Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 1.2 M
Molarity of diluted solution (M₂) = 0.5 M
Volume of diluted solution (V₂) = 100 mL
Volume of stock solution needed (V₁) =?The volume of stock solution needed can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂1.2 × V₁ = 0.5 × 100
1.2 × V₁ = 50
Divide both side by 1.2
V₁ = 50 / 1.2
V₁ ≈ 42 mLThus, 42 mL of the stock solution is needed.
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Answer:
They need 41.7 mL of the original stock solution.
Explanation:
We can use the following equation for dilutions:
Cc x Vc = Cd x Vd
Where Cc and Vc are the concentration and volume values in the concentrated condition, whereas Cd and Vd are the concentration and volume values in the diluted condition.
The concentrated solution is the original stock solution, and it has:
Cc = 1.2 M
The diluted solution must be:
Cd = 0.500 M
Vd = 100 mL
So, we have to calculate Vc. For this, we replace the data in the equation:
[tex]V_{c} = \frac{C_{d} V_{d} }{C_{c} } = \frac{(0.500 M)(100 mL)}{1.2 M} = 41.7 mL[/tex]
Therefore, 41.7 mL of 1.2 M original stock solution are required to make 100 mL of a diluted solution with a concentration of 0.500 M.
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:
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.
Tema: Métodos de Separação de Misturas – Homogêneas e Heterogêneas;
1. Capa (0,5 ponto)
2. Índice ou Sumário (0,5 ponto)
3. Texto do trabalho
a) Introdução (1,0 ponto)
b) Objetivos (0,5 ponto)
c) Método (0,5 ponto)
d) Desenvolvimento: Fundamentação Teórica (5,0 pontos)
e) Conclusão (1,0 ponto)
4. Bibliografia (1,0 ponto)
Answer:
fjskeowkcnekvo Dee five votes come vote for dog even r
Acid-catalyzed hydrolysis of a nitrile to give a carboxylic acid occurs by initial protonation of the nitrogen atom, followed by nucleophilic addition of water. Draw curved arrows to show the movement of electrons in this step of the reaction mechanism.
Answer:
See explanation and image attached
Explanation:
The acid-catalyzed hydrolysis of a nitrile to give a carboxylic acid occurs by initial protonation of the nitrogen atom. This step is shown in the image attached.
The next step is the nucleophilic addition of water. The task is to show the movement of electrons in this step of the reaction mechanism. This was clearly shown in the image attached to this answer.
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)
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.
In what order do electrons fill orbitals?
A. Orkjtals s, p, and then d fill in one energy level before starting the
next level.
B. Before pairing, 1 electron occupies each sand porbital.
C. Electrons fill orbitals in order of increasing orbital energy.
D. The p orbitals fill before the s orbitals in an energy level.
SUBMIT
Answer:
c......................
Answer:
electrons will fill the lowest energy orbitals first and then move up to higher energy orbitals only after the lower energy orbitals are full
Balance the following chemical equation.
CCl4 -> ___ C+ ___ Cl2
Answer:
Explanation:
CCl4 => C + 2Cl2
Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)5(OH)+(aq) + H3O+(aq) for which Ka = 1.0 x 10-8. Calculate the pH of 0.10 M Cu(NO3)2(aq).
Answer: The pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.
Explanation:
Given: Initial concentration of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] = 0.10 M
[tex]K_{a} = 1.0 \times 10^{-8}[/tex]
Let us assume that amount of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] dissociates is x. So, ICE table for dissociation of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] is as follows.
[tex]Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}[/tex]
Initial: 0.10 M 0 0
Change: -x +x +x
Equilibrium: (0.10 - x) M x x
As the value of [tex]K_{a}[/tex] is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.
And, (0.10 - x) will be approximately equal to 0.10 M.
The expression for [tex]K_{a}[/tex] value is as follows.
[tex]K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}[/tex]
Hence, [tex][H_{3}O^{+}] = 3.2 \times 10^{-5}[/tex]
Formula to calculate pH is as follows.
