Photophosphorylation of sugar moiety produces ATP (adenosine triphosphate) which stores chemical energy that can be used for the Calvin cycle.
What is Calvin cycle?The Calvin cycle, often known as the Calvin-Benson cycle after its discoverers, is a series of chemical processes that collect carbon dioxide.
It is also referred to as the C3 cycle alone. C3 plants are defined as those that only fix carbon through the Calvin cycle. In the stroma of chloroplasts, carbon dioxide diffuses and mixes with the five-carbon sugar ribulose 1, 5-biphosphate (RuBP).
Rubisco, a large molecule that may be the most prevalent organic compound on Earth, is the name of the enzyme that catalyzes this reaction.
The three-carbon chemical 3-PGA is changed into another three-carbon compound termed G3P by ATP and NADPH using their stored energy. A reduction reaction is the name for this kind of reaction.
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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:
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.
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
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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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.
The following change occurs in acidic solution: S^2+ + Cr_2O_7^2+ S + Cr^3+ Complete and balance the foregoing equation. In the balanced equation, for every mole of Cr_2O_7^2 that reacts, moles of H+ are consumed. A. 5 B. 7 C. 8 D. 10 E. 14
Answer : The correct option is, (E) 14
Explanation :
Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.
Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.
Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.
Rules for the balanced chemical equation in acidic solution are :
First we have to write into the two half-reactions.Now balance the main atoms in the reaction.Now balance the hydrogen and oxygen atoms on both the sides of the reaction.If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion [tex](H^+)[/tex] at that side where the less number of hydrogen are present.Now balance the charge.The given chemical reaction is,
[tex]S^{2-}+Cr_2O_7^{2-}\rightarrow S+Cr^{3+}[/tex]
The oxidation-reduction half reaction will be :
Oxidation : [tex]S^{2-}\rightarrow S[/tex]
Reduction : [tex]Cr_2O_7^{2-}\rightarrow Cr^{3+}[/tex]
First balance the main element in the reaction.
Oxidation : [tex]S^{2-}\rightarrow S[/tex]
Reduction : [tex]Cr_2O_7^{2-}\rightarrow 2Cr^{3+}[/tex]
Now balance oxygen atom on both side.
Oxidation : [tex]S^{2-}\rightarrow S[/tex]
Reduction : [tex]Cr_2O_7^{2-}\rightarrow 2Cr^{3+}+7H_2O[/tex]
Now balance hydrogen atom on both side.
Oxidation : [tex]S^{2-}\rightarrow S[/tex]
Reduction : [tex]Cr_2O_7^{2-}+14H^+\rightarrow 2Cr^{3+}+7H_2O[/tex]
Now balance the charge.
Oxidation : [tex]S^{2-}\rightarrow S+2e^-[/tex]
Reduction : [tex]Cr_2O_7^{2-}+14H^++6e^-\rightarrow 2Cr^{3+}+7H_2O[/tex]
In order to balance the charge, we multiply oxidation reaction by 3. Thus, added both equation, we get the balanced redox reaction.
The balanced chemical equation in acidic medium will be,
[tex]3S^{2-}+Cr_2O_7^{2-}+14H^+\rightarrow 3S+2Cr^{3+}+7H_2O[/tex]
From the balanced chemical equation we conclude that, the number of moles of [tex]H^+[/tex] consumed from every mole of [tex]Cr_2O_7^{2-}[/tex] are 14 moles.
Hence. the correct option is, (E) 14
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.
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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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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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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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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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.
For water ∆H°vap = 40.7 kJ/mol at 100.°C, its boiling point. Calculate ∆S° for the vaporization of 1.00 mol water at 100.°C and 1.00 atm pressure.
Answer : The value of change in entropy for vaporization of water is [tex]1.09\times 10^2J/mol[/tex]
Explanation :
Formula used :
[tex]\Delta S^o=\frac{\Delta H^o_{vap}}{T_b}[/tex]
where,
[tex]\Delta S^o[/tex] = change in entropy of vaporization = ?
