Answer:
Approximately [tex]10.88[/tex].
Explanation:
Equilibrium constant[tex]\rm OBr^{-}[/tex] can act as a weak Bronsted-Lowry base:
[tex]\rm OBr^{-}\; (aq) + H_2O\; (l) \rightleftharpoons HOBr\; (l) + OH^{-}\; (aq)[/tex].
(Side note: the state symbol of [tex]\rm HOBr[/tex] in this equation is [tex]\rm (l)[/tex] (meaning liquid) because [tex]\rm HOBr[/tex] is a weak acid.)
However, the equilibrium constant of this reaction, [tex]K_\text{eq}[/tex], isn't directly given. The idea is to find [tex]K_\text{eq}[/tex] using the [tex]\rm pH[/tex] value at the half-equivalence point. Keep in mind that this system is at equilibrium all the time during the titration. If temperature stays the same, then the same [tex]K_\text{eq}[/tex] value could also be used to find the [tex]\rm pH[/tex] of the solution before the acid was added.
At equilibrium:
[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].
At the half-equivalence point of this titration, exactly half of the base, [tex]\rm OBr^{-}[/tex], has been converted to its conjugate acid, [tex]\rm HOBr[/tex]. Therefore, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should both be equal to one-half the initial concentration of [tex]\rm OBr^{-}[/tex].
As a result, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should be the same. The expression for [tex]K_\text{eq}[/tex] can thus be simplified:
[tex]\begin{aligned}& K_\text{eq} \\&= \frac{\left(\text{half-equivalence $[\rm HOBr\; (l)]$}\right)\cdot \left(\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\right)}{\text{half-equivalence $[\rm OBr^{-}\; (l)]$}}\\ &=\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\end{aligned}[/tex].
In other words, the [tex]K_\text{eq}[/tex] of this system is equal to the [tex]\rm OH^{-}[/tex] concentration at the half-equivalence point. Assume that [tex]\rm p\mathnormal{K}_\text{w}[/tex] the self-ionization constant of water, is [tex]14[/tex]. The concentration of [tex]\rm OH^{-}[/tex] can be found from the [tex]\rm pH[/tex] value:
[tex]\begin{aligned}& \text{half-equivalence $[\rm OH^{-}\; (aq)]$} \\ &= 10^{\rm pH - p\mathnormal{K}_\text{w}}\;\rm mol \cdot L^{-1} \\ &= 10^{7.75 - 14}\; \rm mol \cdot L^{-1}\\ &= 10^{-6.25}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
Therefore, [tex]\begin{aligned} K_\text{eq} &= 10^{-6.25}\end{aligned}[/tex].
Initial pH of the solutionAgain, since [tex]\rm KOBr[/tex] is a soluble salt, all that [tex]0.200\; \rm M[/tex] of [tex]\rm KOBr[/tex] in this solution will be in the form of [tex]\rm K^{+}[/tex] and [tex]\rm OBr^{-}[/tex] ions. Before any hydrolysis takes place, the concentration of [tex]\rm OBr^{-}[/tex] should be equal to that of [tex]\rm KOBr[/tex]. Therefore:
[tex]\text{$[\rm OBr^{-}\; (aq)]$ before hydrolysis} = 0.200\; \rm M[/tex].
Let the equilibrium concentration of [tex][\rm OH^{-}\; (aq)][/tex] be [tex]x\; \rm M[/tex]. Create a RICE table for this reversible reaction:
[tex]\begin{array}{c|ccccccc} & \rm OBr^{-}\; (aq) &+&\rm H_2O\; (l)& \rightleftharpoons & \rm HOBr\; (l)& + & \rm OH^{-}\; (aq) \\ \textbf{I}& 0.200\; \rm M & & & & 0 \; \rm M & & 0\; \rm M \\ \textbf{C} & -x\; \rm M & & & & +x \; \rm M & & +x\; \rm M \\ \textbf{E}& (0.200 + x)\; \rm M & & & & x \; \rm M & & x\; \rm M \end{array}[/tex].
