The boiling point and freezing point 0.10 m NaCl are both giver than those of 0.10 m and the correct option is option B.
Although the concentrations of the two solutions are the same, the electrolyte NaCl dissociates into two ions (Na+ and Cl-) while the non-electrolyte glucose stays as 1 particle.
Boiling point elevation and freezing point depression is higher with more dissolved solute particles, so 0.10 m NaCl would have more boiling point and freezing point as compared with that of glucose.
Thus, the ideal selection is option B.
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for the reaction below: a. estimate the gas phase enthalpy change using bond dissociation enthalpies from the owl table reference, not data from your text. click the references button and then click the tables link on the drop-down that appears. include algebraic sign and units. fill in the blank 1 b. is the reaction exothermic or endothermic? c. is the reaction likely to proceed spontaneously in the direction written? submit answer retry entire group 7 more group attempts remaining
a. The gas phase enthalpy change using bond dissociation enthalpies is - 641 kJ/mol.
b. The reaction is exothermic reaction.
c. Due negative enthalpy is the reaction likely to proceed spontaneously in the direction written.
The chemical equation is as :
Na + Cl--------> NaCl
The enthalpy change for the breaking of the bonds in the reaction is :
= (1 mol Na x 109 kJ/mol Na) + (1 mol Cl x 121 kJ/mol Cl)
= 230 kJ/mol
The enthalpy change for the formation of the bonds in NaCl is:
= 1 mol Na-Cl x (-411 kJ/mol Na-Cl)
= -411 kJ/mol
The overall enthalpy change is as :
ΔH = (enthalpy of products) - (enthalpy of reactants)
ΔH = (-411 kJ/mol) - (230 kJ/mol)
ΔH = -641 kJ/mol
The change enthalpy is -641 kJ/mol.
The reaction is the exothermic reaction because of the enthalpy change is the negative.
The chemical reaction is spontaneously in the direction in the written because of the negative enthalpy change.
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This question is incomplete, the complete question is :
for the reaction below: a. estimate the gas phase enthalpy change using bond dissociation enthalpies from the owl table reference, not data from your text. click the references button and then click the tables link on the drop-down that appears. include algebraic sign and units. fill in the blank 1 b. is the reaction exothermic or endothermic? c. is the reaction likely to proceed spontaneously in the direction written?
Na + Cl--------> NaCl
How many atoms in 4Li2O
Answer:4L + i + 20 add the atoms together and get your answer which would be 7
Explanation:
why is cyanide heap leaching used to extract gold from some deposits?
Cyanide heap leaching is used to extract gold from some deposits because it is a relatively inexpensive and efficient process.
Gold deposits that are located in low-grade ores or rocks cannot be economically extracted through traditional mining methods, such as underground or open-pit mining. Instead, these deposits can be processed using cyanide heap leaching, which involves piling the ore into large heaps and spraying it with a dilute cyanide solution. The cyanide reacts with the gold in the ore to form a soluble gold-cyanide complex, which is then recovered from the heap using activated carbon or other methods.
This process is effective because cyanide has a high affinity for gold, and can selectively dissolve it from the surrounding rock and mineral material. Additionally, cyanide heap leaching is relatively inexpensive compared to other methods, and can be used on a large scale to process large volumes of low-grade ore. However, there are also environmental concerns associated with the use of cyanide, as it can be toxic to wildlife and ecosystems if not properly managed.
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Can someone help me with question one
The balanced redox equations are given below:
(d) MnO4- + 8H+ + 5Cl- → Mn2+ + 5Cl- + 4H2O
The oxidation state of Mn in MnO4-: +7The oxidation state of Mn in Mn2+: +2(e) ClO3- + I- + Cl- + H2O → IO3- + 3Cl- + 2OH-
The oxidation state of Cl in ClO3-: +5The oxidation state of Cl in Cl-: -1The oxidation state of I in I-: -1The oxidation state of I in IO3-: +5What are redox reactions?Redox reactions are oxidation-reduction chemical processes where the reactants' oxidation states change.
In redox or oxidation-reduction processes, the oxidation and reduction reactions usually take place concurrently.
In a chemical reaction, the material that is being reduced is referred to as the reducing agent, while the substance that is being oxidized is the oxidizing agent.
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Galvanized is a term associated most closely with which metal?
Galvanized is a term most closely associated with the metal zinc.
Due to its favorable chemical properties, zinc is used as a structural or catalytic cofactor in a very large number of proteins.
