Answer: 656.6 nm.
Explanation:
Using Rydberg's Equation for hydrogen atom:
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
Where,
[tex]\lambda[/tex] = Wavelength of radiation
[tex]R_H[/tex] = Rydberg's Constant
[tex]n_f[/tex] = Higher energy level = 3 (least energetic for visible series)
[tex]n_i[/tex]= Lower energy level = 2
We have:
[tex]n_f=3, n_i=2[/tex]
[tex]R_H=1.097\times 10^7 m^{-1}[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \left(\frac{1}{2^2}-\frac{1}{3^2} \right )[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \frac{5}{36}[/tex]
[tex]\frac{1}{\lambda}=0.1523\times 10^{7} m[/tex]
[tex]\lambda=6.566\times 10^{-7}m=656.6nm[/tex]
([tex]1 m= 10^9nm[/tex])
The wavelength of the photon emitted when the hydrogen atom undergoes a transition from n = 2 to n = 3 is 656.6 nm.
What is the concentration in molarity of a solution made using 10.0 grams of KCl in 300.0 mL of water?
Please help immediately!!! :(
Answer:
From Molarity=concentration/molar mass
Concentration=10/0.3dm³
33.33g/dm³
Molar mass=H2O=18g/mol
Molarity=1.852mol/dm³
Answer:
0. 446 mol L−1
Explanation:
For Molarity, we must know the following things
• the number of moles of solute present in solution
• the total volume of the solution
we know the mass of one mole of potassium chloride = 74.55 g
Number of moles = Given Mass of substance / Mass of one mole
No of moles = 10 / 74.55
= 0.134 moles
Now we know that molarity is expressed per liter of solution. Since you dissolve 0.134 moles of potassium chloride in 300. mL of solution, you can say that 1.0 L will contain
For 300 ml of solution, no of moles are = 0.134 moles
For 1 ml of solution, no of moles are = 0.134/300 moles
For 1(1000) ml of solution, no of moles are= 0.134/300 x 1000
= 0.446 moles/ L
Answer is =0. 446 mol L−1
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e– → Sn(s). What is the concentration of Sn2+ if Zn2+ is 2.5 × 10−3 M and the cell emf is 0.660 V? The standard reduction potentials are given below Zn+2(aq) + 2 e− Sn2+(aq) + 2 e– →→ Zn(s) Sn(s) E∘red E∘red == −0.76 V −0.136 V Consider a voltaic cell where the anode half-reaction is and the cathode half-reaction is . What is the concentration of if is and the cell emf is 0.660 ? The standard reduction potentials are given below 9.0 × 10−3 M 3.3 × 10−2 M 6.9 × 10−4 M 7.6 × 10−3 M 1.9 × 10−4 M
the concentration of [tex]\(Sn^{2+}\)[/tex] is approximately [tex]\(3.3 \times 10^{-2} \, \text{M}\)[/tex].
To solve this problem, we can use the Nernst equation, which relates the cell potential [tex](\(E_{\text{cell}}\))[/tex] to the standard cell potential [tex](\(E^{\circ}_{\text{cell}}\))[/tex] and the concentrations of the species involved in the redox reaction:
[tex]\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0592}{n} \log \left( \frac{[\text{cathode product}]^m}{[\text{anode product}]^n} \right) \][/tex]
Given the standard reduction potentials, we can determine that the number of electrons involved in the half-reactions is [tex]\(n = 2\)[/tex]. The given cell potential is [tex]\(E_{\text{cell}} = 0.660 \, \text{V}\)[/tex]. We need to find the concentration of [tex]\(Sn^{2+}\)[/tex].
First, let's write the Nernst equation for the given cell:
[tex]\[ 0.660 \, \text{V} = (E^{\circ}_{\text{cell}}) - \frac{0.0592}{2} \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Zn}^{2+}]} \right) \][/tex]
We're given the standard reduction potentials, which are [tex]\(E^{\circ}_{\text{cell}} = -0.76 \, \text{V}\)[/tex] for the zinc half-reaction and [tex]\(E^{\circ}_{\text{cell}} = -0.136 \, \text{V}\)[/tex] for the tin half-reaction.
Plugging in the values and solving for [tex]\([\text{Sn}^{2+}]\)[/tex]:
[tex]\[ 0.660 \, \text{V} = (-0.76 \, \text{V}) - \frac{0.0592}{2} \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]
[tex]\[ 0.660 \, \text{V} = (-0.76 \, \text{V}) - 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]
[tex]\[ 0.660 + 0.76 = 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]
[tex]\[ 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = 0.76 + 0.660 \][/tex]
[tex]\[ \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = \frac{1.42}{0.0296} \][/tex]
[tex]\[ \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = 47.973 \][/tex]
[tex]\[ \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} = 10^{47.973} \][/tex]
[tex]\[ [\text{Sn}^{2+}] = (2.5 \times 10^{-3}) \times 10^{47.973} \][/tex]
[tex]\[ [\text{Sn}^{2+}] \approx 3.3 \times 10^{45} \, \text{M} \][/tex]
Therefore, the concentration of [tex]\(Sn^{2+}\)[/tex] is approximately [tex]\(3.3 \times 10^{-2} \, \text{M}\)[/tex].
Nitrogen is a vital component of proteins and nucleic acids, and thus is necessary for life. The atmosphere is composed of roughly 80% N2, but most organisms cannot directly utilize N2 for biosynthesis. Bacteria capable of "fixing" nitrogen ( ie converting N2 to a chemical form, such as NH3, which can be utilized in the biosynthesis of proteins and nucleic acids) are called diazatrophs. The ability of some plants like legumes to fix nitrogen is due to a symbiotic relationship between the plant and nitrogen-fixing bacteria that live in the plant’s roots.
Assume the hypothetical reaction for fixing nitrogen biologically is N2 (g) + 3H2O (l) → 2 NH3 (aq) + 3/2 O2 (g)
a. Calculate the standard enthalpy change for the biosynthetic fixation of nitrogen at T = 298 K. For NH3 (aq), ammonia dissolved in aqueous solution, ΔHof = - 80.3 kJ mol-1.
b. In some bacteria, glycine is produced from ammonia by the reaction NH3 (aq) + 2CH4 (g) + 5/2 O2 (g) → NH2CH2COOH (s) + 3H2O (l) Calculate the standard enthalpy change for the synthesis of glycine from ammonia. For glycine, ΔHof = - 537.2 kJ mol -1. Assume that T = 298 K.
c. Calculate the standard enthalpy change for the synthesis of glycine from nitrogen, oxygen, and methane.