Assume that external factors (such as temperature) stays the same. The [tex]K_\text{eq}[/tex] found at the half-equivalence point should apply here, as well.
[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].
At equilibrium:
[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x}[/tex].
Assume that [tex]x[/tex] is much smaller than [tex]0.200[/tex], such that the denominator is approximately the same as [tex]0.200[/tex]:
[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x} \approx \frac{x^2}{0.200}[/tex].
That should be equal to the equilibrium constant, [tex]K_\text{eq}[/tex]. In other words:
[tex]\displaystyle \frac{x^2}{0.200} \approx K_\text{eq} = 10^{-6.25}[/tex].
Solve for [tex]x[/tex]:
[tex]x \approx 3.35\times 10^{-4}[/tex].
In other words, the [tex]\rm OH^{-}[/tex] before acid was added was approximately [tex]3.35\times 10^{-4}\; \rm M[/tex], which is the same as [tex]3.35\times 10^{-4}\; \rm mol \cdot L^{-1}[/tex]. Again, assume that [tex]\rm p\mathnormal{K}_\text{w} = 14[/tex]. Calculate the [tex]\rm pH[/tex] of that solution:
[tex]\begin{aligned}\rm pH &= \rm p\mathnormal{K}_\text{w} + \log [\mathrm{OH^{-}}] \approx 10.88\end{aligned}[/tex].
(Rounded to two decimal places.)
A sample of an unknown gas effuses in 11.1 min. An equal volume of H2 in the same apparatus at the same temperature and pressure effuses in 2.42 min. What is the molar mass of the unknown gas
Answer:
Molar mass of the gas is 0.0961 g/mol
Explanation:
The effusion rate of an unknown gas = 11.1 min
rate of [tex]H_{2}[/tex] effusion = 2.42 min
molar mass of hydrogen = 1 x 2 = 2 g/m
molar mas of unknown gas = ?
From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.
from
[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]
where
[tex]R_{h}[/tex] = rate of effusion of hydrogen gas
[tex]R_{g}[/tex] = rate of effusion of unknown gas
[tex]M_{h}[/tex] = molar mass of H2 gas
[tex]M_{g}[/tex] = molar mass of unknown gas
substituting values, we have
[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587
[tex]\sqrt{M_{g} }[/tex] = 0.31
[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol
The molar mass of the unknown gas will be "0.0961 g/mol".
Given:
Effusion rate of unknown gas,
[tex]R_g = 11.1 \ min[/tex]Effusion rate of [tex]H_2[/tex],
[tex]R_h = 2.42 \ min[/tex]Molar mass of hydrogen,
[tex]M_h = 1\times 2[/tex][tex]= 2 \ g/m[/tex]
According to the Graham's law, we get
→ [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]
By substituting the values, we get
→ [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]
→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]
→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]
[tex]\sqrt{M_g} = 0.31[/tex]
[tex]M_g = 0.0961 \ g/mol[/tex]
Thus the above solution is right.
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Diluting sulfuric acid with water is highly exothermic:
(Use data from the Appendix to find for diluting 1.00 mol of H2SO4(l) (d = 1.83 g/mL) to 1 L of 1.00 MH2SO4(aq) (d = 1.060 g/mL). )
Suppose you carry out the dilution in a calorimeter. The initial T is 25.2°C, and the specific heat capacity of the final solution is 3.458 J/gK. What is the final T in °C ?
Answer:
The correct answer is 51.2 degree C.
Explanation:
The standard enthalpy for H₂SO₄ (l) is -814 kJ/mole and the standard enthalpy for H₂SO₄ (aq) is -909.3 kJ/mole.
Now the dHreaction = dHf (product) - dHf (reactant)
= -909.3 - (-814)
dHreaction or q = -95.3 kJ of energy will be used for dissociating one mole of H₂SO₄.
The heat change in calorimetry can be determined by using the formula,
q = mass * specific heat capacity * change in temperature -----------(i)
Based on the given information, the density of H₂SO₄ is 1.060 g/ml
The volume of H₂SO₄ is 1 Liter
Therefore, the mass of H₂SO₄ will be, density/Volume = 1.060 g/ml / 1 × 10⁻³ ml = 1060 grams
The initial temperature given is 25.2 degrees C, or 273+25.2 = 298.2 K, let us consider the final temperature to be T₂.
