Answer:
Option B - reducing; NADH; negative
explanation:
Nicotinamide adenine dinucleotide, or NADH, is a compound, that's more actively used in the transportation of electron chain. It's evident that produced and make use NADH than FADH2 in the process that brings about energy creation.
NADH carries electrons from one reaction to another. Its Cofactor is in two forms in cells: NAD+ accepts electron from other molecule and hence reduced as an oxidizing agent while NADH is formed by this reaction, which can be used to donate electrons because it's a reducing agent. The main function of NAD is to transfer these reactions.
A 27.5 −g aluminum block is warmed to 65.6 ∘C and plunged into an insulated beaker containing 55.5 g water initially at 22.0 ∘C. The aluminum and the water are allowed to come to thermal equilibrium.
Answer:
Final temperature T is 26.197 degrees celcius.
Explanation:
As the system is in thermal equilibrium, the heat lost by aluminium is the heat gained by water.
specific heat capacity of aluminium = 0.9 J/g per degree celcius
specific heat capacity of water = 4.186 J/g per degree celcius
Let the final common temperature attained be "T".
Heat lost by aluminium = [tex](m)(c)(65.6-T)[/tex]
Heat gained by water = [tex](m)(c)(T-22)[/tex]
by law of conservation of energy,
Heat lost by hot body = Heat gained by cold body
[tex](m)(c)(65.6-T)[/tex] = [tex](m)(c)(T-22)[/tex]
[tex](27.5)(0.9)(65.6 - T) = (55.5)(4.186)(T - 22)[/tex]
calculating the value of T comes out as 26.197 degrees celcius.
Which biogeochemical cycle is the only one that does not include the atmosphere?
Answer:
Phosphorus cycle
Explanation:
Biogeochemical cycle, any of the natural pathways by which essential elements of living matter are circlated.
There are four types of biogeochemical cycle, they are ; water cycle,carbon cycle,nitrogen cycle and phosphorous cycle
Carbon cycle is the cycle in which photosynthesis and cellular respiration take place.
Water cycle involves transpiration.
Nitrogen cycle Is the cycle that is dependent upon bacteria for nitrogen fixation and denitrification.
Phosphorus cycle is one of the slowest biogeochemical cycle. It does not stay in the atmosphere, because it is normally in a liquid state at room temperature. It does not include the atmosphere.
Final answer:
The phosphorus cycle is the biogeochemical cycle that does not include an atmospheric component, differentiating it from cycles like water, carbon, and nitrogen.
Explanation:
The biogeochemical cycle that does not include the atmosphere is the phosphorus cycle. Unlike the water, carbon, and nitrogen cycles, the phosphorus cycle does not involve the atmospheric pathway. Phosphorus mainly cycles through rocks, soil, water, and living organisms. It is essential for life as part of DNA, RNA, and ATP molecules, which are vital for genetic information and energy transfer in cells. However, phosphorus doesn't enter the atmosphere in a gaseous phase; hence, it is distinct amongst biogeochemical cycles.
A geological sample is found to have a Pb-206/U-238 mass ratio of 0.337/1.00. Assuming there was no Pb-206 present when the sample was formed, how old is it? The half-life of U-238 is 4.5 × 109 years.
Answer:
[tex]Age=2.52*10^9 years[/tex]
Explanation:
To determine the age, we need to know how much U have become Pb, so first we have to calculate the Pb moles present in a sample:
[tex]n_{Pb}=\frac{0.337g}{206g/mol}=0.00164 mol[/tex]
[tex]n_{U}=\frac{1 g}{238g/mol}=0.0042 mol[/tex]
The percentage of U degradation:
[tex]P=\frac{n_{Pb}}{n_{Pb}+n_{U}}[/tex]
[tex]P=\frac{0.00164}{0.00164+0.0042}=0.28[/tex]
Assuming that the life time is linear:
[tex]Age=\frac{4.5*10^9 years}{0.5 life time}*0.28 life time[/tex]
[tex]Age=2.52*10^9 years[/tex]
Answer:
7.1 x 10^9 years
Explanation:
This is the answer according to Mastering Chemistry.
It takes 208.4 kJ of energy to remove 1 mole of electrons from 1 mole of atoms on the surface of rubidium metal. How much energy does it take to remove a single electron from an atom on the surface of solid rubidium?
Answer:
3.46*10⁻²² kJ
Explanation:
By the Avogadro's number, 1 mole of electrons at 1 mole of atoms correspond to 6.02x10²³ electrons. So it's necessary 208.4 kJ to remove 6.02x10²³ electrons. To remove a single electron:
6.02*10²³ electrons ---------------- 208.4 kJ
1 electron ---------------- x
By a simple direct three rule:
6.02*10²³ x = 208.4
x = 3.46*10⁻²² kJ
What is the polymerase chain reaction (PCR)?
a. A method to amplify a fragment of DNA.
b. A method to propagate a gene in bacteria.
c. A method to join two fragments of DNA together.
d. A method to determine the sequence of bases in a gene.
Answer: D.) A method to determine the sequence of bases in a gene.