Despite the apparent abundance of this metal in all cell types, the intracellular pool of loosely bound zinc ions available for biological exchanges is in the picomolar range and nearly all zinc is tightly bound to proteins. In addition, to limit bacterial growth, some zinc-sequestering proteins are produced by eukaryotic hosts in response to infections.
Therefore, to grow and multiply in the infected host, bacterial pathogens must produce high affinity zinc importers, such as the ZnuABC transporter which is present in most Gram-negative bacteria. Studies carried in different bacterial species have established that disruption of ZnuABC is usually associated with a remarkable loss of pathogenicity.
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What is the Gibbs free energy at 298 K for this reaction?
C(s) + 2H₂(g) -> CH(g)
ΔH = -74.85 kJ/mol
ΔS = -0.08084 kJ/(mol-K)
Is the reaction spontaneous at 298 K (room temperature)?
The reaction is spontaneous at 298K room temperature with the Gibb's free energy of -50.76KJ/Mol.
How to calculate Gibb's free energy?
The maximum amount of reversible work that a system may perform under constant temperature and pressure circumstances is determined using the thermodynamic function known as Gibbs free energy (G).
The change in Gibbs free energy is zero when a system is in equilibrium. The formula: can be used to assess the overall spontaneity of a response or bodily change.
ΔG= Δ H − T Δ S
When the specified values are substituted, the reaction's change in Gibbs free energy is:
ΔG = [tex]\frac{-74.85KJ/Mol*-298K}{-8.084 *10^{-2}KJ/Mol*K }[/tex]
ΔG = -50.76 KJ/mol
At 298K, the reaction has a negative G value. Because it is exergonic, the reaction will happen on its own.
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There are 62 g of CO2 in 129L of solution. What is the molarity of the solution?
There are the 62 g of CO₂ in the 129L of the solution. The molarity of the solution is 0.010 M.
The mass of the CO₂ = 62 g
The volume of the solution = 129 L
The number of moles =mass / molar mass
Where,
The mass = 62 g
The molar mass = 44 g/mol
The number of moles = 62 / 44
The number of moles = 1.40 mol
The molarity is expressed as :
The Molarity of the solution = moles / volume in L
The Molarity of the solution = 1.40 / 129
The Molarity of the solution = 0.010 M
The molarity of the solution is 0.010 M in the volume of the 129 L.
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media containing some ingredients of unknown chemical composition are called __________ media.
Media containing some ingredients of unknown chemical composition are called Composite media.
Composite media are media that contain ingredients of unknown chemical composition. These media are used in a variety of applications, including environmental sampling, water and wastewater treatment, bioremediation, food processing, and other industrial processes.
Composite media are often used when specific components are not yet known or when their presence is not known. Composite media are composed of a mixture of known and unknown chemicals, and their composition can vary depending on the application.
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Refer to Animation: Chemical Synapses. Which answer choice determines whether or not an action potential is triggered in the postsynaptic neuron?
the overall net change in membrane potential caused by the combined EPSPs and IPSPs
the magnitude of the depolarizing excitatory postsynaptic potentials (EPSPs)
the magnitude of the hyperpolarizing inhibitory postsynaptic potential (IPSP)
The overall net change in membrane potential caused by the combined EPSPs and IPSPs determines whether or not an action potential is triggered in the postsynaptic neuron.
EPSPs are depolarizing and IPSPs are hyperpolarizing. When they occur together, their combined effect on the membrane potential of the postsynaptic neuron determines whether or not the neuron reaches the threshold for firing an action potential.
If the net effect is depolarizing and reaches threshold, an action potential will be triggered. If the net effect is hyperpolarizing and moves the membrane potential further from threshold, an action potential will not be triggered. Therefore, it is the overall net change in membrane potential caused by the combined EPSPs and IPSPs that determines whether or not an action potential is triggered in the postsynaptic neuron.
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calculate the current needed to deposit 2.45 g of ni in 3 hours and 15 minutes from a ni(no3)2 solution
The current needed to deposit 2.45 g of ni in 3 hours and 15 minutes from a ni(no3)2 solution is approximately 0.084 amperes.
The first step in solving this problem is to determine the amount of charge required to deposit 2.45 g of ni. This can be done using Faraday's law, which states that the amount of charge required to deposit a given amount of a substance is proportional to its atomic weight and the number of electrons involved in the reaction.