ΔT = T₂ -T₁ = T₂ - 298.2 K
Now putting the values in equation (i) we get,
95.3 kJ = 1060 grams × 3.458 j/gK (T₂ - 298.2 K) (the specific heat capacity of the final solution is 3.458 J/gK)
(T₂ - 298.2 K) = 95300 J / 1060 × 3.458 = 26 K
T₂ = 298.2 K + 26 K
T₂ = 324.2 K or 324.2 - 273 = 51.2 degree C.
Which sample is most likely to experience the smallest temperature change upon observing 55KJ of heat? 
Answer:
100 g of water: specific heat of water 4.18 J/g°C
Explanation:
To know the correct answer to the question, we shall determine the temperature change in each case.
For 100 g of water:
Mass (M) = 100 g
Specific heat capacity (C) = 4.18 J/g°C
Heat absorbed (Q) = 55 KJ = 55000 J
Change in temperature (ΔT) =..?
Q = MCΔT
55000 = 100 x 4.18 x ΔT
Divide both side by 100 x 4.18
ΔT = 55000/ (100 x 4.18)
ΔT = 131.6 °C
Therefore the temperature change is 131.6 °C
For 50 g of water:
Mass (M) = 50 g
Specific heat capacity (C) = 4.18 J/g°C
Heat absorbed (Q) = 55 KJ = 55000 J
Change in temperature (ΔT) =..?
Q = MCΔT
55000 = 50 x 4.18 x ΔT
Divide both side by 50 x 4.18
ΔT = 55000/ (50 x 4.18)
ΔT = 263.2 °C
Therefore the temperature change is 263.2 °C
For 50 g of lead:
Mass (M) = 50 g
Specific heat capacity (C) = 0.128 J/g°C
Heat absorbed (Q) = 55 KJ = 55000 J
Change in temperature (ΔT) =..?
Q = MCΔT
55000 = 50 x 0.128 x ΔT
Divide both side by 50 x 0.128
ΔT = 55000/ (50 x 0.128)
ΔT = 8593.8 °C
Therefore the temperature change is 8593.8 °C.
For 100 g of iron:
Mass (M) = 100 g
Specific heat capacity (C) = 0.449 J/g°C
Heat absorbed (Q) = 55 KJ = 55000 J
Change in temperature (ΔT) =..?
Q = MCΔT
55000 = 100 x 0.449 x ΔT
Divide both side by 100 x 0.449
ΔT = 55000/ (100 x 0.449)
ΔT = 1224.9 °C
Therefore the temperature change is 1224.9 °C.
The table below gives the summary of the temperature change of each substance:
Mass >>> Substance >> Temp. Change
100 g >>> Water >>>>>> 131.6 °C
50 g >>>> Water >>>>>> 263.2 °C
50 g >>>> Lead >>>>>>> 8593.8 °C
100 g >>> Iron >>>>>>>> 1224.9 °C
From the table given above we can see that 100 g of water has the smallest temperature change.
A vehicle travels 2345 meter in 35 second toward the evening sun in the West. What is its speed? A. 47 m/s West
Explanation:
Speed = 2345 ÷ 35 = 67m/s
Determine the volume occupied by 10 mol of helium at 27 ° C and 82 atm
please.
Answer:
3.00 L
Explanation:
Convert the pressure to Pascals.
P = 82 atm × (101325 Pa/atm)
P = 8,308,650 Pa
Convert temperature to Kelvins.
T = 27°C + 273
T = 300 K
Use ideal gas law:
PV = nRT
(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)
V = 0.00300 m³
If desired, convert to liters.
