Answer:
hola como estas
Explanation:
A silver nitrate (AgNO3) solution is 0.150 M. 100.0 mL of a diluted silver nitrate solution is prepared using 10.0 mL of the more concentrated solution (which is then diluted with distilled water to 100.0 mL). Use the dilution equation to find the molarity of silver nitrate in the diluted solution.
Answer: The molarity of silver nitrate in the diluted solution is 0.015 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of concentrated solution = 0.150 M
[tex]V_1[/tex] = volume of concentrated solution = 10.0 ml
[tex]M_2[/tex] = molarity of diluted solution = ?
[tex]V_2[/tex] = volume of diluted solution = 100.0 ml
[tex]0.150\times 10.0=M_2\times 100.0[/tex]
[tex]M_2=0.015M[/tex]
Thus the molarity of silver nitrate in the diluted solution is 0.015 M
Final answer:
The molarity of the diluted silver nitrate solution, prepared by diluting 10.0 mL of a 0.150 M solution to a final volume of 100.0 mL, is 0.015 M when calculated using the dilution equation.
Explanation:
The question is asking how to calculate the molarity of a diluted silver nitrate solution using the dilution equation. When 10.0 mL of a 0.150 M silver nitrate solution is diluted to 100.0 mL, the molarity of the diluted solution can be found using the equation M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
Using the dilution equation:
To find M2 (the molarity of the diluted solution), rearrange the equation to M2 = (M1V1) / V2.
The molarity of the diluted silver nitrate solution would be 0.015 M (calculated as (0.150 M * 10.0 mL) / 100.0 mL).
At a certain temperature the rate of this reaction is first order in HI with a rate constant of :7.21/s2HI(g) = H2(g) + I2(g)Suppose a vessel contains HI at a concentration of 0.440M. Calculate the concentration of HI in the vessel 0.210 seconds later. You may assume no other reaction is important.
Answer:
C HI = 0.0968 M
Explanation:
2HI(g) → H2(g) + I2(g)- ra = K(Ca)∧α = - δCa/δt∴ a: HI(g)
∴ order (α) = 1
∴ rate constant (K) = 7.21/s
∴ Initial concenttration HI(g) (Cao) = 0.440 M
∴ t = 0.210 s ⇒ Ca = ?
⇒ - δCa/δt = KCa
⇒ - ∫δCa/Ca = K*∫δt
⇒ Ln(Cao/Ca) = K*t
⇒ Cao/Ca = e∧(K*t)
⇒ Cao/e∧(K*t) = Ca
⇒ Ca = (0.440)/e∧((7.21*0,21))
⇒ Ca = 0.440/4.54533
⇒ Ca = 0.0968 M
Treatment of ethyl acetoacetate with NaOEt (2 equiv) and BrCH2CH2Br forms compound X. This reaction is the first step in the synthesis of illudin-S, an antitumor substance isolated from the jack-o'-lantern, a saffron-colored mushroom. What is the structure of X
Answer:
See explanation below
Explanation:
In this reaction we have the ethyl acetoacetate which is reacting with 2 eq of sodium etoxide. The sodium etoxide is a base and it usually behaves as a nucleophyle of many reactions. Therefore, it will atract all the acidics protons in a molecule.
In the case of the ethyl acetoacetate, the protons that are in the methylene group (CH3 - CO - CH2 - COOCH2CH3) are the more acidic protons, therefore the etoxide will substract these protons instead of the protons of the methyl groups. This is because those hydrogens (in the methylene group) are between two carbonile groups, which make them more available and acidic for any reaction. As we have 2 equivalents of etoxide, means that it will substract both of the hydrogen atoms there, and then, reacts with the Br - CH2CH2 - Br and form a product of an aldolic condensation.
The mechanism of this reaction to reach X is shown in the attached picture.
Why does a higher concentration make a reaction faster?
There are more collisions per second and the collisions are of greater energy.
Collisions occur with greater energy.
There are more collisions per second.
Higher concentration make a reaction faster due to faster collisions. The correct answer is "There are more collisions per second."
When the concentration of reactants is increased, it leads to a higher number of particles or molecules in the reaction mixture. As a result, there are more frequent collisions between the reactant particles per unit time.
In a chemical reaction, the reactant particles must collide with sufficient energy and proper orientation for a successful reaction to occur. By increasing the concentration, the chances of successful collisions increase because there are more particles available to collide with each other.
However, it's important to note that while a higher concentration increases the frequency of collisions, it does not necessarily guarantee that all collisions will result in a reaction. The energy of the collisions and proper orientation are also essential factors for a successful reaction.
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COLLEGE CHEMISTRY 35 POINTS
Determine the percent yield when 12 g of CO2 are formed experimentally from the reaction of 0.5 moles of C8H18 reacting with excess oxygen according to the following balanced equation:
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(l)
Please show your work.
6.81% is the percent yield when 12 g of CO2 are formed experimentally from the reaction of 0.5 moles of C8H18 reacting with excess oxygen.