Explanation:Polymerase chain reaction (PCR) is a method widely used in molecular biology to make several copies of a specific DNA segment. Using PCR, copies of DNA sequences are exponentially amplified to generate thousands to millions of more copies of that particular DNA segment
The energy change, ∆H, associated with the following reaction is +81 kJ. NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g) What is the expected energy change for the reverse reaction of nine moles of HOBr and two moles of NH3?
Answer:
162 kJ
Explanation:
The reaction given by the problem is:
NBr₃(g) + 3 H₂O(g) → 3 HOBr(g) + NH₃(g) ∆H = +81 kJIf we turn it around, we have:
3 HOBr(g) + NH₃(g) → NBr₃(g) + 3 H₂O(g) ∆H = -81 kJIf we think now of HOBr and NH₃ as our reactants, then now we need to find out which one will be the limiting reactant when we have 9 moles of HOBr and 2 moles of NH₃:
When we have 1 mol NH₃, we need 3 mol HOBr. So when we have 2 moles NH₃, we need 6 moles HOBr. We have more than 6 moles HOBr so that's our reactant in excess, thus NH₃ is our limiting reactant.-81 kJ is our energy change when there's one mol of NH₃ reacting, so we multiply that value by two when there's two moles of NH₃ reacting. The answer is 81*2 = 162 kJ.
Final answer:
The energy change for the reverse reaction of NBr3 and H2O to form HOBr and NH3 is -81 kJ. When considering nine moles of HOBr and two moles of NH3, thrice the stoichiometric amounts, the expected energy change would be three times as much, resulting in -243 kJ.
Explanation:
The energy change, ΔH, for a reaction indicates how much energy is absorbed or released during the reaction. In the given reaction, NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g), the energy change is +81 kJ, meaning 81 kJ of energy is absorbed in the forward reaction. Since energy is a state function and the magnitude of energy change is the same for a reaction whether it goes in the forward or reverse direction, the energy change for the reverse reaction would be -81 kJ. The reaction in reverse would release 81 kJ of energy.
However, the question asks for the expected energy change for reversing the original reaction with nine moles of HOBr and two moles of NH3. The energy change provided (+81 kJ) is for the stoichiometric amounts as per the original equation. For nine moles of HOBr and two moles of NH3, which are thrice the stoichiometric amounts indicated by the original reaction, the energy released would be three times as much. Therefore, the expected energy change for the reverse reaction under these conditions would be 3 × (-81 kJ) = -243 kJ.
Choose the molecule or compound that exhibits dipole-dipole forces as its strongest intermolecular force. a) BCl3 b) N2 c) H2 d) SO2 e) CBr4.
Answer:
SO₂
Explanation:
The dipole-dipole force is not only determined by the electron density around each atoms in the molecule (dependent of electronegativity difference), but also how the atoms in the molecules are arranged. In general, the more symmetric a molecule is, the less dipole force it exerts as each dipole moments cancels each other out.
Now let's examine each answer
b and c, N₂ and H₂ are composed of same type of atoms, therefore, no dipole moment occurs, and no dipole-dipole forces are exerted a and e, BCl₃ and CBr₄ are composed of atoms with different electronegativity, but are symmetric. With BCl₃ having trigonal planar structure and CBr₄ has tetragonal structure, each B-Cl and C- Br bond cancels out each other dipole moment, and thus, no dipole moments were generated.d, SO₂ structure is not linear, but is a little bent, this allows net dipole moment to occurs in this molecule as dipole moment from each S=O bond do not cancels each other outThe flashing of fireflies is the result of a chemical reaction, and the rate of flashing can be described by the Arrhenius equation. A certain batch of fireflies were observed to flash at a rate of 17.0 times per minute at 25°C and at a rate of 5.0 times per minute at 15°C. Use these data to find the apparent activation energy for the reaction that causes the flies to flash.
Answer:
-87.4 kJ/mol
Explanation:
We can rewrite Arrhenius' equation as:
ln(k₂/k₁) = -Ea/R * (1/T₂ - 1/T₁)
Where k₂ and k₁ are the rates of the reaction at temperatures T₂ and T₁, Ea is the activation energy and R is the universal gas constant.
For this problem:
k₂ = 5
k₁ = 17
T₂ = 15°C = 288.16 K
T₁ = 25°C = 298.16 K
R = 8.314 J/mol·K
We put the data in the equation and solve for Ea:
ln(5/17) = -Ea/8.314 J/mol·K * (1/288.16K - 1/298.16K)
-1.224 = -Ea/8.314 J/mol·K * 1.1639x10⁻⁴K⁻¹
Ea = -87416.78 J/mol ≅ -87.4 kJ/mol
The following sets of quantum numbers, listed in the order n, l, ml, and ms were written for the last electrons added to an atom. Identify which sets are valid and classify the others by the rule or principle that is violated.
choices for these answers are: "other violation," "pauli violation," and "valid."
a. 5 0 0 +1/2
5 0 0 -1/2
b. 4 1 -1 +1/2
4 1 0 +1/2
4 1 +1 +1/2
c. 3 2 -1 +1/2
3 2 0 +1/2
3 2 +1 +1/2
3 2 0 +1/2
3 2 +2 +1/2
d. 3 1 -1 +1/2
3 1 0 +1/2
3 3 +1 +1/2
Answer:
a. 5 0 0 +1/2
5 0 0 -1/2
Valid
b. 4 1 -1 +1/2
4 1 0 +1/2
4 1 +1 +1/2
Valid
c. 3 2 -1 +1/2
3 2 0 +1/2
3 2 +1 +1/2
3 2 0 +1/2
3 2 +2 +1/2
Pauli violation
d. 3 1 -1 +1/2
3 1 0 +1/2
3 3 +1 +1/2
other violation
Explanation:
The four quantum number and possible values are:
n = 1,2,3.....
l = 0 , (n-1), (n-2).....
m = +l , 0 , -l
s = [tex]+\frac{1}{2}[/tex] or [tex]-\frac{1}{2}[/tex]
Pauli's exclusion principle: No two electrons in an atom can have all the four quantum numbers same.