The atomic weight of ni is 58.69 g/mol, and each ni ion requires two electrons to be deposited. Therefore, the amount of charge required to deposit 2.45 g of ni can be calculated as follows:
(2.45 g)/(58.69 g/mol) x (2 electrons/ni) x (1 mole/6.022 x 10^23 electrons) = 2.501 x 10^20 electrons
The next step is to determine the time required to deposit this amount of charge. This can be done using the equation:
Q = It
where Q is the amount of charge (in coulombs), I is the current (in amperes), and t is the time (in seconds).
Since we are given the time in hours and minutes, we need to convert it to seconds:
3 hours and 15 minutes = (3 x 60 x 60) + (15 x 60) = 11,700 seconds
Substituting the values we have calculated into the equation, we get:
2.501 x 10^20 electrons = I x 11,700 seconds
Solving for I, we get:
I = 2.501 x 10^20 electrons/11,700 seconds = 0.084 amperes
Therefore, the current needed to deposit 2.45 g of ni in 3 hours and 15 minutes from a ni(no3)2 solution is approximately 0.084 amperes.
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if 60.0 ml of a 1.5 m hcl solution is put into a flask and diluted with water to make 2.0 l of solution, what is the molarity of the final solution?
To find the molarity of the final solution, we can use the formula:
M1V1 = M2V2
Where M1 is the initial molarity of the solution (1.5 M), V1 is the initial volume of the solution (60.0 mL), M2 is the final molarity of the solution (unknown), and V2 is the final volume of the solution (2.0 L).
First, we need to convert the initial volume from milliliters to liters:
V1 = 60.0 mL = 0.0600 L
Now we can plug in the values and solve for M2:
M1V1 = M2V2
(1.5 M)(0.0600 L) = M2(2.0 L)
0.0900 mol = 2.0 M2
M2 = 0.045 M
Therefore, the molarity of the final solution is 0.045 M.
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which is a better electrolyte- an ionic compound with one cation with +3 charge and three anions with -1 charge or an ionic compound with 2 cations with +1 charge and one anion with -2 charge
An electrolyte is a substance that conducts electricity when dissolved in a solvent, usually water. The effectiveness of an electrolyte depends on its ability to dissociate into ions and the mobility of these ions.
Comparing the two ionic compounds provided, one with a cation with a +3 charge and three anions with -1 charge, and the other with two cations with +1 charge and one anion with -2 charge, we can evaluate which is a better electrolyte.
In general, electrolytes with higher charges and greater dissociation tend to have better conductivity. The first ionic compound, with one cation carrying a +3 charge and three anions with -1 charge, has a greater overall charge and more ions per formula unit when compared to the second ionic compound. The higher charge and increased number of ions can result in stronger electrostatic interactions, which may improve the conductivity of the electrolyte.
However, other factors such as the size of the ions and their mobility can also affect the electrolyte's performance. Smaller and more mobile ions typically lead to better conductivity. Without additional information about the specific compounds being compared, it is challenging to determine which electrolyte would perform better.
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Calculate the concentration of bicarbonate ion, HCO3-, in a 0.030 M H2CO3 solution that has the stepwise dissociation constants Ka1 = 4.3 × 10^-7 and Ka2 = 5.6 × 10^-11.
A) 1.1 × 10^-4 M
B) 4.3 × 10^-7 M
C) 1.3 × 10^-8 M
D) 5.6 × 10^-11 M
Therefore, the concentration of bicarbonate ion ([tex]HCO_3[/tex]-) in the 0.030 M H2CO3 solution is 1.1 × [tex]10^{-4[/tex]M. The correct answer is A.
The concentration of bicarbonate ion (([tex]HCO_3[/tex]-) in a 0.030 M ([tex]HCO_3[/tex] solution, we need to use the stepwise dissociation constants Ka1 and Ka2.
The dissociation reaction for [tex]H_2CO_3[/tex] can be written as follows:
[tex]H_2CO_3[/tex] ⇌ H+ + ([tex]HCO_3[/tex]-
Ka1 = [H+][([tex]HCO_3[/tex]-] / [[tex]H_2CO_3[/tex]] = 4.3 × [tex]10^{-7[/tex]
Similarly, the dissociation can be written as:
[tex]H_2CO_3[/tex]- ⇌ H+ + [tex]H_2CO_3[/tex]-
Ka = [H+][32 CO-] / [([tex]HCO_3[/tex]-] = 5.6 × [tex]10^{11}[/tex]
We can use the first dissociation reaction to calculate the concentration of HCO3- in the solution:
Ka1 = [H+][([tex]HCO_3[/tex]-] / [[tex]H_2CO_3[/tex]]
[H+][([tex]HCO_3[/tex]-] = Ka1[[tex]H_2CO_3[/tex]]
[[tex]H_2CO_3[/tex]-] = Ka1[[tex]H_2CO_3[/tex]] / [H+]
At equilibrium, the concentration of H+ is equal to the concentration of -, [tex]HCO_3[/tex] so:
[([tex]HCO_3[/tex]-] = Ka1[[tex]H_2CO_3[/tex]] / [H+] = Ka1[[tex]H_2CO_3[/tex]] / [[tex]H_2CO_3[/tex]-]
[([tex]HCO_3[/tex]-]^2 = Ka1[[tex]H_2CO_3[/tex]]
[([tex]HCO_3[/tex]-] = [tex]\sqrt{(Ka1[H_2CO_3])}[/tex]
[([tex]HCO_3[/tex]-] = [tex]\sqrt{(4.3 * 10^{-7} * 0.030)}[/tex]
[([tex]HCO_3[/tex]-] = [tex]1.1 * 10^{-4} M[/tex]
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when you have finished the experiment and are cleaning up, where are you supposed to place the product alkene?
The product alkene should be placed in the appropriate waste container for organic compounds.
When cleaning up after an experiment, it is important to properly dispose of all materials used. Organic compounds, such as alkenes, should not be disposed of in regular trash or poured down the drain. Instead, they should be placed in a designated waste container for organic compounds. This helps to ensure that the chemicals are disposed of safely and do not harm the environment. Be sure to follow any specific disposal instructions provided by your institution or laboratory.
During the cleanup process, it's essential to handle the alkene product carefully, as it may be flammable or hazardous. Follow your lab's guidelines and safety protocols, which typically involve placing the alkene in a designated waste container for organic waste or hazardous chemicals. This ensures proper disposal and prevents any potential harm to the environment or lab users.
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b. what is the hybridization of the central atom in nf3?what are the approximate bond angles in this substance ?
The central atom in NF₃ is nitrogen, which undergoes sp³ hybridization. This means that it forms four hybrid orbitals, which are directed toward the corners of a tetrahedron. The approximate bond angles in NF₃ are around 107 degrees, which is slightly less than the ideal tetrahedral angle of 109.5 degrees.
The hybridization of the central atom in NF₃ (nitrogen trifluoride) is sp³. The central atom in this molecule is nitrogen (N), which forms three single bonds with three fluorine (F) atoms. In an sp³ hybridization, the central atom has four electron pairs, including three bonding pairs and one lone pair. The approximate bond angles in this substance are around 107 degrees, which is slightly less than the ideal tetrahedral angle of 109.5 degrees due to the lone pair of electrons on the nitrogen atom.
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Nitromethane CH3NO2 and methyl nitrite CH3ONO have the same empirical formula. What information regarding the N-O bond length can you obtain by drawing the resonance structures of these two molecules?
A. N-O bonds have same bond length in nitromethane, but different bond length in methyl nitrite
B. N-O bonds have different bond length in both molecules
C. N-O bonds have different bond length in nitromethane, but same bond length in methyl nitrite
D. N-O bonds have same bond length in both molecules
C. N-O bonds have different bond length in nitromethane, but same bond length in methyl nitrite.
By drawing the resonance structures of nitromethane (CH3NO2) and methyl nitrite (CH3ONO), we can observe the following:
1. Nitromethane has one N-O bond and no resonance structures, meaning that the N-O bond length remains different.
2. Methyl nitrite has two N-O bonds and resonance structures, which involve the movement of electrons between the two N-O bonds. Due to this resonance, both N-O bonds have the same bond length as the electrons are distributed equally between the two bonds.
Hence, the correct option is C, as nitromethane has different N-O bond lengths and methyl nitrite has the same N-O bond lengths.
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A gas, in a sealed container, occupies a volume of 7.54 L at a pressure of 7.59
atm and a temperature of 293 K. What volume, in liters, will the gas occupy if
the pressure changes to 3.49 atm and the temperature remains unchanged?
Answer:
16.4
Explanation:
(7.59atm)(7.54L) (3.49atm)(V )
———————— = ——————²
293K 293K
Steps:
7.59×7.54×293= 16767.9798
293×3.49= 1022.57
16767.9798÷1022.57= 16.39787 ➡️sig fig➡️ 16.4
How do I balance __Mg(CIO3)2 → __MgCl +__0₂
To balance the equation, we need to ensure that the same number of each type of atom appears on both sides of the equation. In this case, we have one magnesium atom, two chlorine atoms, and six oxygen atoms on the left side (reactants), and one magnesium atom, two chlorine atoms, and 2 oxygen atoms on the right side (products).
To balance the number of chlorine atoms, we need to ensure that there are two chlorine atoms on both sides of the equation. This is achieved by placing a coefficient of 1 in front of Mg(ClO3)2 and a coefficient of 1 in front of MgCl2:
1Mg(ClO3)2 → 1MgCl2 + _O2
note coefficients of one do not need to be emphasized and can be left outNow we have two chlorine atoms on both sides of the equation, but the number of oxygen atoms is not balanced. We need three oxygen atoms on the right side of the equation to balance the six oxygen atoms on the left side. This is achieved by placing a coefficient of 3 in front of O2:
Mg(ClO3)2 → MgCl2 + 3O2
Now the equation is balanced, as there are one magnesium atom, two chlorine atoms, and six oxygen atoms on both sides of the equation. Following the law of conservation of mass which states that mass can not be created nor destroyed in a chemical reaction.
when considering rate of a reaction, you only consider the concentrations of the products. true or false.
When we consider the rate of reaction, we also consider the concentrations of both the reactants and the products, as both influence the reaction apart from considering of only products.
The rate of a chemical reaction is influenced by various factors, including the concentrations of the reactants and products. When considering the rate of a reaction, both the concentrations of the reactants and the products are taken into account. The rate of a reaction is influenced by the frequency and effectiveness of collisions between the reactant molecules, which in turn depends on their concentrations. The higher the concentration of reactants, the more frequent the collisions, and the greater the chance of effective collisions leading to product formation. Therefore, a higher concentration of reactants generally leads to a higher reaction rate.
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A crystal has a face-centered cubic unit cell and a volume of 9.32 × 10−23 cm3. What is the atomic radius of the atoms in the unit cell in units of cm?
The atomic radius of the atoms in the face-centered cubic unit cell is 1.24 × 10^-8 cm.
The face-centered cubic unit cell has atoms at each corner and in the center of each face. Therefore, the total number of atoms in the unit cell is 4 (8 corners × 1/8 atom per corner) + 6 (6 faces × 1/2 atom per face) = 4 + 3 = 7.
The volume of the unit cell can be calculated using the formula: V = a^3/4, where a is the edge length of the unit cell. Therefore, a = (4V)^1/3 = (4 × 9.32 × 10^-23 cm^3)^1/3 = 3.54 × 10^-8 cm.
Since the atomic radius is half of the edge length for face-centered cubic structures, the atomic radius is 1/2 × 3.54 × 10^-8 cm = 1.77 × 10^-8 cm.
However, this value is the distance between the centers of two adjacent atoms. Since we want the radius of a single atom, we need to divide this value by 2. Therefore, the atomic radius of the atoms in the face-centered cubic unit cell is 1.24 × 10^-8 cm.
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Calculate the concentration of OH⁻ in a solution that contains 3.9 × 10^-4 M H3O⁺ at 25°C. Identify the solution as acidic, basic, or neutral.
A) 2.6 × 10^-11 M, acidic
B) 2.6 × 10^-11 M, basic
C) 3.9 × 10^-4 M, neutral
D) 2.7 × 10^-2 M, basic
E) 2.7 × 10^-2 M, acidic
The concentration of OH⁻ in a solution that contains 3.9 × 10^-4 M H3O⁺ at 25°C is 2.6 × 10⁻¹¹ M and the solution is acidic. The correct answer is option A.
To calculate the concentration of OH⁻ in a solution containing 3.9 × 10^-4 M H3O⁺ at 25°C, you can use the ion product of water (Kw), which is given by:
Kw = [H3O⁺] × [OH⁻]
At 25°C, the value of Kw is 1.0 × 10^-14. Given the concentration of H3O⁺ (3.9 × 10^-4 M), you can solve for the concentration of OH⁻: 1.0 × 10^-14 = (3.9 × 10^-4) × [OH⁻]
To get [OH⁻], divide both sides of the equation by (3.9 × 10^-4):
[OH⁻] = (1.0 × 10^-14) / (3.9 × 10^-4) = 2.564 × 10^-11 M
This value is close to 2.6 × 10^-11 M. Since the concentration of H3O⁺ is greater than the concentration of OH⁻, the solution is acidic.
Therefore, the correct answer is: A) 2.6 × 10^-11 M, acidic
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A girl leaves a history classroom and walks 20 meters north to a drinking fountain then she turns and walks 30 meter south to an art classroom what is the girl’s total displacement from the history classroom to the art classroom
Answer: 50 meters total
Explanation:
20 + 30 = 50
List the following elements in order of decreasing atomic radii (largest to smallest) B N Li He. Li, B, N, He.
The order of decreasing atomic radii (largest to smallest) for the given elements is Li, B, N, He.
The positions of these elements in the periodic table.
Li (lithium) is in Group 1 and Period 2.
B (boron) is in Group 13 and Period 2.
N (nitrogen) is in Group 15 and Period 2.
He (helium) is in Group 18 and Period 1.
the general trends in atomic radii across the periodic table.
Atomic radius generally decreases across a period from left to right due to an increase in effective nuclear charge.
Atomic radius generally increases down a group due to an increase in the number of electron shells.
these trends to the elements in question.
Within Period 2, Li has the largest atomic radius, followed by B and then N.
He is in Period 1, which is above Period 2, so its atomic radius is smaller than the others.
Arrange the elements in order of decreasing atomic radii.
Li, B, N, He
So, the elements B, N, Li, and He in order of decreasing atomic radii (largest to smallest) are: Li, B, N, He.
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in the reaction 2h2o2(aq) > 2h2o(l) o2(g), the initial concentration of h2o2 is 0.745 m and, 1min. 25 seconds later, the concentration of h2o2 is 0.218 m. what is the average rate of reaction in mh2o2/s over this time interval?group of answer choices0.003100.04220.1240.00620
For a chemical reaction 2H₂O₂ (aq) -> 2H₂O (l) + O₂ (g), the average rate of reaction in over this time interval is equals to the 0.0062 M/s. So, option(d) is right one.
A reaction, 2H₂O₂ (aq) -> 2H₂O (l) + O₂ (g),
The initial concentration of H₂O₂
= 0.745 M
After 1 min. 25 seconds, the final concentration of H₂O₂ = 0.218 M
We have to determine average rate of reaction over this time interval. Average rate of reaction is defined as the change in the concentration of reactants or products over a period of time in the course of the reaction. In formula form,
Average rate = -∆[R]/∆t or ∆[P]/∆t
Change in concentration of reactant, ∆[R] = 0.218 M - 0.745 M = - 0.527 M
Change in time interval, ∆t = 1 min 25 sec - 0 = 85 seconds
So, average rate of reaction = -( -0.527)/85 = 0.527/85
= 0.0062 M/sec
Hence, required value is 0.0062 M/s.
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Which Gas Law’s formula could be used to determine a desired value using the information in the above graph?
Note that the Gas law formula that could be used to determined a desired value using the information given in the above graph is: the Ideal Gas Law.
What do we mean by this?The Idea gas law speaks to the pressure, volume, temprature and number of moles of a gas and expresses their relationships in the formula:
PV = nRT, where
P = Pressure
V = Volume
n = Number of Moles and
R is the GAs constant.
Hence, it is correct to state that the he Gas law formula that could be used to determined a desired value using the information given in the above graph is: the Ideal Gas Law.
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SOMEONE PLEASE I NEED HELP WITH CHEMISTRY
Draw a diagram for Copper(ll) nitrate & Cu(NO3)2 in a 250.0 mL of aqueous solution to show how to make the solution. Information to include…
Molarity of solution - 0.1176
She then draws 30.0 mL of the solution into a pipet. (MOLES OF CU(NO3)2 - 0.00352)
THEN : Mrs. Mandochino empties the 30.0 mL into an empty volumetric flask and fills it to the 240.0 mL mark with distilled water.
What is the molarity of this new solution?
Make sure to have 5 ACCURATE steps drawn. Your drawing should only be 1 picture but include 5 steps.
According to the question Measure 250.0 mL of distilled water into a volumetric flask.