Explanation:
Data given:
mass of carbon dioxide formed = 12 grams (actual yield)
atomic mass of CO2 = 44.01 grams/mole
moles of [tex]C_{8} H_{18}[/tex] = 0.5
Balanced chemical reaction:
2 [tex]C_{8} H_{18}[/tex] + 25 [tex]O_{2}[/tex] ⇒ 16 C[tex]O_{2}[/tex] + 18 [tex]H_{2} _O{}[/tex]
number of moles of carbon dioxide given is
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
number of moles= [tex]\frac{12}{44}[/tex]
number of moles of carbon dioxide gas = 0.27 moles
from the reaction 2 moles of [tex]C_{8} H_{18}[/tex] reacts to produce 16 moles of C[tex]O_{2}[/tex]
So, when 0.5 moles reacted it produces x moles
[tex]\frac{16}{2}[/tex] = [tex]\frac{x}{0.5}[/tex]
x = 4
4 moles of carbon dioxide formed, mass from it will give theoretical yield.
mass = number of moles x molar mass
mass = 4 x 44.01
= 176.04 grams
percent yield =[tex]\frac{actual yield}{theoretical yield}[/tex] x 100
percent yield = [tex]\frac{12}{176.04}[/tex] x 100
percent yield = 6.81 %
Determine the ground-state electron configuration and bond order for each of the Period 2 diatomic molecules. Specify which MO is the HOMO and which is the LUMO. List the molecules in order of increasing dissociation and list them in order of bond length. You may need to THINK a bit to get this correct
Answer:
Explanation:
check the attachment below for correct explanations.
Define the end point of a titration. Select one: a. It is a synonym for equivalence point. b. It is when a change that indicates equivalence is observed in the analyte solution. c. It is the point at which the pH no longer changes. Next
Answer:
b. It is when a change that indicates equivalence is observed in the analyte solution.
Explanation:
The end point of a titration is the point at which the indicator undergoes the change noticeable by our senses. Ideally, the equivalence point and end point coincide; but this does not usually happen in practice, because the indicator does not always change perceptibly at the same moment in which the equivalence point is reached and also for the change of the indicator, some of the reagent used in the evaluation is usually necessary .
The difference between the end point and the equivalence point of a valuation is called the valuation error or end point error.
At this point the pH changes the color of the indicator. nvhdurn
Final answer:
The end point of a titration is the moment observed (option b) when a change, due to an indicator or sensor, suggests that stoichiometric equivalence has been reached. It is an experimental determination and serves as the best estimate of the theoretical equivalence point.
Explanation:
The end point of a titration is not a synonym for the equivalence point. Rather, it represents the moment during a titration when a change indicating equivalence is observed in the analyte solution (option b). This point is typically detected through a change in color due to an indicator or through a change detected by a sensor. It is important to note that the end point is an experimental value, our best estimate of the equivalence point, and that any difference between the end point and the theoretical equivalence point is a source of determinate error.
In an acid-base titration, the end point is often observed when a pH indicator changes color, signifying that a stoichiometric amount of titrant has been added to the analyte. In redox and complexation titrations, it can be detected by changes in the solution conditions that are measurable by indicators and sensors.
What is the full ionic equation for CuCl2 and (NH4)2SO4?
Answer:
Explanation:
No reaction if there is no solid then the ions just stay there brah. All the ions remain in solution. We are all gonna make it!
For most solids at room temperature, the specific heat is determined by oscillations of the atom cores in the lattice (each oscillating lattice site contributes 3kT of energy, by equipartition), as well as a contribution from the mobile electrons (if it's a metal). At room temperature the latter contribution is typically much smaller than the former, so we will ignore it here. In other words, you can reasonably estimate the specific heat simply by counting the number of atoms! Use this fact to estimate the specific heat of copper (atomic mass = 63.6), given that the specific heat of aluminum (atomic mass = 27.0) is 900 J/kg-K.
Answer:
The specific heat of copper is [tex]C= 392 J/kg\cdot ^o K[/tex]
Explanation:
From the question we are told that
The amount of energy contributed by each oscillating lattice site is [tex]E =3 kT[/tex]
The atomic mass of copper is [tex]M = 63.6 g/mol[/tex]
The atomic mass of aluminum is [tex]m_a = 27.0g/mol[/tex]
The specific heat of aluminum is [tex]c_a = 900 J/kg-K[/tex]
The objective of this solution is to obtain the specific heat of copper
Now specific heat can be defined as the heat required to raise the temperature of 1 kg of a substance by [tex]1 ^o K[/tex]
The general equation for specific heat is
[tex]C = \frac{dU}{dT}[/tex]
Where [tex]dT[/tex] is the change in temperature
[tex]dU[/tex] is the change in internal energy
The internal energy is mathematically evaluated as
[tex]U = 3nk_BT[/tex]
Where [tex]k_B[/tex] is the Boltzmann constant with a value of [tex]1.38*10^{-23} kg \cdot m^2 /s^2 \cdot ^o K[/tex]
T is the room temperature
n is the number of atoms in a substance
Generally number of atoms in mass of an element can be obtained using the mathematical operation
[tex]n = \frac{m}{M} * N_A[/tex]
Where [tex]N_A[/tex] is the Avogadro's number with a constant value of [tex]6.022*10^{23} / mol[/tex]
M is the atomic mass of the element
m actual mass of the element
So the number of atoms in 1 kg of copper is evaluated as
[tex]m = 1 kg = 1 kg * \frac{10000 g}{1kg } = 1000g[/tex]
The number of atom is
[tex]n = \frac{1000}{63.6} * (6.0*0^{23})[/tex]
[tex]= 9.46*10^{24} \ atoms[/tex]
Now substituting the equation for internal energy into the equation for specific heat
[tex]C = \frac{d}{dT} (3 n k_B T)[/tex]
[tex]=3nk_B[/tex]
Substituting values
[tex]C = 3 (9.46*10^{24} )(1.38 *10^{-23})[/tex]
[tex]C= 392 J/kg\cdot ^o K[/tex]
g Identify which of the following statements about human glycogen debranching enzyme are true based on the HPLC results. The (α‑1→6) glucosidase catalytic center is in the C‑terminal half. The transferase catalytic center can hydrolyze α‑1,6 glycosidic bonds. The oligo‑(α1→4)‑(α1→4) glucanotransferase catalytic center is in the C‑terminal half. Oligo‑(α1→4)‑(α1→4) glucanotransferase activity creates a substrate for (α‑1→6) glucosidase. Based on the peaks in the HPLC charts, what do you think is the most likely substrate for the oligo‑(α1→4)‑(α1→4) glucanotransferase catalytic center of glycogen debranching enzyme? maltotetraosyl and glucosyl residues 6‑O‑α‑glucosyl cyclomaltoheptaose maltosyl and maltotriosyl residues maltoheptaosyl and maltooctaosyl residues cyclomaltoheptaose (β‑cyclodextrin)
The most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center of glycogen debranching enzyme are maltosyl and maltotriosyl residues.