[tex]\Delta H^o_{vap}[/tex] = change in enthalpy of vaporization = 40.7 kJ/mol
[tex]T_b[/tex] = boiling point temperature of water = [tex]100^oC=273+100=373K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S^o=\frac{\Delta H^o_{vap}}{T_b}[/tex]
[tex]\Delta S^o=\frac{40.7kJ/mol}{373K}[/tex]
[tex]\Delta S^o=1.09\times 10^2J/mol[/tex]
Therefore, the value of change in entropy for vaporization of water is [tex]1.09\times 10^2J/mol[/tex]
The entropy change ∆S° for the vaporization of 1.00 mol of water at its boiling point is calculated to be 109.09 J/K·mol using the given enthalpy of vaporization.
Explanation:To calculate ∆S° for the vaporization of 1.00 mol water at 100.°C and 1.00 atm pressure when ∆H°vap = 40.7 kJ/mol, we can use the thermodynamic relation ∆G° = ∆H° - T∆S° and the fact that at the boiling point, ∆G° for the phase change is zero. This would mean the ∆S° can be calculated as ∆S° = ∆H°vap/T. Plugging in the known values,
∆S° = 40.7 kJ/mol / 373.15 K
∆S° = 0.10909 kJ/K·mol
Since 1 kJ = 1000 J, we convert this to J:
∆S° = 109.09 J/K·mol
Therefore, the entropy change ∆S° for the vaporization of 1.00 mol of water at its boiling point is 109.09 J/K·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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Which set of chemical name and chemical formula for the same compound is correct?
A) Ammonium sulfite (NH4)2S
B) Iron(lll) phosphate, FePO4
C) lithium carbonate, LiCO3
D) magnesium dichromate, MgCrO4
Answer:
B) Iron(lll) phosphate, FePO4
Option.B-Iron(lll) phosphate, FePO4 is the correct answer.
A monoclinic crystal powder with the chemical formula FePO4, ferric phosphate is sometimes referred to as ferric phosphate and ferric orthophosphate.
Chemical Formula and Chemical name of Iron(lll) phosphate, FePO4With the chemical formula FePO4, iron(III) phosphate, commonly known as ferric phosphate, is an inorganic substance.Four polymorphs of FePO4 and two polymorphs of the dihydrate FePO4(H2O)2 are two examples of related materials.These materials are found in the mineral world and have minimal technical uses.The most prevalent type of FePO4 adopts a -quartz structure. As a result, the substance comprises of phosphate and tetrahedral Fe(III) sites.As a result, the molecular geometry of the P and Fe is tetrahedral. A phase shift to a denser structure with octahedral Fe centers takes place at high pressures.Additionally known are two orthorhombic structures and a monoclinic phase.The Fe center is octahedral with two cis water ligands in the two polymorphs of the dihydrate.In the production of steel and metals, iron(III) phosphate can be employed. Iron phosphate stops further metal oxidation when it is adhered to a surface made of metal.The iron pillar of Delhi's resistance to corrosion is in part due to its existence.In order to promote adhesion to the iron or steel substrate and avoid corrosion, which can result in the premature failure of following coating processes, iron phosphate coatings are frequently used as a pre-treatment for painting or powder coating.Additionally, it can be applied to surfaces made of iron or steel to bond textiles, wood, and other materials.To know more about similar question click here-
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The 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
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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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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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]
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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]
Fireworks that contain metallic salts such as sodium, strontium, and barium can generate bright colors. A technician investigates what colors are produced by the metallic salts by performing flame tests. During a flame test, a metallic salt is heated in the flame of a gas burner. Each metallic salt emits a characteristic colored light in the flame. 3. State how bright-line spectra viewed through a spectroscope can be used to identify the metal ions in the salts used in the flame tests. 4. Explain, in terms of electrons, how a strontium salt emits colored light. 5. Explain why the electron configuration of 2-7-1-1 represents a sodium atom in an excited state.