Let us check each case:
a. 5 0 0 +1/2
5 0 0 -1/2
Valid
b. 4 1 -1 +1/2
4 1 0 +1/2
4 1 +1 +1/2
Valid
c. 3 2 -1 +1/2
3 2 0 +1/2
3 2 +1 +1/2
3 2 0 +1/2
3 2 +2 +1/2
Pauli violation
The two electrons have same four quantum numbers
3 2 0 +1/2
3 2 0 +1/2
d. 3 1 -1 +1/2
3 1 0 +1/2
3 3 +1 +1/2
other violation
As mentioned above in the condition the value of "l" can be only less than "n"
So for 3 3 +1 +1/2 : n = 3 and l= 3, which is not valid.
The arrangement of electrons in orbitals are showed by four sets of quantum numbers.
According to the Pauli exclusion theory, no two electrons in an atom should have all the four quantum numbers as the same. According to this principle, the spin quantum number of electrons in an atom must differ even if they are in the same orbital.
For the first set;
5 0 0 +1/2
5 0 0 -1/2
This set, correctly corresponds to the 5s orbital so it is valid.
For the second set;
4 1 -1 +1/2
4 1 0 +1/2
4 1 +1 +1/2
This should have corresponding to the 4p orbital so it is valid.
For the third set;
3 2 -1 +1/2
3 2 0 +1/2
3 2 +1 +1/2
3 2 0 +1/2
3 2 +2 +1/2
This set should correspond to a 3d orbital but we can see that are two electrons in the set that has exactly the same quantum numbers of 3 2 0 +1/2. This violates the Pauli exclusion theory so we should mark "Pauli violation".
For the fourth set:
3 1 -1 +1/2
3 1 0 +1/2
3 3 +1 +1/2
This set should have corresponded to a 3p orbital but remember that the values of l only range from 0 to (n - 1). This means that we can not have n =3, l=3 so the arrangement 3 3 +1 +1/2 is not possible. We should mark "other violation".
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Aerosol cans have a label that warns the user not to use them above a certain temperature and not to dispose of them by incineration. Even an empty can contains residual gaseous propellant. For example, the residual pressure in a can is 1.11 atm when it is sitting on a shelf at 23 ∘C. If the can is placed on top of the furnace where the temperature reaches the boiling point of water, what is the pressure inside the can?
Answer:
1.40 atm
Explanation:
To answer this question we can use Gay-Lussac's law, which states:
P₁T₂=P₂T₁When volume and number of moles remain constant.
P₁ = 1.11 atmT₁ = 23°C ⇒ 23+273.16 = 296.16 KT₂ = Boiling point of water = 100 °C ⇒ 100+273.16 = 373.16 KP₂ = ?We put the known data in the equation and solve for P₂:
1.11 atm * 373.16 K = P₂ * 296.16 KP₂ = 1.40 atmUsing the ideal gas law, the pressure inside an aerosol can increases proportionately with temperature. If the can's temperature is raised to the boiling point of water (100°C or 373 K), the pressure increases from 1.11 atm to 1.40 atm.
Explanation:The pressure inside an aerosol can, like this one which has a residual propellant, can be explained by the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature when the volume and amount of gas are held constant. If the can is heated to the boiling point of water (100°C or 373 K), we can expect the pressure to increase proportionately.
To find out the new pressure, we need to convert the initial temperature to Kelvin (23°C = 296 K), and then use the formula P1/T1 = P2/T2 where P1 is the initial pressure, T1 is the initial temperature, P2 is the pressure at elevated temperatures, and T2 is the temperature at the boiling point of water.
So, P2 = P1×T2/T1 = 1.11 atm × 373 K / 296 K = 1.40 atm. Therefore, the pressure inside the can at the boiling point of water is 1.40 atm.
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One difference between a chemical reaction and a nuclear reaction is that in a nuclear reaction:
a) atoms retain their identity.
b) only the valence electrons are involved.
c) atoms often change from one element to another.
d) only small amounts of energy are absorbed or emitted.
Answer:
C
Explanation:
A nuclear reaction is one in which 2 atoms combine or an atom divides to give rise to entirely different atoms. This is basically the difference it has with a chemical reaction. In a chemical reaction, only new compounds are formed but not new elements. In the case of a nuclear reaction however, even if new elements are not formed, a different atom would be formed.
You sealed an Erlenmeyer flask that was determined to have a volume of 272.2 mL with a stopper. The density of air at 25°C is 0.001185 g/mL. Air has an effective molar mass of 28.96 g/mol. What amount of air was sealed in the flask?
