Answer:199mL
Explanation:
Let V ml be the volume of blood in animal.
When 1.0 ml sample is added, total volume becomes V+1.0ml. Its activity is 1000 dpm.
After equilibrium, 2.0 ml of the sample had activity of 10 dpm.
Hence, after equilibrium, the activity of V+1.0 ml of blood sample will be 10/2 (V+1.0ml)=1000dpm
Hence, V=199ml.
A certain first-order reaction has a rate constant of 2.15×10−2 s−1 at 20 ∘C. What is the value of k at 55 ∘C if Ea = 72.0 kJ/mol ?
Answer:
[tex]k_2=0.504s^{-1}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the rate constant at 55 °C by using the temperature-variable version of the Arrhenius equation:
[tex]ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]
Thus, we plug in the temperatures, activation energy and universal constant of gases in consistent units to obtain:
[tex]ln(\frac{k_2}{0.0215s^{-1}} )=-\frac{72000\frac{J}{mol}}{8.3145\frac{J}{mol*K}}(\frac{1}{55+273} -\frac{1}{20+273} ) \\\\ln(\frac{k_2}{0.0215s^{-1}} )=3.154\\\\k_2=0.0215s^{-1}exp(3.154)\\\\k_2=0.504s^{-1}[/tex]
Regards!
An endothermic reaction will start when the required
energy is received from the environment or solution.
AH
activation
thermal
kinetic
Answer:
A: ΔH
Explanation:
Endothermic reactions are this that occur as a result of absorption of heat energy from the surroundings by the reactants to form new products.
Thus, we can say it is one with an increase in enthalpy (ΔH) of the system.
Thus, option A is correct.
The equilibrium concentrations for the reaction between SO2 and O2 to form
SO3 at a certain temperature are given in the table below. Determine the
equilibrium constant and whether the reaction favors reactants, products, or
neither at this temperature.
O(g) +250 (9)
2250 (9)
[02]
[SO2)
[S03)
1.2 M
0.80 M
1.9 M
A. K = 4.7; product favored
B. K = 0.51; product favored
C. K = 0.51; reactant favored
o
D. K= 4.7; reactant favored
Answer:
A. K = 4.7; product favored
Explanation:
Step 1: Write the balanced reaction at equilibrium
O₂(g) + 2 SO₂(g) ⇄ 2 SO₃(g)
Step 2: Calculate the concentration equilibrium constant (Kc)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.
Kc = [SO₃]² / [SO₂]² × [O₂]
Kc = 1.9² / 0.80² × 1.2 = 4.7
When Kc > 1, the products are favored.
Tema: Métodos de Separação de Misturas – Homogêneas e Heterogêneas;
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4. Bibliografia (1,0 ponto)
Answer:
fjskeowkcnekvo Dee five votes come vote for dog even r
the carbon tetrachloride molecule CCI 4 is
Answer:
is a nonpolar molecule with polar bonds
Identify the oxidation half-reaction for this reaction:
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
A. Fe2+ + 2e → Fe(s)
O B. H2(g) → 2H+ + 2e
O C. Fe(s) → Fe2+ + 2e
O D. 2H+ + 2e → H2(9)
Answer:
Fe(s)->Fe2+2e-
Explanation:
A.p.e.x
The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct
What is Oxidation reaction ?Oxidation reaction is a chemical reaction which can be described as follows ;
Addition of oxygen Removal of hydrogen Loss of ElectronAddition of electronegative atomRemoval of Electropositive elementIn the given reaction ;
Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)
Fe at RHS got converted to Fe²⁺ state at LHS which shows the gain of electron by Fe with in the reaction.
Therefore,
The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct
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What is the correct order for the reactions that produce the following transformation. a. (1) H2/Lindlar (2) CH3CO2OH b. (1) H2/Lindlar (2) O3, Zn, HCl c. (1) H2/Pd (2) CH3CO2OH d. (1) Na, NH3 (2) CH3CO2OH
Answer:
Explanation:
Can you provide a picture? I can outline the reactions though. a) will make a Z double bond from a triple bond and then peroxyacid can do epoxidation. b) will make the Z double bond then ozonolysis to double bond will create to aldehyde compounds. c) is essentially useless unless there is a ketone or aldehyde in the compound already since H2/Pd will fully reduce the alkyne (which I am assuming is present) and so the peroxyacid can't do epoxidation and can only do baeyer villiger oxidation, and d) reduces the alkyne to an E alkene and then do epoxidation to give an epoxide (with trans steroechemistry)
When perchloric acid (HClO4) reacts with tetraphosphorus decaoxide, phosphoric acid and dichlorine heptaoxide are produced.
a. Trei
b. False
Answer:
я не знаю ответа :(
Explanation:
Which statement is true with respect to standard reduction potentials?
