Answer:
A)coupling between 1H nuclei attached to adjacent C atoms.
Explanation:
The word ‘vicinal’ in chemistry means three bonds from the functional groups. The two functional groups are in a relationship with the atoms in adjacent position to them.
The 1H nuclei consists of two Hydrogen nucleus which acts as the functional groups. They are however attached and in a relationship with the adjacent C atoms. This makes option A the right choice.
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppose 0.963 g of methane is mixed with 7.5 g of oxygen. Calculate the minimum mass of methane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
[tex]left\ over=0g[/tex]
Regards.
2. Which one is the odd one
out and why?
o Water
• Hydrogen
Chlorine
o Aluminum
Answer:
Reaction of Chlorine with Hydrogen Chlorine and Hydrogen mixed together explodes when exposed to sunlight, which produces Hydrogen Chloride. In the dark away from sunlight, no reaction occurs, so light energy is required for a reaction. Cl2 + H2 = 2 HCl Reaction of Chlorine with Non-Metals Chlorine directly combines with most non-metals.
Explanation:
I hope this helps bro
Which of the following contains a nonpolar covalent bond?
O A. Co
B. NaCl
O C. 02
O D. HE
Answer:
The answer is o2
Explanation:
I took the test
What compound is formed when methyloxirane (1,2-epoxypropane) is reacted with ethylmagnesium bromide followed by treatment with aqueous acid
Answer:
Pentan-2-ol
Explanation:
On this reaction, we have a Grignard reagent (ethylmagnesium bromide), therefore we will have the production of a carbanion (step 1). Then this carbanion can attack the least substituted carbon in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the treatment with aqueous acid, when we add acid the hydronium ion ([tex]H^+[/tex]) would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be attacked by the negative charge produced in the second step to produce the final molecule: "Pentan-2-ol".
See figure 1
I hope it helps!
19. For the following unbalanced equation: S (s) O2 (g) H2O (l) ----> H2SO4 (aq) At what temperature (K) does O2 have to be if the volume of the gas is 5.01 L with a pressure of 0.860 atm to produce 17.55 grams of H2SO4
Answer:
195.5 K
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2S + 3O2 + 2H2O → 2H2SO4
Next, we shall determine the number of mole in 17.55 g of H2SO4. This can be obtained as follow:
Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 + 64 = 98 g/mol
Mass of H2SO4 = 17.55 g
Mole of H2SO4 =...?
Mole = mass /Molar mass
Mole of H2SO4 = 17.55/98
Mole of H2SO4 = 0.179 mole.
Next, we shall determine the number of mole of O2 needed for the reaction. This can be obtained as illustrated below:
From the balanced equation above,
3 moles of O2 reacted to produce 2 moles of H2SO4.
Therefore, Xmol of O2 will react to produce 0.179 moleof H2SO4 i.e
Xmol of O2 = (3 x 0.179) / 2
Xmol of O2 = 0.2685 mole
Therefore, 0.2685 mole of O2 is needed for the reaction.
Finally, we shall determine the temperature of O2. This can be obtained by using the ideal gas equation as follow:
Volume (V) of O2 = 5.01 L
Pressure (P) = 0.860 atm
Number of mole (n) of O2 = 0.2685 mole
Gas constant (R) = 0.0821 atm.L/Kmol
Temperature (T) =..?
PV =nRT
0.860 x 5.01 = 0.2685 x 0.0821 x T
Divide both side by 0.2685 x 0.0821
T = (0.860 x 5.01) /(0.2685 x 0.0821)
T = 195.5 K
Therefore, the temperature of O2 must be 195.5 K.
The density of concentrated nitric acid (HNO3) is 1.413 g/mL. What volume in liters would be occupied by a mass of 47.2 g?
Answer:
The volume that a mass of 47.2 g would occupy is 0.0334 L
Explanation:
Density is the property that matter, whether solid, liquid or gas, has to compress into a given space. Density is defined as the amount of mass it has per unit volume, that is, the ratio between the mass of a body and the volume it occupies:
[tex]Density=\frac{mass}{volume}[/tex]
This indicates that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.