Explanation:The glycogen debranching enzyme has two activities; it acts as a glucosidase and a glucanotransferase. The (α‑6) glucosidase activity hydrolyzes α-1,6 glycosidic bonds, whereas the oligo-(α1→4)-(α1→4) glucanotransferase activity shifts α-1,4-linked glucose chains from one branch to another, often creating a substrate that the glucosidase can act upon. Based on the mechanism of glycogenolysis, the enzyme glycogen phosphorylase releases glucose units from the linear chain until a few are left near the branching point, and it is here that the glucan transferase action is relevant. The glucan transferase shifts the remaining α-1,4 linked glucose units, leaving a single α-1,6 linked glucose that the glucosidase can then release. Hence, the most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center are maltosyl and maltotriosyl residues, as these are the short α-1,4 linked glucose chains that are left after phosphorylase action.
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The pH at 25 °C of an aqueous solution of the sodium salt of p-monochlorophenol (NaC6H4ClO) is 11.05. Calculate the concentration of C6H4ClO- in this solution, in moles per liter. Ka for HC6H4ClO is equal to 6.6×10-10.
Answer:
Approximately [tex]8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}[/tex].
Explanation:
The [tex]K_a[/tex] in this question refers the dissociation equilibrium of [tex]\rm HC_6H_4ClO[/tex] as an acid:
[tex]\rm HC_6H_4ClO\, (aq) \rightleftharpoons H^{+} \, (aq) + C_6H_4ClO^{-}\, (aq)[/tex].
[tex]\displaystyle K_a\left(\mathrm{HC_6H_4ClO}\right) = \frac{\left[\mathrm{H^{+}}\right] \cdot \left[\mathrm{C_6H_4ClO^{-}}\right]}{\left[\mathrm{HC_6H_4ClO}\right]}[/tex].
However, the question also states that the solution here has a [tex]\rm pH[/tex] of [tex]11.05[/tex], which means that this solution is basic. In basic solutions at [tex]\rm 25\;^\circ C[/tex], the concentration of [tex]\rm H^{+}[/tex] ions is considerably small (typically less than [tex]10^{-7}\;\rm mol \cdot L^{-1}[/tex].) Therefore, it is likely not very appropriate to use an equilibrium involving the concentration of [tex]\rm H^{+}[/tex] ions.
Here's the workaround: note that [tex]\rm C_6H_4ClO^{-}\, (aq)[/tex] is the conjugate base of the weak acid [tex]\rm HC_6H_4ClO\, (aq)[/tex]. Therefore, when [tex]\rm C_6H_4ClO^{-}\, (aq)[/tex] dissociates in water as a base, its [tex]K_b[/tex] would be equal to [tex]\displaystyle \frac{K_w}{K_a} \approx \frac{10^{-14}}{K_a}[/tex]. ([tex]K_w[/tex] is the self-ionization constant of water. [tex]K_w \approx 10^{-14}[/tex] at [tex]\rm 25\;^\circ C[/tex].)
In other words,
[tex]\begin{aligned} & K_b\left(\mathrm{C_6H_4ClO^{-}}\right) \\ &= \frac{K_w}{K_a\left(\mathrm{HC_6H_4ClO}\right)} \\ &\approx \frac{10^{-14}}{6.6 \times 10^{-10}} \\ & \approx 1.51515 \times 10^{-5}\end{aligned}[/tex].
And that [tex]K_b[/tex] value corresponds to the equilibrium:
[tex]\rm C_6H_4ClO^{-}\, (aq) + H_2O\, (l) \rightleftharpoons HC_6H_4ClO\, (aq) + OH^{-}\, (aq)[/tex].
[tex]\displaystyle K_b\left(\mathrm{C_6H_4ClO^{-}}\right) = \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{\left[\mathrm{C_6H_4ClO^{-}}\right]}[/tex].
The value of [tex]K_b[/tex] has already been found.
The [tex]\rm OH^{-}[/tex] concentration of this solution can be found from its [tex]\rm pH[/tex] value:
[tex]\begin{aligned}& \left[\mathrm{OH^{-}}\right] \\ &= \frac{K_w}{\left[\mathrm{H}^{+}\right]} \\ & = \frac{K_w}{10^{-\mathrm{pH}}} \\ &\approx \frac{10^{-14}}{10^{-11.05}} \\ &\approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1} \end{aligned}[/tex].
