree C} * 75.0 \textdegree C[/tex]
[tex]75.0 \textdegree C *\frac {2.70 \ L}{25.0 \textdegree C}= V_2[/tex]
The units of degrees Celsius (° C) cancel.
[tex]75.0 *\frac {2.70 \ L}{25.0}= V_2[/tex]
[tex]75.0 *0.108 \ L = V_2[/tex]
[tex]8.1 \ L = V_2[/tex]
The original measurements have 3 significant figures, so our answer must have the same. Currently, the answer has 2. If we add another 0, the value of the answer does not change, but the number of sig figs does.
[tex]8.10 \ L = V_2[/tex]
The volume of this gas sample at 75.0 degrees Celsius is 8.10 Liters.
Ethylene glycol (C2H6O2) is mixed with water to make auto engine coolants. How many grams of C2H6O2 are in 5.00 L of a 6.00 M aqueous solution
Answer:
1860g.
Explanation:
It is known that the molar mass of C2H6O2 is 62.08 g/mol.,
Now to solve for the number of moles of solute, one must multiply both
sides by the volume:
moles of solute = (6.00 M)(5.00 L) = 30.0 mol
Notice since the definition of molarity is mol/L, the
product M × L gives mol, a unit of amount.
Use the molar mass of C3H8O3, one can convert mol to g:
Mass m =30 mol × 62.08 g/mol
m = 1860g.
Hence, there are 1,860 g of C2H6O2 in the specified amount of
engine coolant.
from kinatic point of view explain the change from solid to liquied based on the effect of change of tempreture.
Answer:
Temperature affects the kinetic energy in a gas the most, followed by a comparable liquid, and then a comparable solid. The higher the temperature, the higher the average kinetic energy, but the magnitude of this difference depends on the amount of motion intrinsically present within these phases.
Explanation:
Liquids have more kinetic energy than solids. When a substance increases in temperature, heat is being added, and its particles are gaining kinetic energy. Because of their close proximity to one another, liquid and solid particles experience intermolecular forces. These forces keep particles close together.
A 2.584 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 5.874 g CO2 and 2.404 g H2O. What mass of oxygen is contained in the original sample?a. 0.7119 g.b. 0.8463 g.c. 0.29168 g.d. 0.1793 g.e. 0.6230 g.
Answer:
a. 0.7119 g
Explanation:
To solve this question we need to know that all carbon of the compound will react producing CO2 and all Hydrogen producing H2O.
Thus, we can find the mass of C and the mass of H and by difference regard to the 2.584g of the compound we can find the mass of oxygen as follows:
Moles CO2 = Moles C -Molar mass: 44.01g/mol-
5.874g CO2 * (1mol/44.01g) = 0.1335 moles CO2 = 0.1335 moles C
Mass C -Molar mass: 12.01g/mol-:
0.1335 moles C * (12.01g /mol) = 1.6030g C
Moles H2O -Molar mass: 18.01g/mol-
2.404gH2O * (1mol / 18.01g) = 0.1335 moles H2O * (2mol H / 1mol H2O) = 0.267 moles H
Mass H -Molar mass: 1g/mol-
0.267 moles H * (1g/mol) = 0.2670g H
Mass Oxygen =
Mass O = 2.584g compound - 1.6030g C - 0.2670g H
Mass O = 0.714g O ≈
a. 0.7119 gLeaming Task 1:
Distinguish the process as spontaneous or non-spontaneous process. Write S it spontaneous and NSi non-spontaneous
on the bionk.
1. Melling ofice
2 Ruisting of ton
3. Marble going down the spiral.
4. Going up the
& Keeping the food fresh from spolage
Solution :
Spontaneous Process
A spontaneous process is defined as the process that occurs without the help of any external aid or inputs. A spontaneous process is a natural process which occurs naturally in the environment.
Non Spontaneous process
A non spontaneous process is a process which does not occur naturally. Some inputs are provided for the process to occur. Energy from external source is applied into the process to start the process.
The following processes are :
1. Melling of ice ---- Spontaneous
2 Rusting of iron --- Spontaneous
3. Marble going down the spiral. --- spontaneous
4. Going up the hill ---- Non spontaneous
5. Keeping the food fresh from spoilage --- Non spontaneous
In aqueous solution the Ni2" ion forms a complex with four ammonia molecules. Write the formation constant expression for the equilibrium between the hydrated metal ion and the aqueous complex. Under that, write the balanced chemical equation for the first step in the formation of the complex K,=________.
Answer:
The correct equation is "[tex]\frac{[Ni(H_2O)_3 (NH_3)]^{2+}}{[Ni(H_2O)_4]^{2+} [NH_3]}[/tex]".
Explanation:
According to the question,
Throughout an aqueous solution, [tex]Ni^{2+}[/tex] exist as [tex][Ni(H_2O)_4]^{2+}[/tex]
So,
⇒ [tex][Ni(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Ni(NH_3)_4]^{2+} + H_2O[/tex]
⇒ [tex]K_f = \frac{[Ni(NH_3)_4]^{2+}}{[Ni(H_2O)_4^{2+}] [NH_3]^4}[/tex]
Here, we have excluded [tex][H_2O][/tex] as concentration of water will be const.
Now,
This formation of [tex][Ni(NH_3)_4]^{2+}[/tex] proceeds via several steps,
Step 1:
⇒ [tex][Ni(H_2O)_4]^{2+}+NH_3 \rightleftharpoons [Ni(H_2O)_3 (NH_3)]^{2+} + H_2O[/tex]
⇒ [tex]K_1 = \frac{[Ni(H_2O)_3 (NH_3)]^{2+}}{[Ni(H_2O)_4]^{2+} [NH_3]}[/tex]
Is ribose a reducing or non reducing sugar?
Ribose is a reducing sugar. A reducing sugar is a carbohydrate that can undergo a redox reaction, in which it donates electrons to another chemical species.
This is usually observed when the sugar opens its ring structure to form an aldehyde or ketone functional group.
Ribose, a five-carbon sugar, can form an open-chain structure with an aldehyde functional group. In this form, it can donate electrons and act as a reducing agent in certain chemical reactions, such as the reduction of other compounds like Benedict's reagent during laboratory tests for reducing sugars.
Learn more about Ribose, here:
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When heated, carboxylic salts in which there is a good leaving group on the carbon beta to the carbonyl group undergo decarboxylation/elimination to give an alkene. Draw the structures of the products expected when this compound is heated.
Answer:
i dont know mate
Explanation:
A hydrocarbon contains only the elements____?
Explanation:
elements are carbons and hydrogen
Answer:
Carbon and Hydrogen.
Explanation:
It’s in the name Hydro (H) Carbon (C)