In this case:
density= 1.413 g/mLmass= 47.2 gvolume=?Replacing:
[tex]1.413 \frac{g}{mL}=\frac{47.2 g}{volume}[/tex]
Solving:
[tex]volume=\frac{47.2 g}{1.413\frac{g}{mL} }[/tex]
volume=33.40 mL
Being 1,000 mL= 1 L:
volume= 0.0334 L
The volume that a mass of 47.2 g would occupy is 0.0334 L
How many valence electrons must two atoms share to form a single covalent bond? answers A.2 B.4 C.3 D.1
Answer:
2
Explanation:
A single covalent bond is formed when two electrons are shared between the same two atoms, one electron from each atom.
Answer:
the answer is 2
Explanation:
A compound is found to contain 30.45 % nitrogen and 69.55 % oxygen by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 46.01 g/mol. What is the molecular formula for this compound?
Answer:
Empirical formulae is NO2
Molecular Formulae is NO2
If the amount of radioactive iodine-123, used to treat thyroid cancer, in a sample decreases from 3.2 to 0.4 mg in 39.6 h, what is the half-life of iodine-123?
Answer:
Half life = 13.197 hour
Explanation:
Given:
Old amount (A₀) = 3.2
New amount (A) = 0.4
Radiation decay time (t) = 39.6 hour
Half life = T(1/2)
Find:
Half life = T(1/2) = T
Computation:
A = A₀[tex]e^{-(\frac{0.693t}{T} )}[/tex]
[tex]e^{-(\frac{0.693t}{T} )}[/tex] = 0.4 / 3.2
-[27.4428 / T] = In (0.125)
-[27.4428 / T] = -2.0794
[27.4428 / T] = 2.0794
T = 13.197
Half life = 13.197 hour
A student completed the experiment but found that the total amount of material recovered weighed more than the original sample. What is the most likely source of error and how may it be corrected?
Answer:
This is due to the water moisture present in the recovered sample.
Explanation:
The total amount of material recovered isn’t meant to weigh more than the original sample. However when this happens then it means there is the presence of water moisture in the recovered sample.
The recovered samples however needs to be heated to make it dry and eliminate the water moisture through evaporation.
Why are cells important to an organisms survival
Answer:
Cells are the basic structures of all living organisms. Cells provide structure for the body, take in nutrients from food and carry out important functions. ... These organelles carry out tasks such as making proteins?, processing chemicals and generating energy for the cell
Answer: I absolutely love this question! Biology is so interesting, so I always love to answer the curiosity of others regarding biology, such as that!
Cells are simply the basic structures of all organisms, that are living, of course! Cells provide structure for the body, and they also take in nutrients that your body needs from food and they carry out important functions. These organelles carry out tasks such as making proteins, processing chemicals, and generating energy for the cell. Isn’t that cool?
Hope this helped! <3
An atom of 108Te has a mass of 107.929550 amu. Calculate the binding energy per MOLE in kJ. Use the values: mass of 1H atom
Answer:
The binding energy = 8.64972649×10¹⁰ kJ/mole
Explanation:
Given that:
An atom of 108Te has a mass of 107.929550 amu.
In a 108 Te atom, there are 52 protons and 56 neutrons
where;
mass of proton= 1.007825 amu
mass of neutron= 1.008665 amu
Similarly; The atomic number of Te = 52
the mass of 52 protons = 52 × 1.007825 amu
the mass of 52 protons = 52.4069 amu
the mass of 56 neutrons = 56 × 1.008665 amu
the mass of 56 neutrons = 56.48524 amu
The total mass can now be = the mass of 52 protons + the mass of 56 neutrons
The total mass = 52.4069 amu + 56.48524 amu
The total mass = 108.89214 amu
Recall : it is given that An atom of 108Te has a mass of 107.929550 amu.
Therefore, the mass defect will be = 108.89214 amu - 107.929550 amu
the mass defect = 0.96259amu
where 1 amu = 1.66× 10⁻²⁷ kg
Therefore, 0.96259amu = (0.96259 × 1.66× 10⁻²⁷) kg
= 1.5978994 × 10⁻²⁷kg
The binding energy = mass defect × (speed of light)²
where;
speed of light c = 2.99792 × 10⁸ m/s
The binding energy = 1.5978994 × 10⁻²⁷kg × 2.99792 × 10⁸ m/s
The binding energy = 1.43611597 × 10⁻¹⁰ J
The binding energy = 1.43611597 × 10⁻¹³ kJ/atom
since 1 mole = 6.023 × 10²³ atom (avogadro's constant)
Then;
The binding energy = ( 1.43611597 × 10⁻¹³ )× (6.023 × 10²³) kJ/mole
The binding energy = 8.64972649×10¹⁰ kJ/mole
How many atoms of oxygen are in one molecule of water (H2O)? one two four three
Answer:
there is one atom of oxygen and two atoms of hydrogen
Explanation:
The combination of a carbonyl group and a hydroxyl group on the same carbon atom is called a ________ group.