Answer:
The amount of air is 11.1 mmol
Explanation:
Density shows the relation between mass and volume, so you can know the mass which has been sealed in the Erlenmayer.
Density = mass/ volume
Density . volume = mass
0.001185 g/ml . 272.2 ml = 0.322 g
We have the mass now, so the molar weight determinates the mols
Mass/molar weight = mol
0.322g/ 28.96 g/m = 0.0111 mol
To get a better number we can inform on milimols
Mols . 1000 = milimols
The amount of air sealed in the 272.2 mL Erlenmeyer flask is approximately 0.0111 mol, calculated using the given density and molar mass of air at 25°C.
Explanation:To determine the amount of air sealed in the flask, we need to use the density of the air (given as 0.001185 g/mL at 25°C), the volume of the flask and the molar mass of air.
First, using the density and the volume of the flask, we can calculate the mass of the air in the flask with the formula density = mass/volume. We multiply the given density of 0.001185 g/mL by the volume of 272.2 mL to find the mass of the air to be around 0.3224 g.
Next, to convert this mass to moles, we use the given molar mass of air (28.96 g/mol). Moles = mass / molar mass, hence the air in the flask is approximately 0.0111 mol. Thus, there is approximately 0.0111 mol of air sealed in the Erlenmeyer flask.
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Consider the titration of 30.0 ml of 0.050 m nh3 with 0.025 mhcl. Calculate the ph after the following volumes of titrant have been added.
Explanation of pH changes in a titration between NH₃ and HCl.
The pH in a titration is determined by the amount of acid and base before and after the equivalence point. In the given scenario, titrating NH₃ with HCl, the initial pH is determined by the NH₃ concentration before any HCl is added.
(a) 0.00 mL: Before any HCl is added, NH₃ is in excess, making the solution basic with a pH >7.
(b) 25.00 mL: At the equivalence point, the pH is determined by the salt formed during the reaction. For NH₃ titrated with HCl, the salt formed is NH₄Cl, making the pH acidic.
The stopcock connecting a 4.87 L bulb containing hydrogen gas at a pressure of 4.38 atm, and a 4.87 L bulb containing argon gas at a pressure of 2.90 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is _____atm
Answer:
P = 3,63 atm
Explanation:
Using PV=nRT and knowing the temperature in the problem remains constant it is possible to rewrite this formula, thus:
PV = kn, where k is RT (Constant), P is pressure, V is volume and n are moles.
The first bulb contains:
4,87L×4,38atm = 21,3k moles
And the second bulb:
4,87L×2,90atm = 14,1k moles
When gases are mixed, the total moles are:
21,3k moles + 14,1k moles = 35,4k moles
As total volume is 4,87L + 4,87L = 9,74L
Replacing:
P = kn/V
P = 35,4k moles / 9,74L
P = 3,63 atm
I hope it helps!
A 100 W light bulb is placed in a cylinder equipped with a moveable piston. The light bulb is turned on for 2.0×10−2 hour, and the assembly expands from an initial volume of 0.85 L to a final volume of 5.88 L against an external pressure of 1.0 atm.
a)Use the wattage of the light bulb and the time it is on to calculate ΔU in joules (assume that the cylinder and light bulb assembly is the system and assume two significant figures). Express your answer using two significant figures.
b) Calculate w. Express your answer using two significant figures.
c) Calculate q. Express your answer using two significant figures.
Answer:
(a) ΔU = 7.2x10²
(b) W = -5.1x10²
(c) q = 5.2x10²
Explanation:
From the definition of power (p), we have:
[tex] p = \frac {\Delta W}{\Delta t} = \frac {\Delta U}{\Delta t} [/tex] (1)
where, p: is power (J/s = W (watt)) W: is work = ΔU (J) and t: is time (s)
(a) We can calculate the energy (ΔU) using equation (1):
[tex] \Delta U = p \cdot \Delta t = 100 \frac{J}{s} \cdot 2.0\cdot 10^{-2} h \cdot \frac{3600s}{1h} = 7.2 \cdot 10^{2} J [/tex]
(b) The work is related to pressure and volume by:
[tex] \Delta W = -p \Delta V [/tex]
where p: pressure and ΔV: change in volume = V final - V initial
[tex] \Delta W = - p \cdot (V_{fin} - V_{ini}) = - 1.0 atm (5.88L - 0.85L) = - 5.03 L \cdot atm \cdot \frac{101.33J}{1 L\cdot atm} = -5.1 \cdot 10^{2} J [/tex]
(c) By the definition of Energy, we can calculate q:
[tex] \Delta U = \Delta W + \Delta q [/tex]
where Δq: is the heat transfer
[tex] \Delta q = \Delta U - \Delta W = 7.2 J - (-5.1 \cdot 10^{2} J) = 5.2 \cdot 10^{2} J [/tex]
I hope it helps you!
Draw the intermediate (in the first box) that forms in the reaction of 1-methylcyclopentene with br2 in ccl4, then draw the expected product (in the second box). do not provide stereochemistry. interactive 3d display mode
Answer:
The product of the reaction is 1,2 di-bromo-1-methylcyclopentane.
Explanation:
In the reaction of 1-methylcyclopentene with Br2 in CCl4, a bromonium ion intermediate forms, followed by nucleophilic attack by Cl-, resulting in the product 1-chloromethylcyclopentane.