V = (0.00300 m³) (1000 L/m³)
V = 3.00 L
Answer:
[tex]\large \boxed{\text{3.0 L}}[/tex]
Explanation:
[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]
The equilibrium between carbon dioxide gas and carbonic acid is very important in biology and environmental science. CO2 ( aq) + H2O ( l) H2CO3 ( aq) Which one of the following is the correct equilibrium constant expression (K c) for this reaction?
a) K =[H2CO3]/ [CO2]
b) K=[CO2]/ [H2CO3]
c) K=[H2CO3]/ [CO2][H2O]
d) K=[CO2][H2O]/ [H2CO3]
e) K=1/[H2CO3]
Answer:
Kc = [H₂CO₃] / [CO₂]
Explanation:
Equilibrium constant expression (Kc) of any reaction is defined as the ratio between molar concentrations in equilibrium of products over reactants.
Pure solids and liquids don't affect the equilibrium and you don't have to take its concentrations in the equilibrium.
Also, each specie must be powered to its reactant coefficient.
For example, for the reaction:
aA(s) + bB(aq) ⇄ cC(l) + nD(g) + xE(aq)
The equilibrium constant, kc is:
Kc = [D]ⁿ / [B]ᵇ[E]ˣ
You don't take A nor C species because are pure solids and liquids. b, n and x are the reactant coefficients of each substance. Ratio of products over reactants
Thus, for the reaction:
CO₂(aq) + H₂O(l) ⇄ H₂CO₃(aq)
The Kc is:
Kc = [H₂CO₃] / [CO₂]
what is chemical equation of Braium chloride?
Answer:
BaCl2
Explanation:
Barium = Ba
Chloride => Cl-
Chemical Equation:
Ba + Cl => BaCl2
Note:
The valency of barium is 2 and valency of chloride is 1 (i.e. chlorine). The formula formed by the combination of these elements is BaCl2 (there's exchange of valencies when these two elements combine).
Write electron configurations for the following ion: Cd2 Cd2 . Express your answer in order of increasing orbital energy. For example, the electron configuration of LiLi would be entered in complete form as 1s^22s^1 or in condensed form as [He]2s^1.
Answer:
Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰
Explanation:
Before proceeding to write out the electron configuration of Cd2+, we have to obtain the electron configuration of Cadmium (Cd),
Cadmium has an atomic number of 48, this means that a neutral cadmium atom will have a total of 48 electrons surrounding its nucleus.
The electronic configuration of Cadmium is;
Cd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10
The shorthand notation is given as;
Cd: [Kr] 4d¹⁰5s²
Cd2+ means that it has two less electrons, hence it's electron configuration is given as;
Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰
Draw the structure of 1,4-hexanediamine.
Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced TemplateTowbars. The single bond is active by default. Include all hydrogen atoms.
View Available Hint(s)
Answer:
1,4-hexanediamine contains two [tex]-NH_{2}[/tex] functional groups.
Explanation:
1,4-hexanediamine is an organic molecule which contains two [tex]-NH_{2}[/tex] functional groups at C-1 and C-4 position.
The longest carbon chain in 1,4-hexanediamine contains six carbon atoms.
Molecular formula of 1,4-hexanediamine is [tex]C_{6}H_{16}N_{2}[/tex].
1,4-hexanediamine used as a bidentate ligand in organometallic chemistry.
The structure of 1,4-hexanediamine is shown below.
Phosphorus pentafluoride, PF5, acts as a __________ during the formation of the anion PF−6. Select the correct answer below: A. Lewis acid B. Lewis base C. catalyst D. drying agent
Answer:
Lewis acid
Explanation:
In chemistry, a Lewis acid is any chemical specie that accepts a lone pair of electrons while a Lewis base is any chemical specie that donates a lone pair of electrons.
If we look at the formation of PF6^-, the process is as follows;
PF5 + F^- -----> PF6^-
We can see that PF5 accepted a lone pair of electrons from F^- making PF5 a lewis acid according to our definition above.
Hence in the formation of PF6^-, PF5 acts a Lewis acid.
Solid sodium oxide and gaseous water are formed by the decomposition of solid sodium hydroxide (NaOH) .
Write a balanced chemical equation for this reaction.
Answer:
2NaOH(s) → Na₂O(s) + H₂O(g)
Hope that helps.
For dinner you make a salad with lettuce, tomatoes, cheese, carrots, and
croutons. Your salad would be classified as a(n)
O A. compound
OB. element
OC. homogeneous mixture
D. heterogeneous mixture
A heterogeneous mixture
Un globo lleno de helio tenia un volumen de 8.5 L en el suelo a 20°C y a una presión de 750 torr. Cuando se le soltó, el globo se elevo a una altitud donde la temperatura era de -20°C y la presión de 425 torr, ¿Cuál era el volumen del gas del globo en estas condiciones?
Answer:
El volumen del gas era 12.95 L
Explanation:
Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:
“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”
La ley de Boyle se expresa matemáticamente como: P*V=k
Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:
[tex]\frac{V}{T}=k[/tex]
Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:
[tex]\frac{P}{T}=k[/tex]
Combinado las mencionadas tres leyes se obtiene:
[tex]\frac{P*V}{T} =k[/tex]
Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:
[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]
Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:
P1: 750 torrV1: 8.5 LT1: 20°C= 293°K (siendo 0°C=273°K)P2: 425 torrV2: ?T2: -20°C= 253 °KReemplazando:
[tex]\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}[/tex]
Resolviendo:
[tex]V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}[/tex]
V2= 12.95 L
El volumen del gas era 12.95 L
What is the name of this molecule?
Answer:
[tex]\boxed{Butyne}[/tex]
Explanation:
Triple Bonds => So it is an alkyne
The suffix used will be "-yne"
4 Carbons => The prefix used will be "But-"
Combining the prefix and suffix, we get:
=> Butyne
Answer:
[tex]\boxed{\mathrm{Butyne}}[/tex]
Explanation:
Alkynes have triple bonds ≡. The molecule has one triple bond.
Suffix ⇒ yne
The molecule has 4 carbon atoms and 6 hydrogen atoms.
Prefix ⇒ But (4 carbons)
The molecule is Butyne.
[tex]\mathrm{C_4H_6}[/tex]
This pluton occurs deep in Earth and does not cause any changes to the surface of Earth . True or False
Answer:
The given statement is false.
Explanation:
However, if the pluton exists beneath the ground, this could be conveniently shown in the illustration something from the peak such pluton appears convex in form resembling a lopolith and perhaps diapir, which would be a particular form of statistically significant pluton recognized as the sill.Mostly from the figure it could also be shown that subsurface sheets are lined or curved, throughout the pluton mold. And therefore it is inferred that such a pluton creates adjustment to something like the ground atmosphere by altering the form of the levels above it.So that the given is incorrect.
A molecule of aluminum fluoride has one aluminum atom. How many fluorine atoms are present?
Answer:
3 fluorine atoms will be present
Answer:
3
Explanation:
The chemical formula of aluminum fluoride is AlF3. As you can see, there is a 1:3 ratio of aluminum atoms to fluorine atoms. Therefore, if a molecule of AlF3 has one aluminum atom, you know there must be 3 fluorine atoms present.
If you want further tutoring help in chemistry or other subjects for FREE, check out growthinyouth.org.
A four carbon chain; the second carbon is also single bonded to CH3. Spell out the full name of the compound
Answer:
This description shows a methyl group.
Explanation:
Which of the following elements is in the same family as fluorine?
a. silicon
b. antimony
O c. iodine
O d. arsenic
e. None of these.
Answer:
c iodine
Explanation:
fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride
identify the correct acid/conjugate base pair in this equation:
NaHCO3 + H20 = + H2CO3 + OH
+ Na
H20 is an acid and H2CO3 is its conjugate base.
HCO3 is an acid and OH is its conjugate base.
H20 is an acid and HCO3 is its conjugate base.
H20 is an acid and OH is its conjugate base.
Answer:
H20 is an acid and OH is its conjugate base.
Explanation:
Chemical reactions involving acids and bases occur. An acid is a substance that dissociates in water i.e. lose an hydrogen ion/proton. According to the Bronsted-Lowry acid-base theory, when an acid dissociates in water and loses its hydrogen ion, the resulting substance that forms is the CONJUGATE BASE. A conjugate base is the compound formed as a result of the removal of an H+ ion from an acid.
Based on the chemical reaction in the question, NaHCO3 + H20 = H2CO3 + OH- + Na+
The H20 loses its hydrogen ion (H+) to form an anion OH-. This anion formed is the conjugate base while H20 is its acid.
Which best describes the total mass of a sample of water when it condenses
from a liquid to a gas?
A. The mass is less because the water molecules get closer together
and take up more space.
B. The mass is the same because the decrease in energy equals the
increase in the number of molecules.
C. The mass is the same because water molecules are not created or
destroyed during a phase change.
D. The mass is greater after water condenses because the mass of
the molecules increases.
Answer:
Its C I hopefully help you
The amount of space an object takes up is called _____. gravity weight mass volume
The displacement of a bromine atom by an amine is a substituion reaction. Write out the mechanism of this reaction (2-->3) Why might you expect that the reaction you have performed, using t-BuNH2, to be much slower than the same reaction using methylamine
Answer:
An alkyl halide can undergo SN2 reaction with an amine
Explanation:
The displacement of a bromine atom by an an amine (step 2---> 3) in the reaction sequence is an example of an SN2 reaction in which the amine is the nucleophile.
The nitrogen atom of the amine which bears a lone pair of electrons functions as the nucleophile and attacks the electrophilic carbon atom of the alkyl halide displacing the bromide and creating a new Carbon-Nitrogen bond. An ammonium intermediate is immediately formed and the reaction is completed by the abstraction of a hydrogen by a base (such as excess amine present in the system).
This reaction is slower with t-BuNH2 because of steric hindrance and steric crowding in the transition state. SN2 reactions are faster with methylamine where the alkyl carbon is easily accessible.
The detailed mechanism of this reaction has been attached to this answer.
The NMR spectrum of your final compound will contain extra peaks that were not present in your starting material. For what hydrogen nuclei do those peaks occur?
Answer:
The peaks are registered from tetramethyl silane (TMS)
Explanation:
Tetramethyl silane (TMS) is used as internal reference in proton nmr (H NMR) spectrometry.
Its peak is usually registered at about a 2.0 chemical shift means that the hydrogen atoms which caused that peak need a magnetic field two millionths less than the field needed by TMS to produce resonance. This is not affected by the chemical shift of the sample analysed.
I hope this helped.
What element is primarily used in appliances to make electronic chips
A. Silicon (Si)
B. Nickel (Ni)
C. Copper (Cu)
D. Selenium (Se)
Answer:
Option A
Explanation:
Silicon (Obtained from Sand (SiO2)) is the element that is primarily used in appliances to make electronic chips.
Answer:
A. Silicon (Si)
Explanation:
Silicon (Si) is primarily used as a semiconductor material to make electronic chips.
1. Methanol is a high-octane fuel used in high performance racing engines. 2 CH3OH(l) + 3O2(g) → 2CO2(g) + 4 H20(g) a) Calculate ∆H० and ∆S० using thermodynamic data, and then ∆G
Answer:
The reaction given in the question is:
2CH₃OH (l) + 3O₂ (g) ⇒ 2CO₂ (g) + 4H₂O (g)
The values of ΔH°formation and ΔS° of the reactants and products given in the reaction based on the thermodynamics data is:
ΔH°formation values of CH3OH (l) is -238.4 kJ/mol, CO2(g) is -393.52 kJ/mol, H2O (g) is -241.83 kJ/mol and O2 (g) is 0.
The S° values of CH3OH (l) is 127.19 J/molK, CO2(g) is 213.79 J/molK, H2O (g) is 188.84 J/moleK, and O2 (g) is 205.15 J/molK.