Answer:
Explanation:ANd of course,
Δ
H
∘
f
=
0
for an element (here dixoygen) in its standard state
Answer:
a. ΔHr = 696,8 kJ/mol
b. ΔHr = -1164,7 kJ/mol
c. ΔHr = -467,9 kJ/mol
Explanation:
It is possible to obtain the standard enthalpy change of a reaction with the ΔH°f of products - ΔH°f reactants.
a. For the reaction:
N₂(g) + 3H₂O(l) → 2NH₃(aq) + ³/₂O₂(g)
ΔHr = 2ΔH°f NH₃(aq) + ³/₂ΔH°fO₂(g) - (3ΔH°fH₂O(l) + 2ΔH°f N₂(g))
ΔHr = 2×-80,3 kJ/mol + ³/₂×0 - (3×-285,8 kJ/mol + 0)
ΔHr = 696,8 kJ/mol
b. and c. For the reaction:
NH₃(aq) + 2CH₄(g) + ⁵/₂O₂(g) → NH₂CH₂COOH(s) + 3H₂O(l)
ΔHr = ΔH°fNH₂CH₂COOH(s) + 3ΔH°fH₂O(l) - (ΔH°fNH₃(aq) + 2ΔH°fCH₄(g) + ⁵/₂ΔH°fO₂(g))
ΔHr = -537,2kJ/mol + 3×-285,8 kJ/mol - (-80,3 kJ/mol + 2×-74,8kJ/mol+ ⁵/₂×0)
ΔHr = -1164,7 kJ/mol
c. From nitrogen, methane and oxygen the reaction is the sum of reactions of a and b:
N₂(g) + 3H₂O(l) → 2NH₃(aq) + ³/₂O₂(g)
+ NH₃(aq) + 2CH₄(g) + ⁵/₂O₂(g) → NH₂CH₂COOH(s) + 3H₂O(l)
N₂(g) + 2CH₄(g) + O₂(g) → NH₂CH₂COOH(s) + NH₃(aq)
By Hess's law, the ΔHr will be the sum of the ΔHr of the last two reactions, that means:
ΔHr = -1164,7 kJ/mol + 696.8 kJ/mol
ΔHr = -467,9 kJ/mol
I hope it helps!
Glucose is present in blood plasma at a concentration of approximately 80 mg/dL. A typical 70 kg human has approximately 5 L blood, which is 55% plasma. Using the 44.96 mg/mL glucose concentration of Sprite that you calculated in the previous question, how many mL of Sprite does a human need to drink to reach a plasma glucose concentration of 80 mg/dL, assuming none is mobilized? (Note: all extracellular glucose is about 10 times this level, which is not important for this calculation.(Please pay attention to the units mg/dL)
Answer:
A human need to drink 48.93 mL of sprite to reach a plasma glucose concentration of 80 mg/dL.
Explanation:
Volume of blood in 7 Kg human = 5 L
Percentage of plasma in blood = 55%
Volume of plasma in 5 L blood = [tex]\frac{55}{100}\times 5 L=2.75 L=27.5 dL[/tex] (1 L = 10 dL)
Concentration of glucose in plasma = 80 mg/dL
Amount of glucose present in 27.5 dL : 80 mg /dL × 27.5 dL= 2200 mg
Let the volume of sprite with 2200 mg glucose be x
Concentration of glucose in sprite = 44.96 mg/mL
[tex]x\times 44.96 mg/mL=2200 mg[/tex]
[tex]x=\frac{2200 mg}{44.96 mg/mL}=48.93 mL[/tex]
A human need to drink 48.93 mL of sprite to reach a plasma glucose concentration of 80 mg/dL.
What problem would arise if the excess acetyl‑CoA were not converted to ketone bodies?
A. Although acetyl‑CoA is not toxic, it inhibits D‑β‑hydroxyacyl‑CoA dehydrogenase.
B. Although acetyl‑CoA is not toxic, it prevents transfer of fatty acids into the mitochondria by forming acetyl carnitine.
C. Fatty acid oxidation would stop when all of the CoA is bound as acetyl‑CoA.
D. Formation of acetyl‑CoA requires uptake of a proton, and the increased pH inhibits further oxidation.
Answer:
The correct answer is C fatty acid oxidation would stop when all of the CoA is bound as acetyl CoA.
Explanation:
Acetyl CoA is the principle end product of beta oxidation of even chain fatty acid such as palmitic acid.
When the cellelar label of actyl CoA increases at that time the excess acetyl CoA is converted to ketone bodies by the process called ketogenesis.
According to the question if the excess acetyl CoA is not converted to ketone bodies then it will interfere with the oxidation of fatty acid because fatty acid molecules will not get any CoA SH molecule to activate themselves to initiate a new round of beta oxidation.
As a result fatty acid oxidation will stop.
Answer:
C. Fatty acid oxidation would stop when all of the CoA is bound as acetyl‑CoA.
Explanation:
Hello,
In this case, due to the fact that the mitochondrial pool of the CoA is short, thus, such cofactor must be recycled from acetyl-CoA through the production of ketone-like bodies. Therefore, the operation of the beta-oxidation pathway is performed, as it is necessary for energy production, in such a way one concludes that fatty acid oxidation would stop when all of the CoA is bound as acetyl‑CoA.
Best regards.
Organic Chem Rxn Question
Using the reagents below, list in order (by letter, no period) those necessary to prepare trans-1,2-dibromocyclopentane from cyclopentane.
Note: Not all spaces provided may be needed. Type "na" in any space where you have no reagent. (There are 3 spaces)
a. Br2, heat, light
b. HBr
c. Br2
d. H2O
e. HBr, HOOH
f. EtOH
g. NaOEt, EtOH, heat
Answer:
a, g, c
Explanation:
The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.
The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.
Final answer:
To synthesize trans-1,2-dibromocyclopentane from cyclopentane, the correct reagent is Br2 alone (option c). Other reagents like heat and light could lead to radical mechanisms, which are not appropriate for this specific transformation.
Explanation:
To prepare trans-1,2-dibromocyclopentane from cyclopentane, you need to add bromine (Br2) across the double bond in a trans configuration. The best way to achieve this without any other rearrangements or products is to use Br2 alone (option c). When Br2 is added to a double bond, it will typically add in an anti-fashion, leading to the trans product. Using Br2 with heat/light (option a) or with solvents like EtOH (option f) would likely lead to radical mechanisms or other side products.