[tex]pH = -log [H^{+}][/tex]
Substitute the values into above formula as follows.
[tex]pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49[/tex]
Thus, we can conclude that the pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.
An unknown element, X, reacts with potassium to form the compound K2X. In other compounds this element also can accommodate up to 12 electrons rather than the usual octet. What element could X be
Answer:
Se
Explanation:
First of all, we must note that any element that we must choose is an element that is in group sixteen.
Elements of groups 16 have six electrons in their outermost shell which can be used for bond formation thereby yielding a total of twelve electrons on the valence shell.
However, this is only possible for the heavier members of the group 16 (from sulphur downwards) which are able to expand their octet.
Oxygen can not expand its octet hence it is not the answer.
Si enfriamos mercurio de 100C. Calcular la cantidad de calor que se debe restar sabiendo que la masa de mercurio es de 1800gr
Answer:
I do not speak Spanish.
Explanation:
us
Which of the following is a chemical property?
A. Hardness
B. Flammability
C. Malleability
D. Melting point
Reset Selection
Identify the solute and solvent in each solution:
a. 6mL of ethanol and 35mL of water
b. 300 g of water containing 8g of NaHCO3
c. 0.005LofCO2and2LofO2
Answer:
a. 6mL of ethanol and 35mL of water: the solute is ethanol (smallest volume) and the solvent is water (greater volume).
b. 300 g of water containing 8g of NaHCO3: the solute is NaHCO3 (smallest mass) and the solvent is water (greater mass).
c. 0.005L of CO2 and 2L of O2: the solute is CO2 (smallest volume) and the solvent is O2 (greater volume).
Explanation:
Hello there!
In this case, according to the given problem, it turns out possible for us to solve these questions by bearing to mind the fact that in a solution, we can find two substances, solute and solvent, whereas the former is in a smaller proportion in comparison to the latter; in such a way, we infer the following:
a. 6mL of ethanol and 35mL of water: the solute is ethanol (smallest volume) and the solvent is water (greater volume).
b. 300 g of water containing 8g of NaHCO3: the solute is NaHCO3 (smallest mass) and the solvent is water (greater mass).
c. 0.005L of CO2 and 2L of O2: the solute is CO2 (smallest volume) and the solvent is O2 (greater volume).
Regards!
2.50 L of 0.700 M phosphoric acid reacts with 5.25 moles of sodium hydroxide. How many moles of hydrogen ions will completely neutralize the moles of hydroxide ions present in this amount of sodium hydroxide? a) 0.583 b) 1.75 c) 3.00 d) 15.75 e) 5.25
Answer:
5.25 moles of protons. Option e
Explanation:
Reaction between phosphoric acid and sodium hydroxide is neutralization.
We can also say, we have an acid base equilibrium right here:
H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
Initially we have 5.25 moles of base.
We have data from the acid, to state its moles:
M = mol/L, so mol = M . L
mol = 1.75 moles of acid
If we think in the acid we know:
H₃PO₄ → 3H⁺ + PO₄⁻³
We know that 1 mol of acid can give 3 moles of protons (hydrogen ions)
If we have 1.75 moles of acid, we may have
(1.75 . 3) /1 = 5.25 moles of protons
These moles will be neutralized by the 5.25 moles of base
H₃O⁺ + OH⁻ ⇄ 2H₂O Kw
In a titration of a weak acid and a strong base, we have a basic pH
Which of the following are examples of physical properties of ethanol? Select all that apply.
The boiling point is 78.37°C
It is a clear, colorless liquid
It is flammable
It is a liquid at room temperature
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!
How are all atoms similar?
Answer:
Atoms are similar in the way that their nuclei contain only protons and neutrons.
Answer:
All things are made of atoms, and all atoms are made of the same three basic particles - protons, neutrons, and electrons. But, all atoms are not the same. The difference in the number of protons and neutrons in atoms account for many of the different properties of elements.