Answer:
See explanation below
Explanation:
3. Each atom has unique electron energy level when compared to other, when electron relaxes from an excited energy level, it emits energy in the form of electromagnetic waves with energy equals to the difference between energy of ground and excited state. As electromagnetic wave wavelength obeys planck relation
[tex]c = \lambda f[/tex]
And the energy of each photon in electromangetic wave of specific frequency is
[tex]E=hf[/tex]
Where h is planck's constant
We can see that the wavelength, which determines color and position of bright line in spectroscope corresponds directly to the energy of electromagnetic wave, and thus, the characteristic energy of atom.
4. From the answer above, you now know that electronic transition can emit electron with specific wavelength. In strontium, emission of colored light occurs by the relaxation of electrons in excited state such that the difference between excited energy level and ground level falls in the energy of visible light spectral range.
5. In Na, ground state configuration is 2-8-1. 2-7-1-1 indicates that one of electron in ground state got excited by external energy to excited state, and thus, indicates that Na atom is in excited state
Final answer:
Bright-line spectra viewed through a spectroscope can be used to identify the metal ions in salts used in flame tests.
Explanation:
Bright-line spectra viewed through a spectroscope can be used to identify the metal ions in the salts used in flame tests. Each metal ion has a unique set of energy levels, and when these ions are heated in a flame, their electrons get excited and jump to higher energy levels. As the electrons return to their ground states, they release energy in the form of light. This emitted light can be observed through a spectroscope, which separates the different wavelengths of light, creating a bright-line spectrum. By comparing the bright-line spectrum of an unknown salt to the known spectra of different metal ions, the metal ions present in the unknown salt can be identified.
A one or two letter abbreviation for a chemical element. True or false
Answer:
True
Explanation:
All the chemical elements have their abbreviations as one or two letters
The first letter in the abbreviation is a capital letter and the second letter is always a small letter
Majority of abbreviations are based on their english names
Final answer:
The statement regarding a chemical symbol being a one- or two-letter abbreviation for a chemical element is true. These symbols are a shorthand for elements, such as 'O' for oxygen or 'Fe' for iron, and are crucial for anyone studying chemistry.
Explanation:
The statement that a chemical symbol is a one- or two-letter abbreviation for a chemical element is true. Chemical symbols provide a concise way to represent elements in scientific writings and discussions. For instance, the symbol 'O' stands for oxygen, while 'Zn' represents zinc. The first letter of a chemical symbol is always capitalized, and if there is a second letter, it is lowercase.
Elements that have been known for a long time often have symbols derived from their Latin names. For example, the symbol for iron is 'Fe', which comes from its Latin name 'ferrum'. It is important for those studying chemistry to become familiar with the chemical symbols of the elements, most of which you can find on the periodic table.
Which of the following is true about halogens? Group of answer choices
they have seven valence electrons
they have two valence electrons
they have one valence electron
they have eight valence electrons
Answer:they have one valence electron
Explanation:halogens are in group seven of the periodi table and they have seven electrons in their outermost shell thus need one electron to complete their octet condition. Thus goes into combination to accept one electrons from another element and that is their combining power which is their valency.
The halogens have seven valence electrons.
• In the periodic table, the elements present in group 7A are known as halogens.
• These are fluorine, chlorine, bromine, iodine, and astatine.
• The halogen term means salt former.
• The major characteristic of halogens is that they have seven valence electrons in their higher energy orbitals.
• They are one electron behind in forming a full octet, so these elements would seem to produce anions exhibiting -1 charges and are known as halides.
Thus, halogens having seven valence electrons is the right statement.
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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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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
Refrigerators are usually kept at about 5°C, while room temperature is about 20°C. if you were to take an empty sealed 2 liter soda bottle at room temperature and place it in the fridge, would you expect it to contact to one fourth it's original volume? options.
(a) yes, because 5 is one fourth of 20.
(b) No, because there is no gas inside the bottle.
(c) No, because Celsius is not an absolute temperature scale.
Answer:
c) No, because Celsius is not an absolute temperature scale
Explanation:
converting 5 oC to kelvin which is the absolute temperature scale gives = 273 + 5 = 278 K
and converting 20 oC to kelvin = 20 + 273 = 293 K
the ratio = 278 / 293 = 0.94 approx 1 not 4
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
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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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
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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.