Therefore, the correct order of reagents to prepare trans-1,2-dibromocyclopentane from cyclopentane would be just: c. The other two spaces would be filled with 'na' as no additional reagents are needed for this transformation.
Thus, the list in order would be: c, na, na.
A student observed that a small amount of acetone sprayed on the back of the hand felt very cool compared to a similar amount of water. Your explanation of this phenomena should be that
A.) All organic compounds do this
B.) Acetone has a lower viscosity and transfers heat quanta better
C.) Water has a higher heat capacity than acetone therefore retaining more heat
D.) The higher vapor pressure of acetone results in more rapid evaporation and heat loss
Answer:
D.) The higher vapor pressure of acetone results in more rapid evaporation and heat loss
Explanation:
Acetone being liquid with very low boiling point and hence can be easily be converted to gaseous state .
And ,
Therefore have higher vapour pressure and as liquid gets converted to gas , the process of evaporation takes place and leads to loss of heat .
Hence , from the given options , the most appropriate option according to given statement of the question is ( d ) .
The cooling sensation felt with acetone is due to its rapid evaporation, which absorbs heat from the skin. This occurs because acetone has a higher vapor pressure and weaker intermolecular forces compared to water. Option D is the correct option.
The phenomenon observed by the student can be explained by the fact that acetone evaporates more rapidly than water. This rapid evaporation leads to a cooling effect because evaporation is an endothermic process that absorbs heat from the surroundings. Option D is the correct explanation: The higher vapor pressure of acetone results in more rapid evaporation and heat loss.
Acetone has weaker intermolecular forces compared to water, primarily dipole-dipole interactions rather than hydrogen bonds. This means acetone molecules can escape into the gas phase much more easily. When acetone evaporates, it requires energy, which it absorbs from the surroundings, leading to a cooling sensation on the skin.
When 18.9 kJ is transferred to a gas sample in a constant volume adiabatic container with a calorimeter constant of 2.22 Kj/K, the temperature of the gas (and the calorimeter) increases by 8.06 K. (a) What is the heat capacity of the sample? (b) If the sample has a mass of 0.5 kilograms, what is the specific heat capacity of the substance? (c) If the sample is Krypton, what is the molar heat capacity at constant volume of Krypton? The molar mass of Krypton is 83.8 grams/mole.
Answer:
(a) Cgas = 0.125 kJ/k
(b) cgas = 0.25kJ/kg.K
(c) cm(gas) = 0.021kJ/mol.K
Explanation:
18.9 kJ is equal to the sum of the heat absorbed by the gas and the heat absorbed by the calorimeter.
Qcal + Qgas = 18.9 kJ [1]
We can calculate the heat absorbed using the following expression.
Q = C . ΔT
where,
C is the heat capacity
ΔT is the change in the temperature
(a) What is the heat capacity of the sample?
From [1],
Ccal . ΔT + Cgas . ΔT = 18.9 kJ
(2.22kJ/K) × 8.06 K + Cgas × 8.06 K = 18.9 kJ
Cgas = 0.125 kJ/k
(b) If the sample has a mass of 0.5 kilograms, what is the specific heat capacity of the substance?
We can calculate the specific heat capacity (c) using the following expression:
[tex]c=\frac{C}{m} =\frac{0.125kJ/K}{0.5kg} =0.25kJ/kg.K[/tex]
(c) If the sample is Krypton, what is the molar heat capacity at constant volume of Krypton? The molar mass of Krypton is 83.8 grams/mole.
The molar heat capacity is:
[tex]\frac{0.25kJ}{kg.K} .\frac{1kg}{1000g} .\frac{83.8g}{mol} =0.021kJ/mol.K[/tex]
Final answer:
The heat capacity of the sample is 2.34 kJ/K. The specific heat capacity of the substance is 4.68 kJ/(kg*K). The molar heat capacity at constant volume of Krypton is 0.0559 kJ/(mol*K).
Explanation:
The heat capacity of a substance is the amount of energy required to increase its temperature by 1 degree. The specific heat capacity is the amount of energy required to increase the temperature of 1 gram of a substance by 1 degree.
(a) To find the heat capacity of the sample, we can use the equation Q = CΔT, where Q is the amount of heat transferred, ΔT is the change in temperature, and C is the heat capacity. Rearranging the equation, C = Q/ΔT. Plugging in the given values, C = 18.9 kJ / 8.06 K
= 2.34 kJ/K.
(b) To find the specific heat capacity of the substance, we need to know the mass of the sample. Given that the mass is 0.5 kilograms, we can use the equation cs = C/m, where cs is the specific heat capacity, C is the heat capacity, and m is the mass. Plugging in the values, cs = 2.34 kJ/K / 0.5 kg
= 4.68 kJ/(kg*K).
(c) To find the molar heat capacity at constant volume of Krypton, we can use the equation Cm = cs / M, where Cm is the molar heat capacity, cs is the specific heat capacity, and M is the molar mass. Plugging in the values, Cm = 4.68 kJ/(kg*K) / 83.8 g/mol
= 0.0559 kJ/(mol*K).
Enthalpy of formation (kJ/mol) C6H12O6(s)-1260 O2 (g)0 CO2 (g)-393.5 H2O (l)-285.8 Calculate the enthalpy of combustion per mole of C6H12O6. C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)
Answer : The enthalpy of combustion per mole of [tex]C_6H_{12}O_6[/tex] is -2815.8 kJ/mol
Explanation :
Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
The equilibrium reaction follows:
[tex]C_6H_{12}O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})][/tex]
We are given:
[tex]\Delta H^o_f_{(C_6H_{12}O_6(s))}=-1260kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol[/tex]
Therefore, the enthalpy of combustion per mole of [tex]C_6H_{12}O_6[/tex] is -2815.8 kJ/mol
Answer:
muahhh
Explanation:
Identify the final concentrations of each species following the addition of 1.0 M KOH to a 2.0 M HF solution. HF ( aq ) + KOH ( aq ) ⟶ KF ( aq ) + H 2 O ( l ) initial 2.0 M 1.0 M 0 M change final ? ? ?
Answer : The final concentration of KOH, HF and KF are 0 M, 1.0 M and 1.0 M respectively.
Explanation :
The given chemical reaction is:
[tex]HF(aq)+KOH(aq)\rightarrow KF(aq)+H_2O(l)[/tex]
Initial conc. 2.0 M 1.0 M 0 M
Final conc. 1.0 M 0 M 1.0 M
When 1.0 M of KOH react with 2.0 M HF then 1.0 M KOH will react with 1.0 M HF to form 1.0 M KF and 1.0 M HF remain unreacted.
So, the final solution contains 1.0 M HF and 1.0 M KF that means the solution contains equal amount of weak acid and salt.
Therefore, the final concentration of KOH, HF and KF are 0 M, 1.0 M and 1.0 M respectively.
The final concentration of KOH, HF and KF are 0 M, 1.0 M and 1.0 M respectively.
Chemical reaction:HF ( aq ) + KOH ( aq ) ⟶ KF ( aq ) + H₂O ( l )
Initial conc. 2.0 M 1.0 M 0 M
Final conc. 1.0 M 0 M 1.0 M
When 1.0 M of KOH react with 2.0 M HF then 1.0 M KOH will react with 1.0 M HF to form 1.0 M KF and 1.0 M HF remain unreacted. So, the final solution contains 1.0 M HF and 1.0 M KF that means the solution contains equal amount of weak acid and salt.
Therefore, the final concentration of KOH, HF and KF are 0 M, 1.0 M and 1.0 M respectively.
Find more information about Final concentration here:
brainly.com/question/24595796
The specific heats and densities of several materials are given below: Material Specific Heat (cal/g·°C) Density (g/cm3) Brick 0.220 2.0 Concrete 0.270 2.7 Steel 0.118 7 Water 1.00 1.00 Calculate the change in temperature produced by the addition of 1 kcal of heat to 100 g of steel.
Answer: The change in temperature is 84.7°C
Explanation:
To calculate the change in temperature, we use the equation:
[tex]q=mc\Delta T[/tex]
where,
q = heat absorbed = 1 kCal = 1000 Cal (Conversion factor: 1 kCal = 1000 Cal)
m = mass of steel = 100 g
c = specific heat capacity of steel = 0.118 Cal/g.°C
[tex]\Delta T[/tex] = change in temperature = ?
Putting values in above equation, we get:
[tex]1000cal=100g\times 0.118cal/g^oC\times \Delta T\\\\\Delta T=\frac{1000cal}{100g\times 0.118cal/g^oC}\\\\\Delta T=84.7^oC[/tex]
Hence, the change in temperature is 84.7°C
When 1.98g of a hydrocarbon is burned in a bomb calorimeter, the temperature increases by 2.06∘C. If the heat capacity of the calorimeter is 69.6 J∘C and it is submerged in 944mL of water, how much heat (in kJ) was produced by the hydrocarbon combustion?
Answer:
8.3 kJ
Explanation:
In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:
q for water:
q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g
c: specific heat of water = 4.186 J/gºC
ΔT : change in temperature = 2.06 ºC
so solving for q :
q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J
For calorimeter
q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC
ΔT : change in temperature = 2.06 ºC
q calorimeter = 69.60J x 2.06 ºC = 143.4 J
Total heat released = 8,140 J + 143.4 J = 8,2836 J
Converting into kilojoules by dividing by 1000 we will have answered the question:
8,2836 J x 1 kJ/J = 8.3 kJ
The heat produced by the combustion of 1.98g of the hydrocarbon is 0.143 kJ.
The volume of water was not needed for this particular calculation.
To determine the heat produced by burning 1.98g of a hydrocarbon in a bomb calorimeter, we can use the formula :q = Ccal × ΔTWhere:
Ccal is the heat capacity of the calorimeterΔT is the change in temperatureGiven:
Mass of hydrocarbon: 1.98gTemperature increase: 2.06°CHeat capacity of calorimeter: 69.6 J/°CVolume of water: 944 mL (not needed for this calculation)Let's plug in the values:
q = 69.6 J/°C × 2.06°Cq = 143.376 JSince we need the heat in kJ :q = 143.376 J / 1000 = 0.143376 kJThe heat produced by the hydrocarbon combustion is 0.143 kJ.Pyruvate dehydrogenase is a large, highly integrated complex containing many copies of three distinct enzymes. There are five coenzymes involved in its catalytic activity: NAD , FAD, coenzyme A, lipoamide, and thiamine pyrophosphate (TPP or TDP). The coenzymes can be classified depending on how they participate in an enzymatic reaction.
A coenzyme prosthetic group is tightly bound to the enzyme and remains bound during the catalytic cycle. The original coenzymes are regenerated during the catalytic cycle.
On the other hand, a coenzyme cosubstrate is loosely bound to an enzyme and dissociates in an altered form as part of the catalytic cycle. Its original form is regenerated not by the cycle, but by another enzyme.
Which are coenzyme prosthetics?
1. NAD+
2. TPP or TDP
3. lipoamide
4. FAD
5. coenzyme A
Answer:
The correct answer is a coenzyme cosubstrate is loosely bound to an enzyme and dissociates in an altered form as the part of the catalytic cycle.Its original form is regenerated not by the cycle but by other enzyme.
Explanation:
The prosthetic group of the enzyme pyruvate dehydrogenase is
2 TPP or Thymine pyrophosphate.
3 Lipomide.
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 665.9 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = + 190.1 k J / m o l what is Δ H for this reaction? X ( s ) + 1 2 O 2 ( g ) + CO 2 ( g ) ⟶ XCO 3 ( s )
Answer:
ΔH of reaction is -856 kJ/mol
Explanation:
The property ΔH is property which can be added to find the net change in enthalpy of reaction.
ΔH of first reaction is -665.9 kJ/mol
ΔH of second reaction is 190.1 kJ/mol
carefully looking at the third equation,
it is first reaction - second reaction.
thus, by Hess's law,
Hess's Law of Constant Heat Summation (or just Hess's Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.
ΔH of third reaction is = [tex]-665.9 - (190.1)[/tex]
ΔH = -856 kJ/mol
A heat flux meter attached to the inner surface of a 3-cm-thick refrigerator door indicates a heat flux of 25 W/m² through the door. Also, the temperatures of the inner and the outer surfaces of the door are measured to be 7°C and 15°C, respectively. Determine the average thermal conductivity of the refrigerator door. Answer: 0.0938 W/m. C
Answer:
0.0938 W/m.°C
Explanation:
The heat is flowing by the refrigerator door by conduction, and can be calculated by the equation:
q = k*ΔT/L
Where q is the heat flux (25 W/m²), k is the thermal conductivity of the refrigerator, ΔT is the variation of the temperature (inner - outer), and L is the thickness of the door (0.03 m).
25 = k*(15-7)/0.03
8k = 0.75
k = 0.0938 W/m.°C
Consider the fructose-1,6-bisphosphatase reaction. Calculate the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 21.3 and the temperature is 37.0 ° C ? Δ G ° ' for the reaction is − 16.7 kJ/mol .
Answer: The Gibbs free energy of the reaction is -8.82 kJ/mol
Explanation:
The equation used to Gibbs free energy of the reaction follows:
[tex]\Delta G=\Delta G^o+RT\ln K_{eq}[/tex]
where,
[tex]\Delta G[/tex] = free energy of the reaction
[tex]\Delta G^o[/tex] = standard Gibbs free energy = -16.7 kJ/mol = -16700 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.314J/K mol[/tex]
T = Temperature = [tex]37^oC=[273+37]K=310K[/tex]
[tex]K_{eq}[/tex] = Ratio of concentration of products and reactants = 21.3
Putting values in above equation, we get:
[tex]\Delta G=-16700J/mol+(8.314J/K.mol\times 310K\times \ln (21.3))\\\\\Delta G=-8816.7J/mol=-8.82kJ/mol[/tex]
Hence, the Gibbs free energy of the reaction is -8.82 kJ/mol
Find the equilibrium partial pressures of A and B for each of the following different values of Kp.?Consider the following reaction:A(g) = 2B(g)Find the equilibrium partial pressures of A and B for each of the following different values of Kp. Assume that the initial partial pressure of B in each case is 1.0 atm and that the initial partial pressure of A is 0.0 atm. Make any appropriate simplifying assumptions.Kp = 1.4?Kp = 2.0 * 10^-4?Kp = 2.0 * 10^5?
Answer:
For Kp = 1,4; [tex]P_{[A] = 0,22[/tex], [tex]P_{[B] = 0,56atm[/tex]
For Kp = 2,0x10⁻⁴; [tex]P_{[A] = 0,495[/tex], [tex]P_{[B] = 0,01atm[/tex]
For Kp = 2,0x10⁵; [tex]P_{[A] = 5x10^{-6}[/tex], [tex]P_{[B] = 0,99999atm[/tex]
Explanation:
For the reaction:
A(g) ⇄ 2B(g)
kp is: [tex]kp = \frac{P_{[B]}^2}{P_{[A]}}[/tex]
If initial pressure of B is 1,0atm and initial pressure of A is 0,0atm the equilibrium pressures are:
[tex]P_{[A] = 0,0atm + X[/tex]
[tex]P_{[B] = 1,0atm - 2X[/tex]
Replacing for Kp= 1,4:
[tex]1,4 = \frac{(1-2X)^2}{X}[/tex]
1,4X = 4X² - 4X + 1
0 = 4X² - 5,4X + 1
Solving for X:
X = 0,22 -Right answer-
X = 1,13
Replacing:
[tex]P_{[A] = 0,22[/tex]
[tex]P_{[B] = 1,0atm - 0,44atm = 0,56atm[/tex]
For Kp= 2,0x10⁻⁴:
[tex]2,0x10^{-4} = \frac{(1-2X)^2}{X}[/tex]
2,0x10^{-4}X = 4X² - 4X + 1
0 = 4X² - 4,0002X + 1
Solving for X:
X = 0,495atm
Replacing:
[tex]P_{[A] = 0,495atm[/tex]
[tex]P_{[B] = 1,0atm - 0,99atm = 0,01atm[/tex]
For Kp= 2,0x10⁵:
[tex]2,0x10^5 = \frac{(1-2X)^2}{X}[/tex]
2,0x10^5X = 4X² - 4X + 1
0 = 4X² - 2,00004x10^5X + 1
Solving for X:
X = 5x10⁻⁶ -Right answer-
Replacing:
[tex]P_{[A] = 5x10^{-6}[/tex]
[tex]P_{[B] = 1,0atm - 0,00001atm = 0,99999atm[/tex]
I hope it helps!
To determine the equilibrium partial pressures of A and B for the given reaction, one must solve a quadratic equation derived from the equilibrium expression involving the given Kp and initial conditions, considering the stoichiometry and the relationship between the changes in partial pressures of A and B as the reaction reaches equilibrium.
Explanation:When considering the equilibrium partial pressures of A and B for the reaction A(g) = 2B(g), we need to calculate how the initial conditions and the equilibrium constant (Kp) affect the final partial pressures at equilibrium. Starting with an initial partial pressure of B as 1.0 atm and A as 0.0 atm, and using the equilibrium expression Kp = (PB)^2/PA, we can plug in different values of Kp to solve for the unknowns. Assuming the reaction has proceeded to equilibrium, we can express any changes in partial pressure of A as -x and of B as +2x because for each mole of A reacting, 2 moles of B are formed.
For example, with Kp = 1.4, let's assume 'x' is the change in partial pressure of B; we then construct the equation Kp = (1 + 2x)^2/(0.0 + x). Solving the quadratic equation derived would give us the value of 'x', which in turn provides the equilibrium partial pressures PA and PB. We would need to solve a similar quadratic equation for different values of Kp such as 2.0 * 10^-4 and 2.0 * 10^5 following the same approach. It's important to note that some values might yield non-physical solutions, like negative pressures; those are dismissed since partial pressures must be positive.
What is the pOH of a 0.0092 M CsOH solution?
A) 2.04
B) 16.04
C) 9.31
D) 4.69
E) 11.96
Answer:
2.04 is the pOH of a 0.0092 M CsOH solution.
Explanation:
Molarity of cesium hydroxde = [CsOH]= 0.0092
[tex]CsOH(aq)\rightarrow Cs^+(aq)+OH^-(aq)[/tex]
1 mole of cesium hydroxide gives 1 mole of cesium ion and 1 mole of hydroxide ion.
Then 0.0092 M cesium hydroxide will give :
[tex][OH^-]=1\times [CsOH]=1\times 0.0092 M=0.0092M [/tex]
The pOH of the solution is the negative logarithm of concentration of hydroxide ions in a solution.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log [0.0092 M]=2.0362\approx 2.04 [/tex]
2.04 is the pOH of a 0.0092 M CsOH solution.
Final answer:
The pOH of a 0.0092 [tex]M CsOH[/tex] solution is calculated using the negative log of the hydroxide ion concentration, which results in a pOH of 2.04. Thus, the correct answer is (A) 2.04.
Explanation:
The student is asking about how to calculate the pOH of a [tex]CsOH[/tex] solution. Since CsOH is a strong base, it will dissociate completely in water.
The concentration of hydroxide ions (OH-) in a 0.0092 [tex]M CsOH[/tex] solution will be the same as the concentration of CsOH itself, which is 0.0092 M.
To find the pOH, we take the negative logarithm (base 10) of the hydroxide ion concentration:
[tex]pOH = -log[OH-][/tex]
[tex]pOH = -log(0.0092)[/tex]
[tex]pOH = -(-2.036)[/tex]
[tex]pOH = 2.036[/tex]
Therefore, rounding to two decimal places, the pOH of a 0.0092 [tex]M CsOH[/tex] solution is 2.04. Hence, the correct answer is (A) 2.04.
A stock solution contains a mixture of ~100 ppm chloride, fluoride, nitrite, bromide, nitrate and phosphate anions. In order to prepare 1 L of 100 ppm nitrite stock solution, you weigh out 150.0 mg of NaNO2. The actual concentration of nitrite would be:______.
Answer:
[tex]150~ppm[/tex]
Explanation:
The first step is to draw the ionization reaction of [tex]NaNO_2[/tex], so:
[tex]NaNO_2~->~Na^+~+~NO_2^-[/tex]
The molar ratio between [tex]NaNO_2[/tex] and [tex]NO_2^-[/tex] is 1:1
The next step is the calculation of the concentration. We have to remember that the formula of ppm is:
[tex]ppm=\frac{mg}{L}[/tex]
In this case we will have a mass of 150 mg and a volume of 1 L so:
[tex]ppm=\frac{150~mg}{1~L}=~150ppm[/tex]
Part A Name the complex ion [Fe(CN)6]^3- . The oxidation number of iron is +3. Part B Name the complex ion [Cu(NH3)2(H2O)4]^2+ . The oxidation number of copper is +2. Part C Name the complex CrCl2(en)2 . The oxidation number of chromium is +2. Part D Name the salt [Ni(H2O)3(Co)]SO4 . The oxidation number of nickel is +2. Part E Name the salt K4[Pt(CO3)2F2] given that the carbonate ion acts as a monodentate ligand in the complex. The oxidation number of platinum is +2.
The complexes are sodium hexachloroplatinate(IV), potassium trisoxalatoferrate(III), and pentaamminechlorocobalt(III) chloride.
Explanation:(a) There are two Na+ ions, so the coordination sphere has a negative two charge: [PtCl6]². There are six anionic chloride ligands, so -2 = -6 + x, and the oxidation state of the platinum is 4+. The name of the complex is sodium hexachloroplatinate(IV), and the coordination number is six.
(b) The coordination sphere has a charge of 3- (based on the potassium) and the oxalate ligands each have a charge of 2-, so the metal oxidation state is given by -3 = -6+ x, and this is an iron(III) complex. The name is potassium trisoxalatoferrate(III) (note that tris is used instead of tri because the ligand name starts with a vowel). Because oxalate is a bidentate ligand, this complex has a coordination number of six.
(c) In this example, the coordination sphere has a cationic charge of 2+. The NH3 ligand is neutral, but the chloro ligand has a charge of 1-. The oxidation state is found by +2 = -1 + x and is 3+, so the complex is pentaamminechlorocobalt(III) chloride and the coordination number is six.
An atom of 105In has a mass of 104.914558 amu. Calculate the binding energy in MeV per NUCLEON. Enter your answer with 3 significant figures and no units. Use the masses:
mass of 1H atom = 1.007825 amu
mass of a neutron = 1.008665 amu
1 amu = 931.5 MeV
Answer:
8.46
Explanation:
Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.
Also, atomic number of I = 549
Thus, the number of protons = 49
Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.
Mass number = Number of protons + Number of neutrons
105 = 49 + Number of neutrons
Number of neutrons = 56
Mass of neutron = 1.008665 amu
Mass of proton = 1.007825 amu
Calculated mass = Number of protons*Mass of proton + Number of neutrons*Mass of neutron
Thus,
Calculated mass = (49*1.007825 + 56*1.008665) amu = 105.868665 amu
Mass defect = Δm = |105.868665 - 104.914558| amu = 0.954107 amu
The conversion of amu to MeV is shown below as:-
1 amu = 931.5 MeV
So, Energy = 0.954107*931.5 MeV/atom = 888.750671 MeV/atom
Also, 1 atom has 105 nucleons (Protons+neutrons)
So, Energy = 888.750671 MeV/105nucleons = 8.46 MeV/nucleon
Answer:- 8.46
The binding energy per nucleon for an atom of 105In is calculated by determining the mass defect, converting it to energy using Einstein's mass-energy equivalence, and dividing by the number of nucleons.
Explanation:To calculate the binding energy per nucleon in MeV for an atom of 105In with a mass of 104.914558 amu, we must first identify the number of protons and neutrons in the nucleus. Indium-105 has 49 protons (since In is the element with atomic number 49), and subtracting this from the mass number 105 gives us 56 neutons. The mass of the protons is 49 x 1.007825 amu, and the mass of the neutrons is 56 x 1.008665 amu.
Next, we calculate the mass defect by subtracting the atomic mass of the 105In from the combined mass of the protons and neutrons. To convert the mass defect into energy, we multiply by 931.5 MeV/amu, according to Einstein’s mass-energy equivalence principle (E=mc2). Lastly, we divide this energy by the total number of nucleons (protons + neutrons) to find the binding energy per nucleon.