To determine the concentration of [tex]\left[\mathrm{HC_6H_4ClO}\right][/tex], consider the following table:
[tex]\begin{array}{cccccc}\textbf{R} &\rm C_6H_4ClO^{-}\, (aq) & \rm + H_2O\, (l) \rightleftharpoons & \rm HC_6H_4ClO\, (aq) & + & \rm OH^{-}\, (aq) \\ \textbf{I} & (?) & \\ \textbf{C} & -x & & + x& & +x \\ \textbf{E} & (?) - x & & x & & x\end{array}[/tex]
Before hydrolysis, the concentration of both [tex]\mathrm{HC_6H_4ClO}[/tex] and [tex]\rm OH^{-}[/tex] are approximately zero. Refer to the chemical equation. The coefficient of [tex]\mathrm{HC_6H_4ClO}[/tex] and [tex]\mathrm{HC_6H_4ClO}[/tex] are the same. As a result, this equilibrium will produce [tex]\rm OH^{-}[/tex] and [tex]\mathrm{HC_6H_4ClO}[/tex] at the exact same rate. Therefore, at equilibrium, [tex]\left[\mathrm{HC_6H_4ClO}\right] \approx \left[\mathrm{OH^{-}}\right] \approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1}[/tex].
Calculate the equilibrium concentration of [tex]\left[\mathrm{C_6H_4ClO^{-}}\right][/tex] from [tex]K_b\left(\mathrm{C_6H_4ClO^{-}}\right)[/tex]:
[tex]\begin{aligned} & \left[\mathrm{C_6H_4ClO^{-}}\right] \\ &= \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{K_b}\\&\approx \frac{\left(1.1220 \times 10^{-3}\right) \times \left(1.1220 \times 10^{-3}\right)}{1.51515\times 10^{-5}}\; \rm mol \cdot L^{-1} \\ &\approx 8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
Construct a simulated 1H NMR spectrum for 1-chloropropane by dragging and dropping the appropriate splitting patterns into the boxes on the chemical shift baseline, and by dragging integration values into the small box above each signal. Items may be used more than once. Peak heights do not represent integration.
To construct a simulated 1H NMR spectrum for 1-chloropropane, you need to consider the chemical shifts and splitting patterns of the hydrogen atoms. Follow these steps: identify the different types of hydrogen atoms, determine the chemical shifts, assign integration values, and drag and drop splitting patterns.
Explanation:To construct a simulated 1H NMR spectrum for 1-chloropropane, you need to consider the chemical shifts and splitting patterns of the hydrogen atoms. Here's a step-by-step guide:
Identify the different types of hydrogen atoms in 1-chloropropane. In this molecule, there are three types: H on the methyl group, H on the second carbon, and H on the third carbon.Determine the chemical shifts of the hydrogen atoms. The chemical shift values for H on the methyl group, H on the second carbon, and H on the third carbon are typically around 0.9-1.2 ppm, 1.2-1.4 ppm, and 2.5-3.0 ppm, respectively.Assign integration values to the signals. The integration values represent the relative number of hydrogen atoms in each type. For example, the H on the methyl group typically has an integration value of 3, indicating that there are three hydrogen atoms in the methyl group.Drag and drop the appropriate splitting patterns into the boxes on the chemical shift baseline. The splitting pattern depends on the neighboring hydrogen atoms and follows the n+1 rule. For example, if a hydrogen atom has two neighboring hydrogen atoms, it will show a triplet splitting pattern.Learn more about simulated 1H NMR spectrum for 1-chloropropane here:https://brainly.com/question/9812005
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What type of particles will a hot object have if it has more kinetic energy?
Answer:
celulares por que tienen una fuente de energiq
Provide a preliminary design o fa n air stripping column t o remove toluene from ground water. Levels of toluene range from 0.1 to 2.1 mglL and this must be reduced to 50 f.LglL. A hydrogeologic study of the area indicates that a flow rate of 1 1 0 gal/min is required to ensure that contamination not spread. Laboratory investigations have determined the overall transfer constant, KLa = 0.020 s-' . Use a column diameter of 2.0 feet and an air-to-water ratio of 15. Specifically determine: liquid loading rate, stripping factor, and height of the tower. Provide a sketch of the unit indicating all required appurtenances.
Answer:
Explanation:
Find attach the solution
Interpret the following Arterial Blood Gases 1. pH 7.33 PaCO2 60 HCO3 34 A. Normal ABG values B. Respiratory acidosis without compensation C. Respiratory acidosis with partial compensation D. Respiratory acidosis with full compensation 2. pH 7.48 PaCO2 42 HCO3 30 A. Metabolic acidosis without compensation B. Respiratory alkalosis with partial compensation C. Respiratory alkalosis with full compensation D. Metabolic alkalosis without compensation 3. pH 7.38 PaCO2 38 HCO3 24 A. Respiratory alkalosis B. Normal C. Metabolic Alkalosis D. None of the above 4. pH 7.21 PaCO2 60 HCO3 24 A. Normal B. Respiratory acidosis without compensation C. Metabolic acidosis with partial compensation D. Respiratory acidosis with complete compensation 5. pH 7.48 PaCO2 28 HCO3 20 A. Respiratory alkalosis with partial compensation B. Respiratory alkalosis with complete compensation C. Metabolic alkalosis without compensation D. Metabolic alkalosis with complete compensation
Answer: 1) C; 2)D; 3)B; 4)B; 5) A
Explanation:Interpreting the following Arterial Blood gases, we have
1. pH 7.33 PaCO2 60 HCO3 34----Respiratory acidosis with partial compensation----C
2. pH 7.48 PaCO2 42 HCO3 30------. Metabolic alkalosis without compensation----D
3. pH 7.38 PaCO2 38 HCO3 24 ----- Normal---B
4. pH 7.21 PaCO2 60 HCO3 24------ Respiratory acidosis without compensation-----B
5. pH 7.48 PaCO2 28 HCO3 20 ----Respiratory alkalosis with partial compensation
The Arterial blood gas interpretation from analysis shows the pH and the partial pressures of oxygen and carbon dioxide in the arterial blood of an individual which can detect how well the lungs are functioning thereby making a physician make a diagnosis, estimate the severity of a condition and profer treatment.