What is volumetric?Volumetric refers to a measurement of volume, or the three-dimensional space an object occupies. It is most commonly used to measure the size of a container or the capacity of a material. Volumetric measurements are usually expressed in cubic units, such as cubic centimeters, cubic feet, or cubic meters. Volumetric measurements can be used to measure the capacity of a container, the volume of a material, or the total volume of a space. Volumetric measurements are also used in engineering and scientific applications, such as calculating the volume of a liquid in a container, the volume of a gas, or the density of a material.
Measure out 2.652 grams of Copper(II) Nitrate and add to the volumetric flask.
Shake the volumetric flask until the Copper(II) Nitrate is dissolved.
Fill the volumetric flask up to the 250.0 mL mark with distilled water.
The solution is now 0.1176 M of Copper(II) Nitrate (Cu(NO3)2). Take 30.0 mL of this solution and add it to an empty volumetric flask, then fill it to the 240.0 mL mark with distilled water. The new solution is 0.0968 M of Copper(II) Nitrate (Cu(NO3)2).
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For the following exothermic reversible reaction at equilibrium, how will adding more Cl2 affect it? C2H4(8) + Cl2(g) โ C2H4Cl2(8) a. it will shift to form more reactant(s) until the equilibrium is re-established
b. it will shift to form more product(s) until the equilibrium is re-established
c. it will decrease the activation energies of both the forward and reverse reactions d. it will have no effect
The correct option is (b) it will shift to form more product(s) until the equilibrium is re-established.
Adding more Cl₂ to the reaction C₂H₄(g) + Cl₂(g) ↔ C₂H₄Cl₂(g) will cause the equilibrium to shift towards the product(s) side of the reaction.
The reason for this can be explained by Le Chatelier's principle, which states that a system at equilibrium will respond to any stress applied to it in a way that counteracts the stress and restores equilibrium. In this case, adding more Cl₂ will increase the concentration of the reactant(s), causing the system to be out of equilibrium. The reaction will then shift in the forward direction to consume the excess Cl₂, and produce more C₂H₄Cl₂ until the equilibrium is re-established.
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strontium-90 (9038sr) is a particularly dangerous fission product of 23592u because it is radioactive and it substitutes for calcium in bones. what other direct fission products would accompany it in the neutron-induced fission of 23592u if three neutrons are released?
Fission is a process where the nucleus of an atom is split into two or more smaller nuclei, along with the release of a large amount of energy. When certain heavy elements, such as uranium-235, undergo fission, they release radioactive particles called fission products. These fission products can pose serious health risks, as they emit harmful radiation.
One such fission product is strontium-90 (90Sr), which is particularly dangerous because it is radioactive and can substitute for calcium in bones, leading to bone cancer and other health problems. If three neutrons are released during the neutron-induced fission of uranium-235, several other direct fission products will accompany strontium-90. These include:
1. Krypton-92 (92Kr)
2. Barium-141 (141Ba)
3. Xenon-142 (142Xe)
4. Rubidium-94 (94Rb)
These fission products, like strontium-90, are radioactive and can pose serious health risks if they are not properly contained. It is important to handle nuclear materials with extreme caution and to ensure that radioactive waste is disposed of safely and securely. Proper precautions and regulations are necessary to prevent the release of harmful radiation into the environment and to protect the health and safety of the public.
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What molecules do NOT make good leaving groups? Why?
Molecules that are strong bases or nucleophiles, such as hydroxide (OH-) or amines (NH2), do not make good leaving groups.
This is because leaving groups must be able to detach from the parent molecule easily and without a lot of energy input. Strong bases and nucleophiles are too tightly bound to the molecule and require too much energy to remove. Additionally, molecules with large or bulky groups attached to them also do not make good leaving groups because they can be sterically hindered and unable to detach easily.
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in the signal transduction pathway seen here, how many active erk molecules can be produced once the receptor is activated?
The number of active ERK molecules produced depends on the specific signal transduction pathway involved.
In a typical signal transduction pathway, a ligand binds to a cell surface receptor, which activates an intracellular signaling cascade. This cascade often involves the activation of multiple proteins, including kinases like ERK. ERK is usually activated through a series of phosphorylation events. Once activated, ERK can phosphorylate other proteins and regulate various cellular processes, such as gene expression, cell division, and cell survival. The number of active ERK molecules produced in a specific pathway depends on factors such as the efficiency of the signaling cascade, the availability of upstream signaling molecules, and the presence of negative feedback loops that can limit the amplification of the signal.
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