In the reaction of 1-methylcyclopentene with Br2 in CCl4, a bromonium ion intermediate is formed. The bromonium ion is a three-membered ring with a bromine atom attached to two adjacent carbon atoms. Due to the lack of stereochemistry information, the structure of the intermediate is a general representation without specifying the stereochemical arrangement.
Following the formation of the bromonium ion, it undergoes a nucleophilic attack by the chloride ion (Cl-) from the solvent CCl4. The chloride ion opens the bromonium ring, leading to the substitution of bromine by chloride on one of the adjacent carbons. The product of this reaction is 1-chloromethylcyclopentane. The chlorine atom is now attached to the carbon that originally carried the methyl group.
This reaction mechanism is a classic example of halogenation of alkenes, where the bromine adds to the alkene in a concerted manner, forming a cyclic intermediate. The subsequent nucleophilic attack by the chloride ion results in the replacement of bromine with chlorine, leading to the final halogenated product.
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Be sure to answer all parts. Write the equations representing the following processes. In each case, be sure to indicate the physical state of the ions. 1. The electron affinity of S⁻:
2. The third ionization energy of titanium:
3. The electron affinity of Mg²⁺:
4. The ionization energy of O²⁻:
Answer:
1.[tex]S^{-}(g)+e^{-} \rightarrow S^{2-}(g)[/tex]
2.[tex]Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}[/tex]
3.The electron affinity of [tex]Mg^{2+}[/tex] is zero.
4.[tex]O^{2-}(g) \rightarrow O^{-}(g)+e^{-}[/tex]
Explanation:
1.
Electron affinity:
It is defined as the amount of energy change when an electron is added to atom in the gaseous phase.
The electron affinity of [tex]S^{-}[/tex] is as follows.
[tex]S^{-}(g)+e^{-} \rightarrow S^{2-}(g)[/tex]
2.
Ionization energy:
Amount of energy required to removal of an electron from an isolated gaseous atom.
The third ionization energy of Titanium is as follows.
[tex]Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}[/tex]
3.
The electronic configuration of Mg: [tex]1s^{2}2s^{2}2p^{6}3s^{2}[/tex]
By the removal of two electrons from a magnesium element we get [tex]Mg^{2+}[/tex] ion.
[tex]Mg^{2+}[/tex] has inert gas configuration i.e,[tex]1s^{2}2s^{2}2p^{6}[/tex]
Hence, it does not require more electrons to get stability.
Therefore,the electron affinity of [tex]Mg^{2+}[/tex] is zero.
4.
The ionization energy of [tex]O^{2-}[/tex] is follows.
[tex]O^{2-}(g) \rightarrow O^{-}(g)+e^{-}[/tex]
The electron affinity equation for S⁻ is S⁻(g) + e⁻ → S²⁻(g). The third ionization energy equation for titanium is Ti²⁺(g) → Ti³⁺(g) + e⁻. The equation representing the electron affinity of Mg²⁺ is Mg²⁺(g) + e⁻ → Mg³⁺(g). The ionization energy equation for O²⁻ is O²⁻(g) → O³⁻(g) + e⁻.
Explanation:1. The electron affinity of S⁻: The electron affinity represents the energy change that occurs when an electron is added to a neutral atom or ion. For sulfur (S⁻), the equation representing this process is: S⁻(g) + e⁻ → S²⁻(g). Here, the physical state of S⁻ and S²⁻ is gas.
2. The third ionization energy of titanium: Ionization energy is the energy required to remove an electron from an atom or ion. For titanium, the equation representing the third ionization energy is: Ti²⁺(g) → Ti³⁺(g) + e⁻. The physical state of Ti²⁺ and Ti³⁺ is gas.
3. The electron affinity of Mg²⁺: The electron affinity of a cation, such as Mg²⁺, is the energy change when an electron is added to the cation. Here, the equation representing this process is: Mg²⁺(g) + e⁻ → Mg³⁺(g). The physical state of Mg²⁺ and Mg³⁺ is gas.
4. The ionization energy of O²⁻: Ionization energy is the energy required to remove an electron from an atom or ion. For oxygen, the equation representing the ionization energy of O²⁻ is: O²⁻(g) → O³⁻(g) + e⁻. The physical state of O²⁻ and O³⁻ is gas.
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A bath tub is half full of warm water. How could the thermal energy of the water be increased? A) Add hot water to the tub. B) Heat the water in the tub without adding more water. C) Add more water at the same temperature to the tub. D) Transfer the water to a smaller container.
Answer:
A, B, & C
Explanation:
got it right on edu
The thermal energy of the bathtub can be increased by adding hot water, heating the water, and adding more water to the tub at the same temperature. Thus, options A, B, and C are correct.
What is thermal energy?Thermal energy is the increase in the collision and the kinetic energy of particles of matter due to a rise in the temperature of the substance. This movement produces energy called thermal energy.
The thermal energy of the substance can be increased by providing them with the same energy molecules. The addition of the hot water will increase the energy due to increased collision of the particles of the hot water.
Heating the water will increase the kinetic energy and the rate of a collision resulting in increased thermal energy. Also, water with the same temperature can be added to increase energy.