A 1.000 kg sample of nitroglycerine, C3H5N3O9, explodes and releases gases with a temperature of 1985°C at 1.100 atm. What is the volume of gas produced? 4 C3H5N3O9(s) → 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) A 1.000 kg sample of nitroglycerine, C3H5N3O9, explodes and releases gases with a temperature of 1985°C at 1.100 atm. What is the volume of gas produced? 4 C3H5N3O9(s) → 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) 4730 L 5378 L 3525 L 742.2 L
Answer:
742.2 L
Explanation:
First we must find the number of moles of nitroglycerine reacted.
Molar mass of nitroglycerine= 227.0865 g/mol
Mass of nitroglycerine involved = 1×10^3 g
Number of moles of nitroglycerine= 1×10^3g/227.0865 g/mol
n= 4.40361 moles
T= 1985°C + 273= 2258K
P= 1.100atm
R= 0.082atmLmol-1K-1
Using the ideal gas equation:
PV= nRT
V= nRT/P
V= 4.40361× 0.082× 2258/1.1
V= 742 L
Considering the reaction stoichiometry and ideal gas law, the volume of gas produced is 5375.626 L.
The balanced reaction is:
4 C₃H₅N₃O₉(s) → 12 CO₂(g) + 10 H₂O(g) + 6 N₂(g) + O₂(g)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
C₃H₅N₃O₉(s): 4 moles O₂(g): 1 moles CO₂(g): 12 moles H₂O(g): 10 molesN₂(g): 6 molesThen 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas [12 moles of CO₂(g) + 10 moles of H₂O(g) + 6 moles of N₂(g) + 1 mole O₂(g)]
Being the molar mass of nitroglycerine C₃H₅N₃O₉(s) 227 g/mole, then the amount of moles that 1 kg (1000 g) of the compound contains can be calculated as:
[tex]1000 gramsx\frac{1 mole}{227 grams}= 4.405 moles[/tex]
Then you can apply the following rule of three: if by stoichiometry 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas, 4.405 moles of C₃H₅N₃O₉(s) will produce how many moles of gas?
[tex]amount of moles of gas=\frac{4.405 moles of nitroglycerinex29 moles of gas}{4 moles of nitroglycerine}[/tex]
amount of moles of gas= 31.93625 moles
On the other hand, an ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
P= 1.1 atmV= ?n= 31.93625 molesR= 0.082[tex]\frac{atmL}{molK}[/tex]T= 1985 C= 2258 K (being 0 C=273 K)Replacing:
1.1 atm× V= 31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K
Solving:
V= (31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K) ÷ 1.1 atm
V= 5375.625 L
Finally, the volume of gas produced is 5375.626 L.
Learn more:
https://brainly.com/question/4147359?referrer=searchResultsbrainly.com/question/16487206?referrer=searchResults brainly.com/question/14446695?referrer=searchResults brainly.com/question/11564309?referrer=searchResults brainly.com/question/4025026?referrer=searchResults brainly.com/question/18650135?referrer=searchResultsA sealed copper container with a mass of 0.3 kg is filled with 1.5 mole of helium gas. Initially, the helium gas is at a temperature of 124 oC and the copper container is at 20 oC. The helium-copper system is thermally isolated. Note that the specific heat of copper is 386 J/(kgK) and the molar specific heat of helium is 12.5 J/(molK). Find the equilibrium temperature of the system.
Answer:
[tex]T = 34.493\,^{\textdegree}C[/tex]
Explanation:
The equilibrium temperature of the gas-container system is:
[tex]Q_{Cu} = -Q_{g}[/tex]
[tex](0.3\,kg)\cdot\left(386\,\frac{J}{kg\cdot ^{\textdegree}C} \right)\cdot (T-20\,^{\textdegree}C) = (1.5\,mol)\cdot \left(12.5\,\frac{J}{mol\cdot ^{\textdegree}C} \right)\cdot (124\,^{\textdegree}C-T)[/tex]
[tex]\left(115.8\,\frac{J}{^{\textdegree}C} \right)\cdot (T-20\,^{\textdegree}C) = \left(18.75\,\frac{J}{^{\textdegree}C} \right)\cdot (124\,^{\textdegree}C-T)[/tex]
[tex]134.55\cdot T = 4641[/tex]
[tex]T = 34.493\,^{\textdegree}C[/tex]
Nitrate salts (NO3), when heated, can produce nitrites (NO2) plus oxygen (O2). A sample of potassium nitrate is heated, and the 02 gas produced is collected in a 700 mL flask. The pressure of the gas in the flask is 2.7 atm, and the temperature is recorded to be 329 K. The value of R= 0.0821 atm L/(mol K) How many moles of O2 gas were produced? moles After a few hours, the 700 mL flask cools to a temperature of 293K. What is the new pressure due to the O2 gas?
Answer: a) 0.070 moles of oxygen were produced.
b) New pressure due to the oxygen gas is 2.4 atm
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 2.7 atm
V = Volume of gas = 700 ml = 0.7 L
n = number of moles = ?
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature = 329 K
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{2.7atm\times 0.7L}{0.0821 L atm/K mol\times 329K}=0.070moles[/tex]
Thus 0.070 moles of oxygen were produced.
When the 700 mL flask cools to a temperature of 293K.