Now the values of ΔH° and ΔS° are,
ΔH°rxn = 2 * ΔH°formation CO2 (g) + 4 * ΔH°formation H2O (g) - 2*ΔH°formation CH3OH (l)
ΔH°rxn = 2 * (-393.52) + 4 (-241.83) -2 * (-238.4)
ΔH°rxn = -1277.56 kJ/mole
ΔS°rxn = 2 * S° CO2 (g) + 4 * S° H2O (g) - 2*S° CH3OH (l) - 3 * S° O2 (g)
ΔS°rxn = 2 * 213.79 + 4 * 188.84 - 2 * 127.19 - 3*205.15
ΔS°rxn = 313.11 J/mole/K
Now the formula for calculating ΔG°rxn is,
ΔG°rxn = ΔH°rxn - TΔS°rxn
ΔG°rxn = -1277.56 * 1000 J/mole - 298 * 313.11 J/mole
ΔG°rxn = -1370.86 kJ/mol
18. Sucralose contains which two functional groups: (2 points)
A) benzene
B) halogen
C) carboxyl
D) hydroxy!
Answer:
The correct answer is option B and D, that is, halogen (chlorine) and hydroxyl.
Explanation:
An artificial sweetener and sugar substitute is sucralose. It is noncaloric as the majority of the sucralose ingested does not get dissociated within the body. The generation of sucralose takes place by the chlorination of sucrose. It is about 300 to 1000 times sweeter in comparison to sucrose.
The consumption of sucralose is safe for both nondiabetics and diabetics, it is used in various food and beverage components due to non-caloric sweetener characteristics. It does not affect the levels of insulin and does not affect dental health. As it is produced by chlorination of sucrose, thus, the functional groups present in it are a halogen (chlorine) and a hydroxyl.
Which of the following statements about water is not true?
Answer:
Water has a low specific heat capacity and so large bodies of water moderate temperatures on Earth.
Explanation:
Water has a very high specific heat capacity, meaning that it has to absorb a lot of energy to raise the temperature by one degree. Because water has a high specific heat capacity, large bodies of water can moderate the temperature of nearby land.
Hope this helps.
If a radioactive isotope of thorium (atomic number 90, mass number 232) emits 6 alpha particles and 4 beta particles during the course of radioactive decay, what is the mass number of the stable daughter product?
Answer:
The mass number of the stable daughter product is 208
Explanation:
First thing's first, we have to write out the equation of the reaction. This is given as;
²³²₉₀Th → 6 ⁴₂α + 4 ⁰₋₁ β + X
In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.
There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.
Mass Number
Reactant = 232
Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x
since reactant = product
232 = 24 + x
x = 232 - 24 = 208
Atomic Number
Reactant = 90
Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x
since reactant = product
90 = 8 + x
x = 90 - 8 = 82
It takes 242. kJ/mol to break a chlorine-chlorine single bond. Calculate the maximum wavelength of light for which a chlorine-chlorine single bond could be broken by absorbing a single photon. Round your answer to 3 significant digits. single by absorbing a significant digit.
Answer:
495nm
Explanation:
The energy of a photon could be obtained by using:
E = hc / λ
Where E is energy of a photon, h is Planck's constant (6.626x10⁻³⁴Js), c is speed of the light (3x10⁸ms⁻¹) and λ is wavelength.
The energy to break 1 mole of Cl-Cl bonds is 242kJ = 242000J. The energy yo break a single bond is:
242000J/mol ₓ (1mol / 6.022x10²³bonds) = 4.0186x10⁻¹⁹J/bond.
Replacing in the equation:
E = hc / λ
4.0186x10⁻¹⁹J = 3x10⁸ms⁻¹ₓ6.626x10⁻³⁴Js / λ
λ = 4.946x10⁻⁷m
Is maximum wavelength of light that could break a Cl-Cl bond.
Usually, wavelength is given in nm (1x10⁻⁹m / 1nm). The wavelength in nm is:
4.946x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =
495nmWhat is titration? Question 1 options: The process of quickly adding one solution to another until a solid is formed. The process of slowly adding one solution to another until the reaction between the two is complete. The process of mixing equal volumes of two solutions to observe the reaction between the two. The process of combining two solids until the reaction between the two is complete.
Answer:
The process of slowly adding one solution to another until the reaction between the two is complete.
Explanation:
When you perform a titration, you are slowly adding one solution of a known concentration called a titrant to a known volume of another solution of an unknown concentration until the reaction reaches neutralization, in which the reaction is no longer taking place. This is often indicated by a color change.
Hope that helps.