Therefore, in options A, B, and C adding hot water and heating water increases the thermal energy.
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A student carried out a titration using HC2H3O2(aq) and NaOH(aq). The net ionic equation for the neutralization reaction that occurs during the titration is represented above. The NaOH(aq) was added from a buret to the HC2H3O2(aq) in a flask. The equivalence point was reached when a total of 20.0mL of NaOH(aq) had been added to the flask. How does the amount of HC2H3O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compare to the amount of HC2H3O2(aq) in the flask after the addition of 1.0mL of NaOH(aq), and what is the reason for this result?
With every addition of NaOH, the concentration of HC2H3O2 in the solution decreases due to the neutralization reaction. Thus, the quantity of HC2H3O2 after adding 5.0 mL of NaOH will be less than the amount after adding 1.0 mL of NaOH. This is due to the stoichiometry of the reaction, where one mole of added NaOH neutralizes one mole of acetic acid.
Explanation:In this titration analysis, the neutralization reaction involves acetic acid (HC2H3O2) and Sodium Hydroxide (NaOH), creating sodium acetate and water as products. When 1.0 mL of NaOH is added, it reacts with the acetic acid present in the flask, reducing its quantity. When 5.0 mL of NaOH is added, more acetic acid reacts, and therefore, the amount of HC2H3O2 in the flask decreases more significantly compared to when only 1 mL was added. The principle of this result is based upon the stoichiometry of the neutralization reaction, indicating that for each mole of NaOH added, one mole of acetic acid is neutralized. Therefore, as more NaOH is added, more HC2H3O2 is neutralized, hence contributing to a lower concentration of HC2H3O2 in the solution.
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The amount of HC2H3O2(aq) in the flask will be less after the addition of 5.0mL NaOH than after the addition of 1.0mL NaOH. This happens because the NaOH neutralizes the acid directly proportional to its quantity added. More the NaOH added, more the acidic substance HC2H3O2 gets neutralized.
Explanation:In a titration process involving HC2H3O2(aq) and NaOH(aq), following the neutralization equation, each acid molecule reacts with one base molecule. Hence, the quantity of HC2H3O2 will reduce in direct proportion to the amount of NaOH added. So if you add 5.0mL of NaOH, you will neutralize 5 times as much HC2H3O2 as when you add 1.0mL of NaOH.
Therefore, the quantity of HC2H3O2(aq) remaining in the flask after the addition of 5.0mL NaOH will be less than that after the addition of 1.0mL NaOH. The reason is that more NaOH means more HC2H3O2(aq) has been neutralized into water and the salt NaC2H3O2.
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List the 4 major groups of inorganic chemicals common in cells:
Water, Carbon Dioxide, Acids, and Oxygen are the inorganic chemicals common in cells.
What are inorganic chemicals?A substance without both carbon and hydrogen is said to be an inorganic compound.Many inorganic substances, like water (H₂O) and the hydrochloric acid (HCl) secreted by your stomach, do contain hydrogen atoms. In contrast, carbon atoms are only present in a small number of inorganic molecules. An example of one of the few is carbon dioxide (CO₂).Water makes up as much as 70% of an adult's body weight. The cells that make up tissues and organs have water both inside of them and in the spaces between them.A substance that releases hydrogen ions (H⁺) in solution is referred to be an acid. Because it releases all of its H⁺ in the stomach's watery environment, hydrochloric acid (HCl), which is secreted by cells in the lining of the stomach, is a powerful acid. This strong acid helps with digestion and destroys ingested microorganisms.Carbon dioxide (CO₂) is expelled during respiration in exchange for oxygen (O₂) from the surrounding air. At the cellular level, O₂ is utilized to produce CO₂ while converting metabolic energy from foods into adenosine triphosphate (ATP).
Thus, inorganic chemicals like Water, Carbon Dioxide, Acids, and Oxygen
are essentail for cells.
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Both covalent-network solids and ionic solids can have melting points well in excess of room temperature, and both can be poor conductors of electricity in their pure form. However, in other ways their properties are quite different.1.Which type of solid is more likely to dissolve in water?2.Which type of solid can become a considerably better conductor of electricity via chemical substitution?
Ionic solids are more likely to dissolve in water and can become better conductors of electricity via chemical substitution.
Explanation:Both covalent-network solids and ionic solids have different properties. Ionic solids dissolve readily in water and can conduct electricity when melted or dissolved because their ions are free to move. Covalent-network solids, on the other hand, are insoluble in water and are poor conductors of electricity in any state because their particles are electrically neutral.
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The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell:________
Pt(s)
|H2(g,1atm)
|H+(aq,1.0M)
|Au3+(aq,?M)
|Au(s).
What is the concentration of Au3+ in the solution if Ecell is 1.23 V ?