[tex]PV=nRT[/tex]
[tex]P=\frac{nRT}{V}[/tex]
[tex]P=\frac{0.070\times 0.0821\times 293}{0.7}[/tex]
[tex]P=2.4atm[/tex]
The new pressure due to the oxygen gas is 2.4 atm
Mole ratios for a solving a stoichiometry problem are obtained from the
1. Molar Mass
2. Periodic Table
3. Balanced Equation
4. Total Mass of the Products
Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The KspKsp of Zn(OH)2Zn(OH)2 is 3×10−153×10−15 and the KfKf of Zn(OH)2−4Zn(OH)42− is 2×10152×1015.
Answer:
pH = 13.09
Explanation:
Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15
Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15
K = Ksp X Kf
= 3*2*10^-15 * 10^15
= 6
Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M
Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)
Initial: 0 0.3 0
Change: -2x +x
Equilibrium: 0.3 - 2x x
K = Zn(OH)₄²⁻/[OH⁻]²
6 = x/(0.3 - 2x)²
6 = x/(0.3 -2x)(0.3 -2x)
6(0.09 -1.2x + 4x²) = x
0.54 - 7.2x + 24x² = x
24x² - 8.2x + 0.54 = 0
Upon solving as quadratic equation, we obtain;
x = 0.089
Therefore,
Concentration of (OH⁻) = 0.3 - 2x
= 0.3 -(2*0.089)
= 0.122
pOH = -log[OH⁻]
= -log 0.122
= 0.91
pH = 14-0.91
= 13.09
The phosphorylation of glucose to glucose 6-phosphate Group of answer choices is so strongly exergonic that it does not require a catalyst. is an exergonic reaction not coupled to any other reaction. is an endergonic reaction that takes place because it is coupled to the exergonic hydrolysis of ATP. is an exergonic reaction that is coupled to the endergonic hydrolysis of ATP.
Answer:
The phosphorylation of glucose to glucose-6-phosphate is endergonic reaction that is coupled to the exergonic hydrolysis of ATP.
Explanation:
In glycosis, the first reaction that takes place is the phosphorylation of glucose to glucose-6-phosphate by the enzyme hexokinase. This is an exergenic reaction. This is a coupled reaction in which phosphorylation of glucose is coupled to ATP hydrolysis. The free energy of ATP hydrolysis fuels glucose phosphorylation.
After treating cyclohexanone with LDA, he then added tert-butyl bromide to the same reaction vessel, which one would normally do for a one-pot two-step reaction sequence. However, none of the desired compound formed. Instead, he isolated three organic compounds in a 1:1:1 ratio. What is the structure of each of these organic compounds?
Answer:
2-Tert-butyl-cyclohexanone, 2,6-Di-Tert-Butylcyclohexanone and Di-isopropyl-amine.
Explanation:
A particular reaction has an activation energy, Ea, of 117 kJ/mol. If the rate constant for the reaction is 0.00289 s −1 at 590 °C, at what temperature(in°C) would the rate constant be 0.492 s −1?
Answer:
986.9 °C
Explanation:
To find the temperature we can use the Arrhenius equation:
[tex] k = Ae^{(-E_{a}/RT)} [/tex] (1)
Where:
k: is the reaction rate coefficient
A: is the pre-exponential factor
Ea: is the activation energy
R: is the gas constant
T: is the absolute temperature
With the values given:
T = 590 °C = 863 K; Ea = 117 kJ/mol; k = 0.00289 s⁻¹,
we can find the pre-exponential factor, A:
[tex]A = \frac{k}{e^{(-E_{a}/RT)}} = \frac{0.00289 s^{-1}}{e^{(-117 \cdot 10^{3} J*mol^{-1}/(8.314 J*K^{-1}*mol^{-1}*863 K))}} = 3.49 \cdot 10^{4} s^{-1}[/tex]
Now, we can find the temperature by solving equation (1) for T:
[tex] ln(\frac{k}{A}) = -\frac{E_{a}}{RT} [/tex]
[tex] T = -\frac{E_{a}}{R*ln(\frac{k}{A})} = -\frac{117 \cdot 10^{3} J*mol^{-1}}{8.314 J*K^{-1}*mol^{-1}*ln(\frac{0.492 s^{-1}}{3.49 \cdot 10^{4} s^{-1}})} = 1259.9 K = 986.9 ^{\circ} C [/tex]
Therefore, at 986.9 °C, the rate constant will be 0.492 s⁻¹.
I hope it helps you!
Answer:
The temperature is 1259.4 K (986.4°C)
Explanation:
Step 1: Data given
The activation energy Ea, = 117 kJ/mol
the rate constant for the reaction is 0.00289 s −1 at 590 °C
The new rate constant = 0.492 s −1
Step 2: Calculate the temperature
ln(kX / k 590°C) = Ea/R * (1/T1 - 1/T2)
⇒with kX is the rate constant at the new temperature = 0.492 /s
⇒with k 590 °C = the rate constant at 590 °C = 0.00289 /s
⇒with Ea = the activation energy = 117000 J/mol
⇒with R = the gas constant = 8.314 J/mol*K
⇒with T1 = 590 °C = 863 K
⇒ with T2 = the new temperature
ln (0.492 / 0.00289) = 117000/8.314 *(1/863 - 1/T2)
5.137 =14072.6 * (1/863 - 1/T2)
3.65*10^-4 = (1/863 - 1/T2)
3.65*10^-4 = 0.001159 - 1/T2
0,000794 = 1/T2
T2 = 1259.4 K
The temperature is 1259.4 K or 986.4 °C
Dinitrogen monoxide gas is collected at 19.0 °C in an evacuated flask with a measured volume of 30.0 L. When all the gas has been collected, the pressure in the flask is measured to be 0.500 atm . Calculate the mass and number of moles of dinitrogen monoxide gas that were collected. Be sure your answer has the correct number of significant digits.
Answer:
0.625 mol
27.5 g
Explanation:
Given data
Gas: N₂OMolar mass (M): 44.01 g/molTemperature (T): 19.0°C + 273.15 = 292.2 KVolume (V): 30.0 LPressure (P): 0.500 atmWe can find the moles of dinitrogen monoxide gas using the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 0.500 atm × 30.0 L/0.0821 atm.L/mol.K × 292.2 K
n = 0.625 mol
Then, we can find the mass corresponding to 0.625 moles.
0.625 mol × 44.01 g/mol = 27.5 g
Answer:
0.626 moles of dinitrogen monoxide gas was collected, this is 27.6 grams
Explanation:
Step 1: Data given
Temperature of Dinitrogen monoxide gas = 19.0 °C = 292 K
Volume of the flask = 30.0 L
Pressure in the flask = 0.500 atm
Step 2: Calculate moles of Dinitrogen monoxide gas
p*V = n*R*T
⇒with p = the pressure of the gas = 0.500 atm
⇒with V = the volume of the flask = 30.0 L
⇒with n = the number of moles of N2O gas = TO BE DETERMINED
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 292 K
n = (p*V) / (R*T)
n = (0.500 * 30.0) / (0.08206 * 292)
n = 0.626 moles
Step 3: Calculate mass N2O gas
Mass N2O gas = moles N2O * molar mass N2O
Mass N2O = 0.626 moles * 44.013 g/mol
Mass N2O = 27.55 grams ≈ 27.6 grams
0.626 moles of dinitrogen monoxide gas was collected, this is 27.6 grams
For many purposes we can treat ammonia (NH) as an ideal gas at temperatures above its boiling point of -33.C. Suppose the temperature of a sample of ammonia gas is raised from -21.0°C to 23.0°C, and at the same time the pressure is changed. If the initial pressure was 4.6 atm and the volume decreased by 50.0%, what is the final pressure? Round your answer to 2 significant digits. atm OP x 5 ?
The final pressure of the ammonia gas, given that the the volume decreased by 50.0%, is 11 atm
How to calculate the final pressure f the ammonia gas?
To solve this question in the most simplified way, we shall begin by listing out the given data from the question. This is shown below:
Initial temperature of ammonia gas (T₁) = -21 °C = -21 + 273 = 252 KFinal temperature of ammonia gas (T₂) = 23 °C = 23 + 273 = 296 KInitial volume of ammonia gas (V₁) = VInitial pressure of ammonia gas (P₁) = 4.6 atmFinal volume of ammonia gas (V₂) = 50% of V₁ = 0.5VFinal pressure of ammonia gas (P₂) = ?The final pressure of the ammonia gas can be calculated as shown below:
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\\\frac{4.6\ \times\ V}{252} = \frac{P_2\ \times\ 0.5V}{296}\\\\252\ \times\ P_2\ \times\ 0.5V = 4.6\ \times\ V\ \times\ 296\\\\P_2 = \frac{4.6\ \times\ V\ \times\ 296}{252\ \times\ 0.5V} \\\\P_2 = 11\ atm[/tex]
Thus, the final pressure is 11 atm
The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa
Answer:
[tex]\large \boxed{\text{103 kPa}}[/tex]
Explanation:
We can use the Ideal Gas Law — pV = nRT
Data:
V = 66.8 L
m = 77.8 g
T = 25 °C
Calculations:
(a) Moles of N₂
[tex]\text{Moles of N}_{2} = \text{77.8 g N}_{2} \times \dfrac{\text{1 mol N}_{2}}{\text{28.01 g N}_{2}} = \text{2.778 mol N}_{2}[/tex]
(b) Convert the temperature to kelvins
T = (25 + 273.15) K = 298.15 K
(c) Calculate the pressure
[tex]\begin{array}{rcl}pV & =& nRT\\p \times \text{66.8 L} & = & \text{2.778 mol} \times \text{8.314 kPa$\cdot$ L$\cdot$K$^{-1}$mol$^{-1}\times$ 298.15 K}\\66.8p & = & \text{6886 kPa}\\p & = & \textbf{103 kPa}\end{array}\\\text{The pressure in the bag is $\large \boxed{\textbf{103 kPa}}$}[/tex]
Suppose the formation of nitryl fluoride proceeds by the following mechanism:
Step. Elementary reaction. Rate constant
1. NO2(g)+F2(g)=NO2F(g)+F(g). K1
2. F(g)+NO2(g)=NO2F(g). K2
Suppose also k1<
Write the balanced chemical equation for the overall chemical reaction:
Write the experimentally observable rate law for the overall chemical reaction:
Note your answer should not contain concentrations or any intermediates.
Express the rate constant k for the overall chemical reaction in terms of k1, k2, and If necessary the rate constants k-1 and k-2 for the reverse of two elementary reactions in the mechanism.
The balanced chemical equation for the overall chemical reaction is 2NO2(g) + F2(g) → 2NO2F(g). The experimentally observable rate law for the overall chemical reaction is Rate = k [NO2]^2 [F2]. The rate constant k for the overall chemical reaction can be expressed in terms of k1, k2, k-1, and k-2 as k = k1 (k2/k-1).
Explanation:The balanced chemical equation for the overall chemical reaction is:
2NO2(g) + F2(g) → 2NO2F(g)
The experimentally observable rate law for the overall chemical reaction is:
Rate = k [NO2]2 [F2]
The rate constant k for the overall chemical reaction can be expressed in terms of k1, k2, k-1, and k-2 as:
k = k1 (k2/k-1)
Consider the following initial rate data (at 273 K) for the decomposition of a substrate (substrate 1) which decomposes to product 1 and product 2: [Substrate 1] (M) Initial Rate (M/s) 0.4 0.183 0.8 0.183 2 0.183 Determine the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M.
Answer:
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.
Explanation:
A → B + C
The rate law of the reaction will be :
[tex]R=k[A]^x[/tex]
Initial rate of the reaction when concentration of the substrate was 0.4 M:
[tex]0.183 M/s=k[0.4 M]^x[/tex]..[1]
Initial rate of the reaction when concentration of the substrate was 0.8 M:
[tex]0.183 M/s=k[0.8 M]^x[/tex]...[2]
[1] ÷ [2] :
[tex]\frac{0.183 M/s}{0.183 M/s}=\frac{k[0.4 M]^x}{k[0.8 M]^x}[/tex]
x = 0
The order of the reaction is zero.
For the value of rate constant ,k:
[tex]0.183 M/s=k[0.4 M]^x[/tex]..[1]
x = 0
[tex]0.183 M/s=k[0.4 M]^0[/tex]
k= 0.183 M/s
The half life of the zero order kinetics is given by :
[tex]t_{1/2}=\frac{[A_o]}{2k}[/tex]
Where:
[tex][A_o][/tex] = Initial concentration of A
k = Rate constant of the reaction
So, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M:
[tex]t_{1/2}=\frac{2.77 M}{0.183 M/s}=15.1 s[/tex]
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.
The half-life for the decomposition of substrate 1 when the initial concentration is 2.77 M is approximately 18 minutes.
Explanation:The half-life of a reaction is the time required for one-half of a given amount of reactant to be consumed. In this case, we have initial rate data for the decomposition of substrate 1 with varying concentrations. To determine the half-life of substrate 1, we need to find the time it takes for the concentration to decrease to half its initial value.
Based on the given data, we can see that the initial rate of the reaction is constant at 0.183 M/s for different initial concentrations of substrate 1 (0.4 M, 0.8 M, and 2 M). Since the initial concentration of substrate 1 is given as 2.77 M, it will take the same amount of time as the other initial concentrations to reach half its initial concentration.
Therefore, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M is the same as the half-life observed when the initial concentration is 0.4 M, 0.8 M, or 2 M, which is approximately 18 minutes.
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Draw the structure(s) of the organic product(s) predicted when this compound reacts with NaBH4, ethanol then H3O . Use the wedge and dash tools to indicate stereochemistry as appropriate.
Answer:
what subject is this and how can u draw the structure :)
The reaction involves reduction of the given compound using NaBH4, followed by solvolysis with ethanol and protonation with H3O. Organic molecules structures are often simplified using the skeletal structure method. The stereochemistry of compounds can be represented using wedge and dash notation.
Explanation:The given reaction involves the reduction of an organic compound using NaBH4, followed by the treatment with ethanol and then H3O. When a carbonyl compound reacts with NaBH4, the borohydride group of NaBH4 (it is a weak reducing agent with H-) adds to the carbonyl carbon. This process, known as reduction, transforms the carbonyl group to an alcohol. Following reduction, the compound is undergoing solvolysis in ethanol and then protonated with H3O. Due to the restriction of the platform, displaying the structural representation of the organic product is limited.
The wedge and dash notation can be used to represent the stereochemistry of the compounds. Solid wedges represent bonds coming up out of the plane whereas dashed lines represent bonds going into the plane. These techniques assist chemists to define molecular structures in three dimensions. Skeletal structure, a common way of simplifying complex organic structures in chemistry, represents carbons as each end of a line or bend in a line.
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How many moles of gas are present in 1.13 L of gas at 2.09 atm and 291 K?
n = 0.0989 moles
Explanation:n = PV / RT
P = 2.09atm
V = 1.13L
R = 0.08206
T = 291K
Plug the numbers in the equation.
n = (2.09atm)(1.13L) / (0.08206)(291K)
n = 0.0989 moles
Final answer:
Using the ideal gas law PV = nRT, and rearranging for n (moles), we plug in the given values of pressure, volume, and temperature alongside the ideal gas constant to calculate the number of moles of gas present. There are approximately 0.0988 moles of gas present in 1.13 L of gas at [tex]\(2.09 \, \text{atm}\) and \(291 \, \text{K}\).[/tex]
Explanation:
To calculate the number of moles of gas present, we can use the ideal gas law equation:
[tex]\[ PV = nRT \][/tex]
First, we need to convert the given pressure to atm and the volume to liters if they are not already in those units.
Given:
- [tex]\( P = 2.09 \, \text{atm} \)[/tex]
-[tex]\( V = 1.13 \, \text{L} \[/tex]
- [tex]\( T = 291 \, \text{K} \)[/tex]
Now, we can rearrange the ideal gas law equation to solve for \( n \):
[tex]\[ n = \frac{PV}{RT} \][/tex]
[tex]\[ n = \frac{(2.09 \, \text{atm})(1.13 \, \text{L})}{(0.0821 \, \text{atm} \cdot \text{L} / \text{mol} \cdot \text{K})(291 \, \text{K})} \][/tex]
[tex]\[ n = \frac{2.3617}{23.9211} \][/tex]
[tex]\[ n \approx 0.0988 \, \text{mol} \][/tex]
Therefore, There are approximately 0.0988 moles of gas present in 1.13 L of gas at [tex]\(2.09 \, \text{atm}\) and \(291 \, \text{K}\).[/tex]