SRP values that are greater than zero always represent a reduction reaction.
SRP values that are less than zero always represent a reduction reaction.
Half-reactions with SRP values greater than zero are spontaneous.
Half-reactions with SRP values greater than zero are nonspontaneous.
Answer:
C). Half-reactions with SRP values greater than zero are spontaneous.
Explanation:
SRPs or Standard Reduction Potentials are characterized as the ability of a probable distinction among the anode and cathode of a usual/standard cell. It aims to examine the capacity of chemicals to reduce themselves.
The third statement asserts a true claim regarding the SRPs(Standard Reduction Potentials) that the 'half-reactions which take place with the SRP possesses the values higher than zero and they are unconstrained.' The other statements are incorrect as they either show the estimation of SRPs more than 0 or display them as being restricted. Thus, option C is the correct answer.
Alkanes are chains of carbon atoms surrounded by hydrogen atoms.
a. True
b. False
Answer:
a. True
Explanation:
Alkanes are chains of carbon atoms surrounded by hydrogen atoms. TRUE.
Alkanes are hydrocarbons, that is, they are organic compounds formed only by carbon and hydrogen. In alkanes, carbon atoms are bonded to each other through single covalent bonds and they are also bonded to hydrogen atoms through the same type of bonds. Alkanes have the general formula CnH2n+2.
Bond length is the distance between the centers of two bonded atoms. On the potential energy curve, the bond length is the internuclear distance between the two atoms when the potential energy of the system reaches its lowest value.
a. True
b. False
Answer:
True
Explanation:
When two atoms are at infinite distance from each other, the both atoms posses high energy.
However, as they begin to approach each other, the distance between them gradually decreases and so does their energy.
A point is eventually reached when the potential energy curve reaches its minimum value. The internuclear distance between the two atoms at this point is called the bond length of the system.
5. A beam of photons with a minimum energy of 222 kJ/mol can eject electrons from a potassium surface. Estimate the range of wavelengths of light that can be used to cause this phenomenon. Show your calculations with units of measure (dimensional analysis) and briefly explain your reasoning.
Answer: The range of wavelengths of light that can be used to cause given phenomenon is [tex]8.953 \times 10^{21} m[/tex].
Explanation:
Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J
Formula used is as follows.
[tex]E = \frac{hc}{\lambda}[/tex]
where,
E = energy
h = Planck's constant = [tex]6.625 \times 10^{-25} Js[/tex]
c = speed of light = [tex]3 \times 10^{8} m/s[/tex]
Substitute the values into above formula as follows.
[tex]E = \frac{hc}{\lambda}\\222000 J = \frac{6.625 \times 10^{-34}Js \times 3 \times 10^{8} m/s}{\lambda}\\\lambda = 8.953 \times 10^{21} m[/tex]
Thus, we can conclude that the range of wavelengths of light that can be used to cause given phenomenon is [tex]8.953 \times 10^{21} m[/tex].
Calculate percent yield when you start with 0.78 grams of camphor and end with 0.23 grams of iso/borneol. The molecular weight of camphor is 152.23 g/mol, and the molecular weight of iso/borneol is 154.25 g/mol.
Answer:
29.1%
Explanation:
First we convert 0.78 g of camphor to moles, using its molar mass:
0.78 g ÷ 152.23 g/mol = 0.00512 mol camphorThen we convert 0.23 g of isoborneol to moles, using its molar mass:
0.23 g ÷ 154.25 g/mol = 0.00149 mol isoborneolFinally we calculate the percent yield:
0.00149 mol / 0.00512 mol * 100% = 29.1%Hydrogen is manufactured on an industrial scale by this sequence of reactions: Write an equation that gives the overall equilibrium constant in terms of the equilibrium constants and . If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.
The question is incomplete. The complete question is :
Hydrogen is manufactured on an industrial scale by this sequence of reactions:
[tex]$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g ) \ \ \ \ \ \ \ \ \ \ K_1$[/tex]
[tex]$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \ K_2$[/tex]
The net reaction is :
[tex]$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$[/tex]
Write an equation that gives the overall equilibrium constant [tex]K[/tex] in terms of the equilibrium constants [tex]K_1[/tex] and [tex]K_2[/tex]. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.
Solution :
[tex]$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g ) \ \ \ \ \ \ \ \ \ \ K_1$[/tex]
[tex]$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$[/tex] ...............(1)
[tex]$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \ K_2$[/tex]
[tex]$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$[/tex] ...................(2)
[tex]$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$[/tex]
[tex]$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$[/tex]
On multiplication of equation (1) and (2), we get
[tex]$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$[/tex]
[tex]$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$[/tex] .................(4)
Comparing equation (3) and equation (4), we get
[tex]$K=K_1K_2$[/tex]
us
Which of the following is a chemical property?
A. Hardness
B. Flammability
C. Malleability
D. Melting point
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Gaseous ethane CH3CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 0.60 g of ethane is mixed with 3.52 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
Answer:
1.8 g
Explanation:
Step 1: Write the balanced equation
CH₃CH₃(g) + 3.5 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:112.0 = 0.2684:1.
The experimental mass ratio of CH₃CH₃ to O₂ is 0.60:3.52 = 0.17:1.
Thus, the limiting reactant is CH₃CH₃
Step 3: Calculate the mass of CO₂ produced
The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:88.02.
0.60 g CH₃CH₃ × 88.02 g CO₂/30.06 g CH₃CH₃ = 1.8 g
When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a. Write and balance the chemical equation.
Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Liquid nitrogen boils at –195.8°C. Express the boiling point of liquid nitrogen in kelvin
Answer:
[tex]\boxed {\boxed {\sf 77.35 \ K}}[/tex]
Explanation:
The Celsius and Kelvin scales are used to measure the temperature of matter. Their scales and unit differences are the same (1 K increase = 1 °C increase), but they have different starting points.
So, the conversion is quite simple and only requires addition because of the different starting points. The formula is:
[tex]T_K = T_C+ 273.15[/tex]
The boiling point of liquid nitrogen is -195.8 °C. We can substitute this value into the formula.
[tex]T_K= -195.8 + 273.15[/tex]
[tex]T_K= 77.35 K[/tex]
The boiling point of liquid nitrogen is 77.35 Kelvin.
Which of state of matter has no definite shape but does have a definite volume?
A. Liquid
O B. Element
C. Solid
D. Gas
Answer:
A. Liquid .
Explanation:
Hello there!
In this case, according to the given information, it turns possible for us to to infer that the answer is A. Liquid because solids have both definite shape and volume and gases have no definite neither volume nor shape because they depend on the container.
This is this way because the molecules in the liquid are able to vibrate and slowly move around unlike gases and solids whereas molecules readily move and merely vibrate respectively.
Regards!
DATA SHEET p 45. TRIAL 1 TRIAL 2 1. Mass of the ground pretzel 1.00 gram 1.03 g 2. Initial volume of the AgNO3 solution 0.00 mL 9.10 mL 3. Final volume of the AgNO3 solution 9.10 mL 17.25 mL 4. Volume of AgNO3 solution used 9.10 mL 8.15 mL Line 3 – Line 2 5. Volume of AgNO3 solution in liters _____ L _____ L 6. Molarity of AgNO3 solution 0.01 M 0.01 M (given) 7. Number of moles of AgNO3 ______ mol _____ mol (Line 5 × Line 6) 8. Number of mol of NaCl present in pretzel ______ mol _____ mol (Line 7) number of mol NaCl = number of mol AgNO3 9. Mass of NaCl present in the titrated sample ______ gram _____ gram (Line 8) × 58.5 g/mol
Answer:
1. 1.00 gm
2. 50 ml
3. 38.93 ml
4. 11.07 ml
5. 0.01107 L
6. 0.010 moles / L
7. 0.0001107 moles
8. 0.0001107 moles
9. 0.00647042 grams
Explanation:
Silver nitrate can react with various compounds to form different products. The weight of products may be different from the original solution introduced due to combustion reaction, as heat energy is released during the chemical process.
In the process of preparing liquid air for fractional distillation, one fraction will be separated as a solid. What is the chemical name of this fraction?
Answer:
carbon dioxide CO₂
Explanation:
Each gas has a characteristic boiling point. You can separate a random sample of gases by gradually cooling the sample until each component gas liquifies. Some compounds, such as CO₂ never liquify. Instead, they turn directly into solids.
The fraction that will be separated as a solid in the process of liquefaction of air is carbon dioxide.
What is sublimation?
Sublimation is the process of changing the material from its solid to gaseous form without it being liquid, according to physics. An illustration is the evaporation of dry ice, which is frozen carbon dioxide, at typical atmospheric pressure and temperature. Vapour pressure and temperature correlations cause the phenomena.
Food is freeze-dried to preserve it by sublimating water from it while it is frozen under a strong vacuum. Phase is a term used in thermodynamics to describe an amount of matter that is chemically and physically uniform or homogeneous, can be mechanically isolated from a nonhomogeneous mixture, and may consist of a single material or a combination of substances.
The three basic phases of matter are solid, liquid, and gas (vapor), however additional phases, including crystalline, colloid, glassy, amorphous, and plasma, are thought to exist.
Therefore, during the liquefaction of air, the gas that will be separated as a solid is carbon dioxide.
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In a aqueous solution of 4-chlorobutanoic acid , what is the percentage of -chlorobutanoic acid that is dissociated
Answer:
Explanation:
Let assume that the missing aqueous solution of 4-chlorobutanoic acid = 0.76 M
Then, the dissociation of 4-chlorobutanoic acid can be expressed as:
[tex]\mathsf{C_3H_6ClCO_2H }[/tex] ⇄ [tex]\mathsf{C_3H_6ClCO_2^-}[/tex] + [tex]\mathsf{H^+}[/tex]
The ICE table can be computed as:
[tex]\mathsf{C_3H_6ClCO_2H }[/tex] ⇄ [tex]\mathsf{C_3H_6ClCO_2^-}[/tex] + [tex]\mathsf{H^+}[/tex]
Initial 0.76 - -
Change -x +x +x
Equilibrium 0.76 - x x x
[tex]K_a = \dfrac{[\mathsf{C_3H_6ClCO_2^-}] [\mathsf{H^+}]}{\mathsf{[C_3H_6ClCO_2H ]}}[/tex]
[tex]K_a = \dfrac{[x] [x]}{ [0.76-x]}[/tex]
where:
[tex]K_a = 3.02*10^{-5}[/tex]
[tex]3.02*10^{-5} = \dfrac{x^2}{ [0.76-x]}[/tex]
however, the value of x is so negligible:
0.76 -x = 0.76
Then:
[tex]3.02*10^{-5}*0.76 = x^2[/tex]
[tex]x=\sqrt{3.02*10^{-5}*0.76 }[/tex]
x = 0.00479 M
∴
[tex]x = \mathsf{[C_3H_6ClCO_2^-] = [H^+]=}[/tex] 0.00479 M
[tex]\mathsf{C_3H_6ClCO_2H }[/tex] = (0.76 - 0.00479) M
= 0.75521 M
Finally, the percentage of the acid dissociated is;
= ( 0.00479 / 0.76) × 100
= 0.630 M
0.50 g of hydrogen chloride (HCl) is dissolved in water to make 4.0 L of solution. What is the pH of the resulting hydrochloric acid solution
Explanation:
Given the mass of HCl is ---- 0.50 g
The volume of solution is --- 4.0 L
To determine the pH of the resulting solution, follow the below-shown procedure:
1. Calculate the number of moles of HCl given by using the formula:
[tex]number of moles of a substance=\frac{given mass of the substance}{its molecular mass}[/tex]
2. Calculate the molarity of HCl.
3. Calculate pH of the solution using the formula:
[tex]pH=-log[H^+][/tex]
Since HCl is a strong acid, it undergoes complete ionization when dissolved in water.
[tex]HCl(aq)->H^+(aq)+Cl^-(aq)[/tex]
Thus, [tex][HCl]=[H^+][/tex]
Calculation:
1. Number of moles of HCl given:
[tex]number of moles of a substance=\frac{given mass of the substance}{its molecular mass}\\=0.50g/36.5g/mol\\=0.0137mol[/tex]
2. Concentration of HCl:
[tex]Molarity of HCl=\frac{number of moles of HCl}{its molar mass}\\=\frac{0.0137 mol}{4.0 L} \\= 0.003425 M[/tex]
3. pH of the solution:
[tex]pH=-log[H^+]\\=-log(0.003425)\\=2.47[/tex]
Hence, pH of the given solution is 2.47.
Does water mess up carbon dating?
Answer:
THE hard-water effect is a recognized source of error in radiocarbon dating. It causes ages to be over-assessed and arises when the material to be dated, such as mollusc shell or plant, synthesizes its skeleton under water and so uses bicarbonate derived in part from old, inert sources.
https://www.nature.com/articles/240460a0#:~:text=THE%20hard%2Dwater%20effect%20is,part%20from%20old%2C%20inert%20sources.
Answer:
In radiocarbon dating, the hard-water effect has been identified as a cause of inaccuracy. It occurs when the item to be dated, such as a mollusk shell or a plant, synthesizes its skeleton underwater and therefore utilizes bicarbonate obtained in part from ancient, inert sources, resulting in the overestimation of ages.
OAmalOHopeO
A gas mixture, with a total pressure of 300. torr, consists of equal masses of Ne (atomic weight 20.) and Ar (atomic weight 40.). What is the partial pressure of Ar, in torr
Answer:
The partial pressure will be "100 torr".
Explanation:
Given:
[tex]P_{Ar} = 300 \ torr[/tex]
By assuming Ar and Ne having 50 gm each, we get
mol of Ne = [tex]\frac{50}{20}[/tex]
= [tex]2.5 \ mol[/tex]
mol of Ar = [tex]\frac{50}{40}[/tex]
= [tex]1.25 \ mol[/tex]
now,
[tex]n_T= mol.A_r+mol.N_e[/tex]
[tex]=1.25+2.5[/tex]
[tex]=3.75[/tex]
then,
[tex]X_{Ar}=\frac{n_{Ar}}{n_T}[/tex]
[tex]=\frac{1.25}{3.75}[/tex]
[tex]=0.33[/tex]
hence,
The partial pressure of Ar will be:
⇒ [tex]P_{Ar} = P_T\times X_{AT}[/tex]
By substituting the values, we get
[tex]=300\times 0.33[/tex]
[tex]=100 \ torr[/tex]
The partial pressure of Ar in the mixture is 99.9 torr
Let the mass of both gas be 10 g
Next, we shall determine mole of each gas.
For Ne:Mass = 10 g
Molar mass of Ne = 20 g/mol
Mole of Ne =?Mole = mass / molar mass
Mole of Ne = 10 / 20
Mole of Ne = 0.5 mole For Ar:Mass = 10 g
Molar mass of Ar = 40 g/mol
Mole of Ar =?Mole = mass / molar mass
Mole of Ar = 10 / 40
Mole of Ar = 0.25 moleNext, we shall determine the mole fraction of Ar
Mole of Ne = 0.5 mole
Mole of Ar = 0.25 mole
Total mole = 0.5 + 0.25 = 0.75 mole
Mole fraction of Ar =?[tex]mole \: fraction \: = \frac{mole}{total \: mole} \\ \\ mole \: fraction \: of \:Ar = \frac{0.25}{0.75} \\ \\ mole \: fraction \: of \:Ar = 0.333 \\ \\ [/tex]
Finally, we shall determine the partial pressure of Ar
Mole fraction of Ar = 0.333
Total pressure = 300 torr
Partial pressure of Ar =?Partial pressure = mole fraction × total pressure
Partial pressure of Ar = 0.333 × 300
Partial pressure of Ar = 99.9 torrLearn more on partial pressure: https://brainly.com/question/15577259
When butane reacts with Br2 in the presence of Cl2, both brominated and chlorinated products are obtained. Under such conditions, the usual selectivity of bromination is not observed. In other words, the ratio of 2-bromobutane to 1-bromobutane is very similar to the ratio of 2-chlorobutane to 1-chlorobutane. Can you offer and explanation as to why we do not observe the normal selectivity expected for bromination
Answer:
Bromine radical formation is carried out in the presence of Br₂ and Cl₂ causing the normal selectivity not to be observed ( this causes the difference in activation energy to be reduced )
Explanation:
Why the normal selectivity expected for bromination is not observed
On the basis of selectivity and applying the Arrhenius equation the greater the difference between the activation energies the more the selectivity.
as seen in the formation of primary and secondary radicals in the Bromine radical formation. this difference is caused mainly by the propagation step ( exothermic ) . But the main reason why the the usual selectivity of bromination is not observed is because it Bromine radical formation is carried out in the presence of Br₂ and Cl₂ ( this causes the difference in activation energy to be reduced )
Write the symbol for every chemical element that has atomic number greater than 70 and atomic mass less than 185.2
Answer:
HF...Ta... W....Lu...
what will the time for half life of the first order reaction?
Answer:
The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t1/2 = 0.693/k. Radioactive decay reactions are first-order reactions.
Explanation:
The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t1/2 = 0.693/k. Radioactive decay reactions are first-order reactions.
A molecule of acetone and a molecule of propyl aldehyde are both made from 3 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom. The molecules differ in their arrangement of atoms. How do formulas for the two compounds compare? Both compounds have the same molecular formula, but have unique structural formulas. Both compounds have unique molecular formulas and structural formulas. Both compounds have the same structural formula, but have unique molecular formulas.
Explanation:
The structures of both acetone and propanal are shown below:
In the formula of propanal there is -CHO functional group at the end.
In acetone -CO- group is present in the middle that is on the second carbon.
The molecular formula is C3H6O.
Both have same molecular formula but different structural formulas.
give reason why mendeleev's definition of transition elements is no longer acceptable