a. carbamate group
b. carbonate
c. carboxlate
d. carboxyl
Answer:
d. carboxyl
Explanation:
The presence of carbonyl group (>C=O)) and a hydroxyl group ( (−OH) on the same carbon atom is called a "carboxyl" group. A carboxyl group is represented as COOH and acts as the functional group part of carboxylic acids.
For example:
Formic acid or Methanoic acid (H-COOH) Butanoic acid (C3H7-COOH)Hence, the correct option is "d. carboxyl ".
For each of the following reactions calculate the mass (in grams) of both the reactants that are required to form 15.39g of the following products.
a. 2K(s) + Cl2(g) → 2Cl(aq)
b. 4Cr(s) + 302(g) → 2Cr2O3(s)
c. 35r(s) + N2(g) → SraNa(s)
Answer:
a.
[tex]m_K=8.056gK\\ \\m_{Cl_2}=4.028gCl_2[/tex]
b.
[tex]m_{Cr}=10.51gCr\\ \\m_{O_2}=4.851gO_2[/tex]
c.
[tex]m_{Sr}=13.88gSr\\\\m_{N_2}=1.479gN_2[/tex]
Explanation:
Hello,
In this case, we proceed via stoichiometry in order to compute the masses of all the reactants as shown below:
a. [tex]2K+Cl_2\rightarrow 2KCl[/tex]
[tex]m_K=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{2molK}{2molKCl}* \frac{39.1gK}{1molK}=8.056gK\\ \\m_{Cl_2}=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{1molCl_2}{2molKCl}* \frac{70.9gCl_2}{1molCl_2}=4.028gCl_2[/tex]
b. [tex]4Cr+ 3O_2\rightarrow 2Cr_2O_3[/tex]
[tex]m_{Cr}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{4molCr}{2molCr_2O_3}* \frac{52gCr}{1molCr_2O_3}=10.51gCr\\ \\m_{O_2}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{3molO_2}{2molCr_2O_3}* \frac{32gO_2}{1molCr_2O_3}=4.851gO_2[/tex]
c. [tex]3Sr(s) + N_2(g) \rightarrow Sr_3N_2[/tex]
[tex]m_{Sr}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{3molSr}{1molSr_3N_2}* \frac{87.62gSr}{1molSr}=13.88gSr\\\\m_{N_2}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{1molN_2}{1molSr_3N_2}* \frac{28gN_2}{1molN_2}=1.479gN_2[/tex]
Regards.
Will a precipitate (ppt) form when 300. mL of 2.0 × 10 –5 M AgNO 3 are added to 200. mL of 2.5 × 10 –9 M NaI? Answer yes or no, and identify the precipitate if there is one
Answer:
A precipitate will form, AgI
Explanation:
When Ag⁺ and I⁻ ions are in an aqueous media, AgI(s), a precipitate, is produced or not based on its Ksp expression:
Ksp = 8.3x10⁻¹⁷ = [Ag⁺] [I⁻]
Where the concentrations of the ions are the concentrations in equilibrium
For actual concentrations of a solution, you can define Q, reaction quotient, as:
Q = [Ag⁺] [I⁻]
If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form.
Actual concentrations of Ag⁺ and I⁻ are:
[Ag⁺] = [AgNO₃] = 2.0x10⁻⁵ × (300mL / 500.0mL) = 1.2x10⁻⁵M
[I⁻] = [NaI] = 2.5x10⁻⁹ × (200mL / 500.0mL) = 1.0x10⁻⁹M
500.0mL is the volume of the mixture of the solutions
Replacing in Q expression:
Q = [Ag⁺] [I⁻]
Q = [1.2x10⁻⁵M] [1.0x10⁻⁹M]
Q = 1.2x10⁻¹⁴
As Q > Ksp
A precipitate will form, AgIRead the chemical equation. Mg + 2HCl → MgCl2 + H2 How many moles of MgCl2 are produced from 1 mole of HCl? 0.2 0.5 1.0 1.5
Answer:
0.5 mol MgCl₂
Explanation:
Step 1: Write the balanced equation
Mg + 2 HCl → MgCl₂ + H₂
In words, 1 mole of Mg reacts with 2 moles of HCl to form 1 mole of MgCl₂ and 1 mole of H₂.
Step 2: Establish the appropriate molar ratio
The molar ratio of HCl to MgCl₂ is 2:1.
Step 3: Calculate the moles of MgCl₂ produced from 1 mole of HCl
1 mol HCl × (1 mol MgCl₂/2 mol HCl) = 0.5 mol MgCl₂
Answer:
it is 2.0, the above one is wrong
Explanation:
I did the test :
. You have two solutions, both with a concentration of 0.1M. Solution A contains a weak acid with a pKa of 5. ThepH of solution A is 3. Solution B contains a weak acid with a pKa of 9. The pH of solution B is:
Answer:
pH of solution B is 5
Explanation:
A weak acid, HA, is in equilibrium with water as follows:
HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)
Where Ka (10^-pKa = 1x10⁻⁹) is:
Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
Where concentrations of this species are equilibrium concentrations
As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:
[HA] = 0.1M - X
[A⁻] = X
[H₃O⁺] = X
Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.
Replacing in Ka expression:
1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
1x10⁻⁹ = [X] [X] / [0.1 - X]
1x10⁻¹⁰ - 1x10⁻⁹X = X²
1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0
Solving for X:
X = -0.00001 → False solution, there is no negative concentrations.
X = 1x10⁻⁵ → Right solution.
As [H₃O⁺] = X
[H₃O⁺] = 1x10⁻⁵M
And pH = -log[H₃O⁺]
pH = 5
pH of solution B is 5
Calculate LaTeX: \DeltaΔGº for a voltaic cell with Eºcell = +0.24 V if the overall reaction involves a 3 electron reduction.
Answer:
-69 kJ
Explanation:
Step 1: Given data
Standard cell potential (E°cell): +0.24 V
Electrons involved (n): 3 mol
Step 2: Calculate the standard Gibbs free energy change (ΔG°) for the voltaic cell
We will use the following expression.
ΔG° = -n × F × E°cell
where,
F is Faraday's constant (96,485 C/mol e⁻)
ΔG° = -n × F × E°cell
ΔG° = -3 mol e⁻ × 96,485 C/mol e⁻ × 0.24 V
ΔG° = -69 kJ
1500 L has how many significants figures
Answer:
It has 2
Explanation:
The significant figures are 1 and 5!
Hope this helps:)
The compound sodium hydroxide is a strong electrolyte. Write the transformation that occurs when solid sodium hydroxide dissolves in water. Use the pull-down boxes to specify states such as (aq) or (s).
Answer:
Solid sodium hydroxide dissolves in water to form an aqueous solution of ions.
NaOH(s) ⇌ Na+(aq) + OH–(aq) ΔH1 = ?
Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride.
NaOH(s) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH2 = ?
Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride.
Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH3 = ?
For the following reaction, 22.8 grams of diphosphorus pentoxide are allowed to react with 13.5 grams of water . diphosphorus pentoxide(s) water(l) phosphoric acid(aq) What is the maximum mass of phosphoric acid that can be formed
Answer:
[tex]m_{H_3PO_4}=31.5gH_3PO_4[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]P_2O_5(s)+3H_2O(l)\rightarrow 2H_3PO_4(aq)[/tex]
Thus, since the diphosphorus pentoxide to water molar ratio is 1:3 and we are given the mass of both of them, for the calculation of the maximum mass phosphoric acid that is yielded, one could first identify the limiting reactant, for which we compute the available moles of diphosphorus pentoxide (molar mass 142 g/mol):
[tex]n_{P_2O_5}=22.8gP_2O_5*\frac{1molP_2O_5}{142gP_2O_5}=0.161molP_2O_5[/tex]
And the moles of diphosphorus pentoxide that are consumed by 13.5 g of water (molar mass 18 g/mol):
[tex]n_{P_2O_5}^{consumed}=13.5gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molP_2O_5}{3molH_2O} =0.25molP_2O_5[/tex]
Hence, since less moles of diphosphorus pentoxide are available, we sum up it is the limiting reactant, therefore, the maximum mass of phosphoric acid (molar mass 98 g/mol) is computed by considering the 1:2 molar ratio between them as follows:
[tex]m_{H_3PO_4}=0.161molP_2O_5*\frac{2molH_3PO_4}{1molP_2O_5} *\frac{98gH_3PO_4}{1molH_3PO_4} \\\\m_{H_3PO_4}=31.5gH_3PO_4[/tex]
Regards.
Need help with chemistry questions
Answer:
1. oxidation
2. reduction
3. oxidation
4. oxidation
Explanation:
Oxidation and Reduction in terms of hydrogen
Oxidation and Reduction with respect to Hydrogen Transfer. Oxidation is the loss of hydrogen. Reduction is the gain of hydrogen.Oxidation and Reduction in terms of Oxygen
Oxidation and Reduction with respect to Oxygen Transfer. Oxidation is the gain of Oxygen. Reduction is the loss of Oxygen.A spontaneous galvanic cell consists of a Pb electrode in a 1.0 M Pb(NO3)2 solution and a Cd electrode in a 1.0 M Cd(NO3)2 solution. What is the standard cell potential for this galvanic cell
Answer:
0.27 V
Explanation:
Given that the both half cells contain 1.0 molar solutions of their respective electrolytes.
E°Pb= -0.13 V
E°Cd = -0.40 V
Since it is a galvanic cell, the electrode having a more negative electrode potential will serve as the anode and the electrode having a less negative electrode potential will serve as the cathode.
Hence cadmium will serve as the anode and lead will serve as the cathode.
E°cell = E°cathode - E°anode
E°cell = -0.13 - (-0.40)
E°cell = 0.27 V
A student ran the following reaction in the laboratory at 242 K: 2NOBr(g) 2NO(g) Br2(g) When she introduced 0.143 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of NOBr(g) to be 0.108 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc
Answer:
1.84 × 10⁻³
Explanation:
Step 1: Write the balanced equation
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
Step 2: Calculate the initial concentration of NOBr
0.143 moles of NOBr(g) are introduced into a 1.00 liter container. The molarity is:
M = 0.143 mol / 1.00 L = 0.143 M
Step 3: Make an ICE chart
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
I 0.143 0 0
C -2x +2x +x
E 0.143-2x 2x x
Step 4: Find the value of x
The equilibrium concentration of NOBr(g) was 0.108 M. Then,
0.143-2x = 0.108
x = 0.0175
Step 5: Calculate the concentrations at equilibrium
[NOBr] = 0.108 M
[NO] = 2x = 0.0350 M
[Br₂] = x = 0.0175 M
Step 6: Calculate the equilibrium constant (Kc)
Kc = [0.0350]² × [0.0175] / [0.108]²
Kc = 1.84 × 10⁻³
Which of the following compounds is more soluble in a 0.10 M NaCN solution than in pure neutral water? Ca3(PO4)2 AgBr CaCO3 Mg(OH)2 NH4ClO4
Answer:
AgBr
Explanation:
Silver bromide has a very low solubility product constant of about 7.7 ×10^-13 in pure water hence it is not quite soluble in pure water.
However, with NaCN, the AgBr forms the complex [Ag(CN)2]^2- which has a formation constant of about 5.6 ×10^8. This very high formation constant implies that the complex is easily formed leading to the dissolution of AgBr in NaCN.
The equation for the dissolution of AgBr in cyanide is shown below;
AgBr(s) + 2CN^-(aq) ----> [Ag(CN)2]^2-(aq) + Br^-(aq)
Methyl iodide reacts irreversibly with azide ion with rate = k[CH3I][N3–]. CH3I(aq) + N3–(aq) → CH3N3(aq) + I–(aq) The reaction is carried out with an initial concentration of CH3I of 0.01 M. Which statement about the reaction is correct?
Answer:
(D) The reaction cannot take place in a single elementary step
Explanation:
Statements are:
(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I].
B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M.
(C) The reaction rate is significantly smaller if excess I- is added to the solution.
(D) The reaction cannot take place in a single elementary step.
The rate of the reaction is:
rate = k[CH3I][N3–].
That means rate depends of concentration of CH₃I as much as N₃⁻ concentration
(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I]. FALSE. The reaction rate depends of N₃⁻ as much as CH₃I
B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M. FALSE. Reaction is second-order. Half-life is 1/K[A]₀. If initial concentration is 0.1M, to a concentration of 0.005M it takes:
1/K*0.1. If initial concentration is 0.005M it takes 1/K*0.005. That means it takes half to decrease from 0.005M to 0.0025 as it does for it to decrease from 0.01M to 0.005M.