I hope this helps :)
How many grams of Ag2CO3 will precipitate when excess K2CO3 solution is added to 56.0 mL of 0.671 M AgNO3 solution?
Answer:
The mass of silver carbonate precipitated is 5.18 grams.
Explanation:
Molarity of the silver nitrate solution = 0.671 M
Volume of the silver nitrate solution = 56.0 mL
[tex]1 mL = 0.001 L\\56.0 mL = 56.0\times 0.001 L=0.0560 L[/tex]
Moles of silver nitrate = n
[tex]Molarity=\frac{\text{Moles of compound}}{\text{Volume of solution in Liters}}\\\\0.671 M=\frac{n}{0.0560 L}\\n=0.671 M\times 0.0560 L=0.0376 mol[/tex]
Moles of silver nitrate used = 0.0376 mol
[tex]K_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2KNO_3[/tex]
According to the reaction, 2 moles of silver nitrate gives 1 mole of silver carbonate, then 0.0376 moles of silver nitrate:
[tex]=\frac{1}{2}\times 0.0376 mol=0.0188 \text{mol of }Ag_2CO_3[/tex]
Moles of the silver carbonate formed = 0.0188 mol
Molar mass of silver carbonate = 275.7453 g/mol
Mass of silver carbonate :
[tex]=275.7453 g/mol\times 0.0188 mol=5.1840 g\approx 5.18 g[/tex]
The mass of silver carbonate precipitated is 5.18 grams.
write balanced half-reactions describing the oxidation and reduction that happen in this reaction 2Fe(s)+3Pb(NO3)2(aq)=3Pb(s)+2Fe(NO3)3(aq)
Answer:
Oxidation half-reaction: 2 Fe (s) ----> 2 Fe³+ (aq) + 6e-
Reduction half-reaction: 3 Pb²+ (aq) + 6 e- ----> 3 Pb (s)
Explanation:
A redox reaction reaction is one in which oxidation and reduction occur simultaneously and to the same extent.
Oxidation involves a loss of electron, hence, a positive increase in the oxidation number of the atom or ion. The oxidation half-reaction is as follows:
2 Fe (s) ----> 2 Fe³+ (aq)
The metallic element iron, Fe , having an oxidation number of zero, loses three electrons to form the Fe³+ ion with a charge of +3. Since each atom loses three electrons each, The number of moles of electrons lost is six.
2 Fe (s) ----> 2 Fe³+ (aq) + 6e-
Reduction involves a gain of electrons, hence, a decrease in the oxidation number of the atom or ion. The reduction half-reaction is given below:
3 Pb²+ (aq) ---> 3 Pb (s)
The lead (ii) ion, Pb²+ having a charge of +2 gains two electrons each to become the neutral metallic lead atom, Pb, with oxidation number of zero. Since 3 moles of Pb²+ are reacting, 6 moles of electrons are gained.
3 Pb²+ (aq) + 6 e- ----> 3 Pb (s)
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.
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.
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
A 0.4272 g sample of an element contains 2.241 x 10 ^21 atoms . what is the symbol of the element?
Answer:
Likely [tex]\rm In[/tex] (indium.)
Explanation:
Number of atoms: [tex]N = 2.241 \times 10^{21}[/tex].
Dividing, [tex]N[/tex], the number of atoms by the Avogadro constant, [tex]N_A \approx 6.023 \times 10^{23} \; \rm mol^{-1}[/tex], would give the number of moles of atoms in this sample:
[tex]\begin{aligned} n &= \frac{N}{N_{A}} \\ &\approx \frac{2.241 \times 10^{21}}{6.023 \times 10^{23}\; \rm mol^{-1}} \approx 3.72 \times 10^{-3}\; \rm mol \end{aligned}[/tex].
The mass of that many atom is [tex]m = 0.4272\; \rm g[/tex]. Estimate the average mass of one mole of atoms in this sample:
[tex]\begin{aligned}M &= \frac{m}{n} \\ &\approx \frac{0.4272\; \rm g}{3.72 \times 10^{-3}\; \rm mol} \approx 114.82\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
The average mass of one mole of atoms of an element ([tex]114.82\; \rm g \cdot mol^{-1}[/tex] in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass [tex]114.82[/tex]. Indium, [tex]\rm In[/tex], is the closest match.
The symbol of the element is In
StoichiometryFrom the question, we are to determine the identity of the element
First, we will determine the number of moles of sample present
Using the formula
[tex]Number \ of\ moles = \frac{Number\ of \ atoms }{Avogadro's\ constant}[/tex]
Number of moles of the sample = [tex]\frac{2.241\times 10^{21} }{6.022\times 10^{23} }[/tex]
Number of moles of the sample = 0.003721355 mole
Now, we will determine the Atomic mass of the sample
From the formula,
[tex]Atomic\ mass = \frac{Mass}{Number\ of\ moles}[/tex]
Therefore,
Atomic mass of the sample = [tex]\frac{0.4272}{0.003721355}[/tex]
Atomic mass of the sample = 114.8 amu
The element that has an atomic mass of 114.8 amu is Indium. The symbol of Indium is In.
Hence, the symbol of the element is In.
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A clean-burning automobile engine emits about 5 lb of C atoms in the form of CO2 molecules for every gallon of gasoline it consumes. The average American car is driven about 12,000 miles per year. Using this information, check the statement that the average American car releases its own weight in carbon into the atmosphere each year. List the assumptions you make to solve this problem.
Solution :
1 lb = 453.592 g
1 gallon = 3785 g
For every 5 gallon gasoline = 5 lb of C is found
= 5 x 453.592 g of C atoms
= 2267.96 g of C atoms
Assume the consumption of car = 15 miles per kg of gasoline
The amount of gasoline used per year [tex]$=\frac{12000}{15}$[/tex]
= 800 kg
In gallons = [tex]$\frac{800}{3.785} = 211.36\text{ gallons}$[/tex]
5 gallons will produce = 2267.96 g of C atoms
Therefore,
211.36 gallons will produce = [tex]$\frac{211.36 \times 2267.96}{5}$[/tex]
= 95871.21 g
= 95.87 kg
or = 25.32 gallon
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
PLEASE HELP ASAP MOLES TO MOLECULES
Answer:
4.77mol is the correct answer
In order to promote the common ion effect, the concentration of the common ion must first: _____________
a. increase
b. decrease
c. be equal to its equilibrium value
d. depends on the equilibrium
Answer:
a. increase.
Explanation:
Hello!
In this case, according to the given information, it turns out possible for us to tell that the common ion effect decreases the solubility of the ionic solid by firstly increasing the concentration of the common ion, which is further solved for the solubility in order to evidence the aforementioned decrease.
As an example, we can consider the solubility equilibrium for silver chloride:
[tex]Ksp=[Ag^+][Cl^-][/tex]
Which goes to:
[tex]Ksp=s^2[/tex]
Whereas s is the solubility to be solved for. However, when a silver- or chloride-containing solution is added, say 0.1 sodium chloride, the equilibrium expression changes to:
[tex]Ksp=(s)(s+0.1)[/tex]
Which turns out into a smaller value for s.
Regards!
During the postabsorptive state, metabolism adjusts to a catabolic state.
a. True
b. False
Answer:
The postabsorptive state (also called the fasting state) occurs when the food is already digested and absorbed, and it usually occurs overnight, when you sleep (if you skip meals for some days, you will enter in this state).
The catabolic state is the metabolic breakdown of molecules into simpler ones, releasing energy (heat) and utilizable resources.
Now, when you are in a postabsorptive state, the glucose levels start to drop, then the body starts to depend on the glycogen stores, which are catabolized into glucose, this is defined as the start of the postabsorptive state.
So yes, as the postabsorptive states, catabolic processes start to happen, so the statement is true.
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.
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 )