Answer: The concentration of [tex]Au^{3+}[/tex] is [tex]1.096\times 10^{-6}[/tex]
Explanation:
The given chemical cell follows:
[tex]Pt(s)|H_2(g,1atm)|H^+(aq,1.0M)||Au^{3+}(aq,?M)|Au(s)[/tex]
Oxidation half reaction: [tex]H_2(g,1atm)\rightarrow 2H^{+}(aq,1.0M)+2e^-;E^o_{2H^{+}/H_2}=0.0V[/tex] ( × 3)
Reduction half reaction: [tex]Au^{3+}(aq,?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V[/tex] ( × 2)
Net cell reaction: [tex]3H_2(g,1atm)+2Au^{3+}(aq,?M)\rightarrow 6H^{+}(aq,1.0M)+2Au(s)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
Putting values in above equation, we get:
[tex]E^o_{cell}=1.50-(0.0)=1.50V[/tex]
To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = 1.23 V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.50 V
n = number of electrons exchanged = 6
[tex][H^{+}]=1.0M[/tex]
[tex][Au^{3+}]=?M[/tex]
Putting values in above equation, we get:
[tex]1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})[/tex]
[tex][Au^{3+}]=-1.0906\times 10^{-6},1.096\times 10^{-6}[/tex]
Neglecting the negative value because concentration cannot be negative.
Hence, the concentration of [tex]Au^{3+}[/tex] is [tex]1.096\times 10^{-6}[/tex]
Consider the following reaction. N2(g) + O2(g) equilibrium reaction arrow 2 NO(g) If the equilibrium partial pressures of N2, O2, and NO are 0.15 atm, 0.33 atm, and 0.050 atm, respectively, at 2200°C, what is KP?
Explanation:
As the given reaction equation is as follows.
[tex]N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)[/tex]
Hence, expression for [tex]K_{p}[/tex] of this reaction is as follows.
[tex]K_{p} = \frac{P^{2}_{NO}}{P_{N_{2}} \times P_{O_{2}}}[/tex]
Now, putting the given values into the above expression as follows.
[tex]K_{p} = \frac{P^{2}_{NO}}{P_{N_{2}} \times P_{O_{2}}}[/tex]
= [tex]\frac{(0.050)^{2} atm}{(0.15 atm) \times (0.33 atm)}[/tex]
= [tex]5.05 \times 10^{-2}[/tex]
Thus, we can conclude that the value of [tex]K_{p}[/tex] is [tex]5.05 \times 10^{-2}[/tex].
Final answer:
The equilibrium constant, Kp, for the reaction N2(g) + O2(g) = 2 NO(g) at the given conditions is approximately 0.1010 when calculated using the given equilibrium partial pressures.
Explanation:
The question is asking us to calculate the equilibrium constant, KP, for the reaction given the equilibrium partial pressures of the reactants and products at a certain temperature. Using the reaction N2(g) + O2(g) = 2 NO(g), we can express the equilibrium constant KP in terms of the partial pressures of the gases:
KP = (PNO)2 / (PN2 * PO2)
Substituting in the given partial pressures:
KP = (0.050 ATM)2 / (0.15 ATM * 0.33 ATM) = 0.0050 / 0.0495 = 0.1010
So, the equilibrium constant KP at 2200°C for this reaction is approximately 0.1010.
Each molecule of hemoglobin combines with four molecule of oxygen in order to transport oxygen throughout the body. It is observed that 1.51 g of hemoglobin combines with 2.30 mL of O 2 at 37 °C and 743 torr. What is the molar mass of hemoglobin?
Answer:
68325.8 g/mol
Explanation:
Given that:
Pressure = 743 torr
Temperature = 37 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (37 + 273.15) K = 310.15 K
T = 310.15 K
Volume = 2.30 mL = 0.00230 L ( 1 mL = 0.001 L)
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 62.364 L.torr/K.mol
Applying the equation as:
743 torr × 0.00230 L = n × 62.364 L.torr/K.mol × 310.15 K
⇒n = 8.84 × 10⁻⁵ moles
Since, given that:-
4 moles of oxygen gas is transferred by 1 mole of haemoglobin.
8.84 × 10⁻⁵ moles of oxygen gas is transferred by 8.84 × 10⁻⁵/4 moles of haemoglobin.
Moles of haemoglobin = 2.21 × 10⁻⁵ moles
Mass = 1.51 g
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]2.21\times 10^{-5}\ mol= \frac{1.51\ g}{Molar\ mass}[/tex]
[tex]Molar\ mass= 68325.8\ g/mol[/tex]
The question is asking for the molar mass of hemoglobin, which can be calculated using the Ideal Gas Law and the information provided. Hemoglobin, a protein in red blood cells, combines with four oxygen molecules, the fact that is fundamental to this calculation.
Explanation:The molar mass of hemoglobin can be calculated using the Ideal Gas Law (PV=nRT). In this case, the volume of O2 (V) is 2.30 mL or 0.0023 L, the pressure (P) is 743 torr (or approximately 0.977 atm), the gas constant (R) is 0.0821 L⋅atm/mol⋅K, and the temperature (T) is 37 °C or 310 K. We can solve this equation to find the amount of moles of O2 (n). As each molecule of hemoglobin combines with four molecules of O2, we can then find out the amount of moles of hemoglobin, and knowing its mass (1.51 g), finally calculate its molar mass (grams per mole).
To confirm the correctness of this procedure, remember that hemoglobin is a protein found in red blood cells which combines with what? Oxygen molecules - four of them, in fact! This is crucial for the transport of oxygen throughout the body, and it also directly pertains to and informs our calculation of hemoglobin's molar mass.