(C) The reaction rate is significantly smaller if excess I- is added to the solution. FALSE. Reaction rate is independent of I⁻
(D) The reaction cannot take place in a single elementary step. TRUE. As this reaction is a single-replacement reaction implies the formation of 1 C-N bond. But also the rupture of the C-I bond is impossible to explain this kind of reaction in a single elementary step.
A sample of radioactive silver contains two isotopes, 108Ag (denoted A) and 110Ag (denoted B). The second of these (B) has a half life of 24 seconds, whereas the first (A) has a half life of 2.3 minutes. If a sample contains equal numbers of each of these isotopes at the beginning of an experiment that runs for an hour, which of the following statements is correct?
A. At the end of the hour, isotope B has a greater decay constant λ than isotope A
B. At the end of the hour, isotope A has the same decay constant λ as isotope B
C. At the end of the hour, isotope A has a greater decay constant λ than isotope B
Answer:
A : At the end of the hour, isotope B has a greater decay constant λ than isotope A
Explanation:
Firstly, we need to understand that radioactive decay follows a first order rate law.
What this means is that we can calculate the radioactive decay constant using the following formula from the half-life
Mathematically;
[tex]t_{1/2}[/tex] = 0.693/λ
where λ represents the radioactive decay constant.
Rearranging the equation, we can have
λ = 0.693/[tex]t_{1/2}[/tex]
Now, to have a fair level playing ground, it is best that the half-life of both isotopes are in the same unit of time(seconds)
For A, the half-life = 2.3 minutes which is same as 2.3 × 60 = 138 seconds
For B, the half-life is 24 seconds
Thus, at the end of the hour, the decay constant for isotope A will be;
λ = 0.693/138 = 0.0050 [tex]s^{-1}[/tex]
For isotope B, the decay constant will be;
λ = 0.693/24 = 0.028875 [tex]s^{-1}[/tex]
We can see that the decay constant of isotope B is higher than that of A at the end of the experiment
In a buffer solution made of acetic acid and sodium acetate, if a small amount of acid is added, the added acid will react with whome?
Answer:
The acid reacts with the conjugate base producing more weak acid.
Explanation:
A buffer solution is defined as the mixture of a weak acid and its conjugate base or a weak base with its conjugate acid.
The acetic buffer, CH₃COOH/CH₃COO⁻, is in equilibrium with water as follows:
CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺
When an acid HX (Source of H₃O⁺) is added to the buffer, the reaction that occurs is:
CH₃COO⁻ + HX → CH₃COOH
The acid reacts with the conjugate base producing more weak acid.In fact, this is the principle of the buffer:
An acid reacts with the conjugate base producing weak acid. And the weak acid reacts with a base producing conjugate base
If a small amount of acid is added to an acetic acid-sodium acetate buffer, the acid will react with the acetate ion from sodium acetate.
We have a buffer formed by acetic acid and sodium acetate.
What is a buffer?A buffer is a solution used to resist abrupt changes in pH when an acid or a base is added.
How are buffers formed?They can be formed in 1 of 2 ways:
By a weak acid and its conjugate base.By a weak base and its conjugate acid.Our buffer is formed by a weak acid (acetic acid) and its conjugate base (acetate ion from sodium acetate).
When an acid (HX) is added, it is neutralized by the basic component of the buffer. The generic net ionic equation is:
H⁺ + CH₃COO⁻ ⇄ CH₃COOH
If a small amount of acid is added to an acetic acid-sodium acetate buffer, the acid will react with the acetate ion from sodium acetate.
Learn more about buffers here: https://brainly.com/question/24188850
All of the following are examples of allotropes of carbon EXCEPT Choose the one alternative that best completes the statement or answers the question. Choose the one alternative that best completes the statement or answers the question. diamond amorphous carbon quartz graphene all of the above
Answer:
quartz
Explanation:
The correct option would be quartz.
Allotropy is a phenomenon that describes the natural existence of the same element in different forms with different physical characteristics. Allotropes are therefore different forms of the same element.
Carbon as an element has several allotropes which include diamond, graphite, graphene, amorphous carbon, and fullerenes. Quartz is a crystalline solid that is composed of silicon dioxide and not carbon.
Hence, all the options are carbon allotropes except quartz.