Use the following information on Cr to determine the amounts of heat for the three heating steps required to convert 126.3 g of solid Cr at 1760°C into liquid Cr at 2060°C. mp = 1860°C bp = 2672°C Enter in kJ. Useful data: \Delta HΔ Hfus = 20.5 kJ/mol; \Delta HΔ Hvap = 339 kJ/mol; c(solid) 44.8 J/g°C; c(liquid) = 0.94 J/g°C
Answer : The amount of heat required is, 639.3 KJ
Solution :
The conversions involved in this process are :
[tex](1):Cr(s)(1760^oC)\rightarrow Cr(s)(1860^oC)\\\\(2):Cr(s)(1860^oC)\rightarrow Cr(l)(1860^oC)\\\\(3):Cr(l)(1860^oC)\rightarrow Cr(l)(2060^oC)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change or heat required = ?
m = mass of Cr = 126.3 g
[tex]c_{p,s}[/tex] = specific heat of solid Cr = [tex]44.8J/g^oC[/tex]
[tex]c_{p,l}[/tex] = specific heat of liquid Cr = [tex]0.94J/g^oC[/tex]
n = number of moles of Cr = [tex]\frac{\text{Mass of Cr}}{\text{Molar mass of Cr}}=\frac{126.3g}{52g/mole}=2.428mole[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 20.5 KJ/mole = 20500 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=[126.3g\times 44.8J/g^oC\times (1860-(1760))^oC]+2.428mole\times 20500J/mole+[126.3g\times 0.94J/g^oC\times (2060-1860)^oC][/tex]
[tex]\Delta H=639342.4J=639.3KJ[/tex] (1 KJ = 1000 J)
Therefore, the amount of heat required is, 639.3 KJ
To determine the amounts of heat for each heating step, we need to calculate the heat required to raise the temperature from solid Cr at 1760°C to its melting point at 1860°C, the heat of fusion to convert the solid Cr at its melting point to liquid Cr, and the heat required to raise the temperature from liquid Cr at 1860°C to the final temperature of 2060°C.
Explanation:To determine the amounts of heat for each heating step, we need to calculate the heat required to raise the temperature from solid Cr at 1760°C to its melting point at 1860°C, the heat of fusion to convert the solid Cr at its melting point to liquid Cr, and the heat required to raise the temperature from liquid Cr at 1860°C to the final temperature of 2060°C.
Step 1: Calculate the heat required to raise the temperature from 1760°C to 1860°C using the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.Step 2: Calculate the heat of fusion using the formula Q = nΔHfus, where Q is the heat, n is the number of moles, and ΔHfus is the heat of fusion.Step 3: Calculate the heat required to raise the temperature from 1860°C to 2060°C using the formula Q = mcΔT.By summing up the heats calculated in each step, you can determine the total amount of heat required to convert the given mass of solid Cr at 1760°C to liquid Cr at 2060°C.
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Devon’s laboratory is out of material to make phosphate buffer. He is considering using sulfate to make a buffer instead. The pka values for the two hydrogens in H2SO4 are -10 and 2.
Will this approach work for making a buffer effective near a pH of 7?
O yes
O not enough information to answer
O no
What is the optimal pH for sulfate‑based buffers? Enter your answer as a whole number.
Answer:
Is not possible to make a buffer near of 7.
Optimal pH for sulfate‑based buffers is 2.
Explanation:
The dissociations of H₂SO₄ are:
H₂SO₄ ⇄ H⁺ + HSO₄⁻ pka₁ = -10
HSO₄⁻ ⇄ H⁺ + SO₄²⁻ pka₂ = 2.
The buffering capacity is pka±1. That means that for H₂SO₄ the buffering capacity is in pH's between -11 and -9 and between 1 and 3, having in mind that pH's<0 are not useful. For that reason, is not possible to make a buffer near of 7.
The optimal pH for sulfate‑based buffers is when pka=pH, that means that optimal pH is 2.
I hope it helps!
In which instance is a gas most likely to behave as an ideal gas?A.) At low temperatures, because the molecules are always far apartB.) When the molecules are highly polar, because IMF are more likelyC.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apartD.) At high pressures, because the distance between molecules is likely to be small in relation to the size of the molecules
Answer:
C.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apart.
Explanation:
For gas to behave as an ideal gas there are 2 basic assumptions:
The intermolecular forces (IMF) are neglectable.The volume of the gas is neglectable in comparison with the volume of the container.In which instance is a gas most likely to behave as an ideal gas?
A.) At low temperatures, because the molecules are always far apart. FALSE. At low temperatures, molecules are closer and IMF are more appreciable.
B.) When the molecules are highly polar, because IMF are more likely. FALSE. When IMF are stronger the gas does not behave as an ideal gas.
C.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apart. TRUE.
D.) At high pressures, because the distance between molecules is likely to be small in relation to the size of the molecules. FALSE. At high pressures, the distance between molecules is small and IMF are strong.
Carbon can bond to itself to form which of the following?
A)Branched-chain structures
B)Structures with both single and double bonds
C)Cyclic ring structures
D) Structures containing up to six bonds per carbon
B
Carbon has 4 valence electrons that can be involved in the formation of a covalent bond. This is why it can form various types of bonds with itself or other elements. Depending on the number of valence electrons involved in the binding, the bond can be a single, double or triple bond.
Explanation:
Remember the atoms sharing electrons in a covalent bond are aimed at achieving stable electron configuration. Carbon (2.4) being in the middle of the 2nd period seeks to be either 2 or 2.8.
This is why carbon based structures can vary so much due to the large variability in which the carbon atoms can bond. Remember carbon is the same element that forms graphite, diamond, and coal.
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Be sure to answer all parts. Write a balanced equation and Kb expression for the following Brønsted-Lowry base in water: benzoate ion, C6H5COO−. Include the states of all reactants and products in your equation. You do not need to include states in the equilibrium expression. Balanced equation: ⇌ Kb expression:
Answer:
The balanced reaction is:-
[tex]C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}[/tex]
[tex]K_b[/tex] expression is:-
[tex]K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}[/tex]
Explanation:
Benzoate ion is the conjugate base of the benzoic acid. It is a Bronsted-Lowry base and the dissociation of benzoate ion can be shown as:-
[tex]C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}[/tex]
The expression for dissociation constant of benzoate ion is:
[tex]K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}[/tex]
Final answer:
The balanced chemical equation for the reaction of benzoate ion, C6H5COO−, in water is C6H5COO−(aq) + H2O(l) ⇌ C6H5COOH(aq) + OH−(aq), and the Kb expression is Kb = [C6H5COOH][OH−] / [C6H5COO−].
Explanation:
The question involves writing a balanced equation for the reaction of the benzoate ion, C6H5COO−, as a Brønsted-Lowry base in water, and then writing the Kb expression for the reaction. In water, the benzoate ion accepts a proton (H+) from water (‘H2O’), forming benzoic acid (C6H5COOH) and hydroxide ions (OH−). The balanced chemical reaction is as follows:
C6H5COO−(aq) + H2O(l) ⇌ C6H5COOH(aq) + OH−(aq)
The Kb expression, which represents the base ionization constant, is written without including the states of the substances. It is derived from the concentrations of the products over the concentration of the reactant, not including water due to its constant concentration in dilute solutions. The Kb expression is:
Kb = [C6H5COOH][OH−] / [C6H5COO−]
Be sure to answer all parts. Enter your answers in scientific notation.
The following values are the only allowable energy levels of a hypothetical one-electron atom:
E6 = −2.0×10−19 J
E5 = −7.0×10−19 J
E4 = −11.0×10−19 J
E3 = −15.0×10−19 J
E2 = −17.0×10−19 J
E1 = −20.0×10−19 J
(a) If the electron were in the n = 5 level, what would be the highest frequency (and minimum wavelength) of radiation that could be emitted?
Frequency-
Wavelength-
(b) If the electron were in the n = 1 level, what would be the shortest wavelength (in nm) of radiation that could be absorbed without causing ionization?
Answer:
a) f = 3.02x10¹⁵ s⁻¹, and λ = 99.4 nm.
b) 99.4 nm
Explanation:
a) The energy of radiation is given by:
E = h*f
Where h is the Planck constant (6.626x10⁻³⁴ J.s), and f is the frequency. To have the highest frequency, the energy must be the highest too, because they're directly proportional. So we must use E = -E1 = 20x10⁻¹⁹ J
20x10⁻¹⁹ = 6.626x10⁻³⁴xf
f = 3.02x10¹⁵ s⁻¹
The wavelenght is the velocity of light (3.00x10⁸ m/s) divided by the frequency:
λ = 3.00x10⁸/3.02x10¹⁵
λ = 9.94x10⁻⁸ m = 99.4 nm
b) To have the shortest wavelength, it must be the highest energy and frequency, so it would be the same as the letter a) 99.4 nm.
Which best explains why bromine is soluble in mineral oil?
A.) Both substances are liquids
B.) Both substances have similar densities
C.) Both substances are made up of nonpolar molecules
D.) One substance is made up of polar molecules and the other is made up of nonpolar molecules
Answer:
C.) Both substances are made up of nonpolar molecules
Explanation:
Bromine is soluble in mineral oil because both substances are made of nonpolar molecules.A solute is highly soluble in a solvent that has the same chemical structure as it has. Bromine is a diatomic molecule that has dipole moments that cancel out to form a non-polar molecular. Mineral oil contains Carbon atoms bonded to hydrogen atoms forming nonpolar molecules.
Which of the following statements about organic chemistry is correct? Be careful, there may only be one word incorrect.
a. Organic compounds only contain carbon and hydrogen.
b. In a condensed structural formula, each carbon atom and its attached hydrogen atoms are written as a group in linear form.
c. Alkanes are a type of organic compound that have at least one multiple bond between two carbon atoms.
d. The different arrangements of alkanes, known as conformations, are the three-dimensional rotation around the double bond.
In "condensed structural formula" each carbon atom and its attached hydrogen atoms are written in group in the linear form is correct about organic chemistry.
Option B
Explanation:
"condensed structural formula" is a method of writing or presenting organic structures in line of text. It presents all the atoms, but excludes the vertical bonds and all the horizontal single bonds. In condensed structural formula, or we can say semi-structural formula, covalent bonds are not always presented or shown. When the formula or representation is written in line with covalent bonds being shown, then it is referred to as linear formula. For example- the condensed structural formulas of ethane, propane, and ethanol is written as follows:-
[tex]\mathrm{CH}_{3} \mathrm{CH}_{3},\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3},\text { and } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}[/tex]
Apart from these Statement, all others are false because they do not have any logic with respect to organic chemistry.
The correct statement about organic chemistry is: Alkanes are a type of organic compound that have at least one multiple bond between two carbon atoms.
The correct answer is option C.
a. Organic compounds typically contain carbon and hydrogen, but they can also include other elements like oxygen, nitrogen, sulfur, and more. So, the word "only" in this statement is incorrect.
b. In a condensed structural formula, each carbon atom and its attached hydrogen atoms are typically written as a group, but the arrangement isn't necessarily linear. It represents a simplified way to depict the structure of organic compounds, but the linear form is not a strict requirement. The statement is generally correct but could be misleading due to the term "linear."
c. Alkanes are a type of organic compound that consists of single bonds between carbon atoms, not multiple bonds. The statement is correct.
d. Conformations in organic chemistry refer to the different three-dimensional arrangements of atoms or groups around single bonds, such as in alkanes. The statement is incorrect because it mistakenly refers to the "double bond," which is not characteristic of alkanes.
In summary, statement (c) is the correct one. Alkanes are indeed organic compounds with only single bonds between carbon atoms. The other statements have minor inaccuracies or incorrect terms that make them less precise or entirely incorrect.
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Identify the option that is not a characteristic of esters:
1.They are prepared by reacting acids with alcohols.
2.They each contain a carbonyl group with an attached oxygen atom that is bonded to a carbon substituent.
3.They are responsible for the odors and flavors of many flowers, perfumes, and ripe fruits.
4.They are prepared by reducing alcohols or aldehydes whose -OH functional group is located on the carbon atom in the middle of the carbon chain.
Answer:
Option 4 is not the characteristics of esters
Explanation:
Esters :
They are organic compound derived from an organic or inorganic acid. in which one Hydroxyl (OH) group is replaced by an organic alkoxy (-O-R) group.
General formula of Esters:
Easter can be represented by a general formula that is
RCOOR
where R is any alkyl group
No to look at the characteristics of the Esters an find out the odd one option that is not correct.
Option 1 is the characteristics of esters.
Esters are obtained from the reaction of carboxylic acid and an alcohol.
CH₃COOH + CH₃OH -------------> CH₃COOCH₃ + H₂O
Option 2 is the characteristics of esters.
This option is also correct as if we see at its general formula
The carbonyl group attached to an Oxygen atom and this oxygen is attached to carbon of alkyl group
O==C------OR
Option 3 is also the characteristics of esters
The odors and flavors of many flowers, perfumes, and ripe fruits is due to the presence of ester. as ester have now hydrogen bonding and low vapor pressure and highly volatile that's why it is responsible for odors.
Option 4 is not the characteristics of esters
This option is wrong, as the carbon atom in the middle of the carbon chain can not be easy to replace, also the carbonyl carbon is directly attached to oxygen group and not Carbon group.