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Copper, silver, and gold have all been know since ancient times because they appear in nature in____ and were thus discovered thousands of years ago. But the majority of elements__________ and, consequently, are hard to find in nature.
Answer:
elemental form, readily form compounds
Explanation:
Copper, silver, and gold have all been know since ancient times because they appear in nature in_elemental form___ and were thus discovered thousands of years ago. But the majority of elements__readily form compounds_and, consequently, are hard to find in nature.
the reaction of 50 mL of gas with 50 mL of gas via the equation: Cl2(g) + C2H4(g) ➔ C2H4Cl2 (g) will produce a total of __________ mL of products if pressure and temperature are kept constant.
Explanation:
The given data is as follows.
50 ml of [tex]Cl_{2}[/tex], 50 ml of [tex]C_{2}H_{4}[/tex]
And, it is known that at STP 1 mole of a gas occupies 22.4 L. Hence, moles present in 50 ml of gas are as follows.
[tex]\frac{50}{22.4 \times 1000}[/tex] (As 1 L = 1000 ml)
= [tex]2.23 \times 10^{-3}[/tex] moles
So, according to the given equation [tex]2.23 \times 10^{-3}[/tex] moles of [tex]Cl_{2}[/tex] reacts with [tex]2.23 \times 10^{-3}[/tex] moles of [tex]C_{2}H_{4}[/tex].
Hence, moles of [tex]C_{2}H_{4}Cl_{2}[/tex] is equal to the moles of [tex]C_{2}H_{4}[/tex] and [tex]Cl_{2}[/tex].
Therefore, moles of [tex]C_{2}H_{4}Cl_{2}[/tex] = [tex]2.23 \times 10^{-3}[/tex] moles
1 mole of [tex]C_{2}H_{4}Cl_{2}[/tex] = 22.4 L
[tex]2.23 \times 10^{-3}[/tex] moles = [tex]22.4 \times 2.23 \times 10^{-3} moles[/tex]
= 50 ml of product
Thus, we can conclude that 50 ml of products if pressure and temperature are kept constant.
I have very good ability to conduct electricity. I am never found alorie in nature. When I combine with other elements, I usually give up my one valence electron. I am the only element in my group with a one-letter symbol.
Answer:
Potassium
Explanation:
The description here fits potassium. Potassium is not found alone in nature as it is very reactive. Hence, it tends to occur in relation with other elements because of its high reactivity.
Because it is stated that it gives up its single valence electron, this shows that it belongs to group one of the periodic table. We might be tempted to think hydrogen is the correct answer. But it is interesting to note that hydrogen is not even supposed to be placed in group one at all. It is only placed there for convenience
The total energy of all the particles in a substance is called
Answer:
Internal energy
Explanation:
Particles have two main energies that scientists acknowledge.
The energy that is associated with their movement which we call kinetic energy. It is worth noting that particles have different kinetic energy, however the average kinetic energy that they have is directly proportional to temperature.The energy that is due to the interactive forces (attraction or repulsion) between particles is called potential energyWhen these two types of energies are considered collectively, we call that internal energy
Select all that apply. Which two of the following equations represent an exothermic reaction? NH3(g) + 12.0 kcal → ½N2(g) + 3/2H2(g) C(graphite) → C(diamond), ΔH = -0.45 kcal/mole C C + 2S → CS2, ΔH = 27,550 cal CH4 + 2O2 → CO2 + 2H2O + 212,800 cal 2H2 O → 2H2 + O2, ΔH = +58 kcal/mole H2O
Answer:
C(graphite) → C(diamond), ΔH = -0.45
CH4 + 2O2 → CO2 + 2H2O + 212,800 cal
Explanation:
NH3(g) + 12.0 kcal → ½N2(g) + 3/2H2(g) in this reaction we see that energy is added to the reactants so this is an endorthermic process. It takes in energy
C(graphite) → C(diamond), ΔH = -0.45 kcal/mole The change in enthalpy for this reaction is listed and it is negative, telling us that this is an exothermic reaction.
C + 2S → CS2, ΔH = 27,550 cal The change of enthalpy for this reaction is listed and it is positive which tells us that the reaction is endothermic reaction
CH4 + 2O2 → CO2 + 2H2O + 212,800 cal In this reaction, the energy is part of the equation on the products side. It is given out. this tells us that the reaction is exothermic
2H2 O → 2H2 + O2, ΔH = +58 kcal/mole H2O In this reaction, the energy is part of the equation on the products side. It is given out. this tells us that the reaction is exothermic
A possible mechanism for the overall reaction represented above is the following.
NO(g) + NO(g) → N2O2(g) slow
N2O2(g) + O2(g) → 2NO2(g) fast
Which of the following rate expressions agrees best with this possible mechanism?
A) Rate = k[NO]2
(D) Rate = k[NO]2[O2]
(B) Rate = k[NO]
(E) Rate = k[N2O2][O2][O2]
(C) Rate = k[NO]2[O2]
Answer:
A
Explanation:
The slow step of the reaction is the rate determining step. This means to get the rate law for a chemical activity with slow and fast steps, we have to consider the slow step if we are to write the rate law successfully.
Since the compound NO is reacting with itself, we have to raise the value of the concentration in the square bracket by 2. That is why we have the concentration squared.
Answer: it’s c
Explanation: