Answer:
1.93
Explanation:
Moles of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH = 38/1000 × 0.50 = 0.019mol
Moles of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa = Mass/Molar mass = 2.64/144.10 = 0.018321mol
Final pH = pKa + log([C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa]/[C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH]
= -log Ka + log(mols of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COONa]/mols of C[tex]_{6}[/tex]H[tex]_{5}[/tex]COOH
= -log(6.5 × 10^(-5)) + log (0.018321/0.019)=4.17
change in pH = final - initial pH
= 4.17 - 2.24
=1.93
pH is the depiction of the acidity and basicity of the solution and can range from 0 to 14 on the pH scale. The change in pH of benzoic acid is 1.93.
What is pH?pH is the concentration of the hydrogen ion in the solution depicting the acidity and the basicity of the solution.
Calculate moles of Benzoic acid as:
[tex]\begin{aligned}\rm moles &= \dfrac{\rm mass}{\rm molar \;mass}\\\\&=\dfrac{38}{1000}\times 0.50\\\\&= 0.019\;\rm mol\end{aligned}[/tex]
Calculate moles of Sodium benzoate as:
[tex]\begin{aligned}\rm moles &= \dfrac{\rm mass}{\rm molar \;mass}\\\\&=\dfrac{2.64}{144.10}\\\\&= 0.01832\;\rm mol\end{aligned}[/tex]
pH can be calculated as:
[tex]\begin{aligned}\rm pH &= \rm pKa + log\dfrac{([C_{6}H_{5}COONa]}{[C_{6}H_{5}COOH]}\\\\&= \rm -log(6.5 \times 10^{-5}) + log (\dfrac{0.018321}{0.019})\\\\&= 4.17\end{aligned}[/tex]
Change in the pH can be calculated as:
[tex]\begin{aligned} \rm pH &= \rm final - initial \; pH\\\\&= 4.17 - 2.24\\\\&=1.93\end{aligned}[/tex]
Therefore, 1.93 is the change in the pH of benzoic acid.
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Answer:
CaCl2
The charge can be +3 or +2
Explanation:
It takes two atoms of chloride and 1 atom of calcium to make this compound
Answer: chemical formula is CaCl2.
CaCl2------> Ca^2+ + Cl^-1
Explanation:
+2 for calcium ion and -1 for the chlorine ion.
For K2Cr2O7
The oxidation state of chromium=
(+1×2) + 2x + (-2×7) = 0
+2+2x-14= 0
2x= 12
x= 12/2
x=+6.
Therefore, the oxidation number of Chromium, Cr in K2CrO7 is +6
According to your textbook, how many of the following are properties of covalent compounds:
a. *Can be gas, liquid or solid at room temperature
b. *Does not conduct electricity
c. *Exist as an extended array of atoms
d. *Involves the sharing of electrons
Answer:
a, b and d are correct.
Explanation:
There are two types of attractive forces in covalent compounds:
The force that holds the atoms together in a molecule (the covalent bond).The force that operates between molecules, called an intermolecular force.Because intermolecular forces are usually quite weak compared with the forces holding atoms together within a molecule, molecules of a covalent compound are not held together tightly. Consequently, covalent compounds are usually gases, liquids, or low-melting solids.
Most covalent compounds aqueous solutions generally do not conduct electricity, because the compounds are nonelectrolytes.
Iron reacts with copper (II) sulfate to produce copper and iron (II) sulfate. If 5.6 moles of iron react, how many molecules of iron (II) sulfate will be produced?
1 Fe + 1 CuSO4 --> 1 Cu + 1 FeSO4
Answer:
The molecules of Fe formed are 3.37 ₓ10²⁴
Explanation:
The reaction is this one:
Fe + CuSO₄ --> Cu + FeSO₄
And the ratio for the reaction is 1:1
If 5.6 moles of iron react, you will have 5.6 moles of FeSO₄. By the way, you should use NA to calculate the number of molecules.
1 mol ____ has ___ 6.02x10²³
5.6 moles _______ (5.6 x 6.02x10²³) = 3.37 ₓ10²⁴
The beta oxidation pathway degrades activated fatty acids (acyl-CoA) to acetyl-CoA, which then enters the citric acid cycle. Additional enzymes are required to oxidize unsaturated, odd-chain, long-chain, and branched fatty acids.
Which of the following statements are true?
a.A 14-carbon monounsaturated fatty acid with cis configuration yields more ATP than a 14-carbon saturated fatty acid.
b.Enoyl-CoA isomerase, an enzyme that converts cis double bonds to trans double bonds in fatty acid metabolism, bypasses a step that reduces Q, resulting in a higher ATP yield.
c.Even-chain saturated fatty acids are oxidized to acetyl-CoA in the beta oxidation pathway.
d.Complete catabolism of the three-carbon remnant of a 15-carbon fatty acid requires some citric acid cycle enzymes.
e.Trans double bonds in unsaturated fatty acids are not recognized by beta oxidation enzymes.
f.The final round of beta oxidation for a 13-carbon saturated fatty acid yields acetyl-CoA and propionyl-CoA, a three-carbon fragment.
Final answer:
Enoyl-CoA isomerase bypasses a step that reduces Q, resulting in higher ATP yield. Even-chain saturated fatty acids are oxidized to acetyl-CoA in beta oxidation. Complete catabolism of a 15-carbon fatty acid requires some citric acid cycle enzymes.
Explanation:
Statement b is true. Enoyl-CoA isomerase, an enzyme in fatty acid metabolism, bypasses a step that reduces Q, resulting in higher ATP yield.
Statement c is true. Even-chain saturated fatty acids are oxidized to acetyl-CoA in the beta oxidation pathway.
Statement d is true. Complete catabolism of the three-carbon remnant of a 15-carbon fatty acid requires some citric acid cycle enzymes.
How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.
Answer:
V H2O = 170.270 mL
Explanation:
QH2O ( heat gained) = Qcoffe ( heat ceded)⇒ Q = mCΔT
∴ m: mass (g)
∴ C: specific heat
assuming:
δ H2O = δ Coffe = 1.00 g/mLC H2O = C coffe = 4.186 J/°C.g....from literature⇒ Q coffe = (mcoffe)(C coffe)(60 - 95)
∴ m coffe = (180mL)(1.00 g/mL) = 180 g coffe
⇒ Q = (180g)(4.186 J/°C.g)(-35°C) = - 26371.8 J
⇒ Q H2O = 26371.8 J = (m)(4.186 J/°C.g)(60 - 23)
⇒ (26371.8 J)/(154.882 J/g) = m H2O
⇒ m H2O = 170.270 g
⇒ V H2O = (170.270 g)(mL/1.00g) = 170.270 mL
Entry to the citric acid cycle involves the generation of acetyl CoA. The enzyme that catalyzes the production of acetyl CoA is _____. This enzyme is negatively regulated by _____ in a high-energy charge situation.
Answer:
Pyruvate dehydrogenase / Feedback on pyruvate decarboxylase production
Explanation:
In the Krebs cycle, pyruvic acid from glycolysis undergoes an oxidative decarboxylation process through the action of the pyruvate dehydrogenase enzyme found within the mitochondria of eukaryotes, it reacts with coenzyme A (CoA).
The result of this reaction is the production of acetylcoenzyme A (acetylCoA) and a carbon dioxide (CO₂) molecule.
When acetyl CoA and ATP are at high concentrations (in addition to the increased NADH / NAD⁺ ratio), pyruvate carboxylase production is stimulated.
This process will eventually generate oxalacetic acid for gluconeogenesis (conversion of pyruvate to glucose).
By increasing oxalacetic acid concentrations, pyruvate dehydrogenase is eventually inhibited by a negative feedback mechanism.
In other words, as energy levels increase, the higher the pyruvate carboxylase production, and therefore the greater the inhibition of pyruvate dehydrogenase.
When 1 mol CS2(l) forms from its elements at 1 atm and 25°C, 89.7 kJ of heat is absorbed, and it takes 27.7 kJ to vaporize 1 mol of the liquid. How much heat is absorbed when 1 mol CS2(g) forms from its elements at these conditions
Answer:
There is 117.4 kJ of heat absorbed
Explanation:
Step 1: Data given
Number of moles CS2 = 1 mol
Temperature = 25° = 273 +25 = 298 Kelvin
Heat absorbed = 89.7 kJ
It takes 27.7 kJ to vaporize 1 mol of the liquid
Step 2: Calculate the heat that is absorbed
C(s) + 2S(s) → CS2(l) ΔH = 89.7 kJ (positive since heat is absorbed)
CS2(l) → CS2(g) ΔH = 27.7 kJ (positive since heat is absorbed)
We should balance the equations, before summing, but since they are already balanced, we don't have to change anything.
C(s) + 2S(s)---> CS2 (g)
ΔH = 89.7 + 27.7 = 117.4 kJ
There is 117.4 kJ of heat absorbed
Calculate the percent by mass of the solute in each of the following aqueous solutions: (a) 5.50 g of NaBr in 78.2 g of solution, (b) 31.0 g of KCl in 152 g of water, (c) 4.5 g of toluene in 29 g of benzene.
Answer:
For a: The mass percent of NaBr is 7.03 %
For b: The mass percent of KCl is 16.94 %
For c: The mass percent of toluene is 13.43 %
Explanation:
To calculate the mass percentage of solute in solution, we use the equation:
[tex]\text{Mass percent of solute}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100[/tex] .......(1)
For a:We are given:
Mass of NaBr (Solute) = 5.50 g
Mass of solution = 78.2 g
Putting values in equation 1, we get:
[tex]\text{Mass percent of NaBr}=\frac{5.50g}{78.2g}\times 100\\\\\text{Mass percent of NaBr}=7.03\%[/tex]
Hence, the mass percent of NaBr is 7.03 %
For b:We are given:
Mass of KCl (Solute) = 31.0 g
Mass of water (solvent) = 152 g
Mass of solution = (31.0 + 152) g = 183 g
Putting values in equation 1, we get:
[tex]\text{Mass percent of KCl}=\frac{31.0g}{183g}\times 100\\\\\text{Mass percent of KCl}=16.94\%[/tex]
Hence, the mass percent of KCl is 16.94 %
For c:We are given:
Mass of toluene (Solute) = 4.5 g
Mass of benzene (solvent) = 29 g
Mass of solution = (4.5 + 29) g = 33.5 g
Putting values in equation 1, we get:
[tex]\text{Mass percent of toluene}=\frac{4.5g}{33.5g}\times 100\\\\\text{Mass percent of toluene}=13.43\%[/tex]
Hence, the mass percent of toluene is 13.43 %
To calculate the percent by mass of the solute in each aqueous solution, divide the mass of the solute by the mass of the solution and multiply by 100%
Explanation:To calculate the percent by mass of the solute in each aqueous solution, you'll need to use the formula:
Percent by mass = (mass of solute/mass of solution) x 100%
For example, in solution (a) with 5.50 g of NaBr in 78.2 g of solution, the mass of the solute is 5.50 g and the mass of the solution is 78.2 g. Plugging these values into the formula gives:
Percent by mass = (5.50 g / 78.2 g) x 100% = 7.03%
Similarly, you can calculate the percent by mass for solutions (b) and (c) using the same formula.
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Find an expression for the change in entropy when two blocks of the same substance of equal mass, one at the temperature Th and the other at Tc, are brought into contact and allowed to reach equilibrium. Evaluate the change for the two blocks of copper, each of mass 500 grams with Cpcm= 24.4 J KT-1 mol-1, taking Th = 500 K and Tc = 250 K.
Explanation:
Relation between entropy change and specific heat is as follows.
[tex]\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})[/tex]
The given data is as follows.
mass = 500 g, [tex]C_{p}[/tex] = 24.4 J/mol K
[tex]T_{h}[/tex] = 500 K, [tex]T_{c}[/tex] = 250 K
Mass number of copper = 63.54 g /mol
Number of moles = [tex]\frac{mass}{/text{\molar mass}}[/tex]
= [tex]\frac{500}{63.54}[/tex]
= 7.86 moles
Now, equating the entropy change for both the substances as follows.
[tex]7.86 \times 24.4 \times [T_{f} - 250][/tex] = [tex]7.86 \times 24.4 \times [500 -T_{f}][/tex]
[tex]T_{f} - 250 = 500 - T_{f}[/tex]
[tex]2T_{f}[/tex] = 750
So, [tex]T_{f}[/tex] = [tex]375^{o}C[/tex]
For the metal block A, change in entropy is as follows.[tex]\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})[/tex]
= [tex]24.4 log [\frac{375}{500}][/tex]
= -3.04 J/ K mol
For the block B, change in entropy is as follows.[tex]\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})[/tex]
= [tex]24.4 log [\frac{375}{250}][/tex]
= 4.296 J/Kmol
And, total entropy change will be as follows.
= 4.296 + (-3.04)
= 1.256 J/Kmol
Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol and change in entropy of block B is 4.296 J/Kmol.
A 2.85-g sample of an unknown chlorofluorocarbon is decomposed and produces 564 mL of chlorine gas (Cl2) at a pressure of 752 mmHg and a temperature of 298 K. What is the percent chlorine (by mass) in the unknown chlorofluorocarbon
The percentage of chlorine by mass in the unknown chlorofluorocarbon is 56.8%
We'll begin by calculating the number of mole of Cl₂ produced using the ideal gas equation as illustrated below:
Volume (V) = 564 mL = 564 / 1000 = 0.564 LPressure (P) = 752 mmHg = 752 / 760 = 0.989 atmTemperature (T) = 298 KGas constant (R) = 0.0821 atm.L/KmolNumber of mole (n) =?PV = nRT
0.989 × 0.564 = n × 0.0821 × 298
0.557796 = n × 24.4658
Divide both side by 24.4658
n = 0.557796 / 24.4658
n = 0.0228 mole
Next, we shall determine the mass of 0.0228 mole of Cl₂
Mole = 0.0228 moleMolar mass of Cl₂ = 2 × 35.5 = 71 g/molMass of Cl₂ =?Mass = mole × molar mass
Mass of Cl₂ = 0.0228 × 71
Mass of Cl₂ = 1.62 g
Finally, we shall determine the percentage of chlorine in the unknown chlorofluorocarbon.
Mass of Cl₂ = 1.62 gMass of compound = 2.85 gPercentage of chlorine =?Percentage = (mass / total mass) × 100
Percentage of chlorine =(1.62 / 2.85) × 100
Percentage of chlorine = 56.8%
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The percent chlorine (by mass) in the unknown chlorofluorocarbon is 0%.
Explanation:In order to determine the percent chlorine (by mass) in the unknown chlorofluorocarbon, we need to calculate the mass of chlorine gas produced from the 2.85-g sample. From the equation 2.5g H and 7.5g C make up a 10.0g sample, we can calculate the percent composition of hydrogen and carbon to be 25% and 75%, respectively.
Knowing the percent composition of hydrogen and carbon, we can assume that the unknown chlorofluorocarbon consists of only hydrogen, carbon, and chlorine. Since chlorine gas is produced when the unknown chlorofluorocarbon is decomposed, the percent chlorine (by mass) can be calculated by subtracting the percent composition of hydrogen and carbon from 100%. Therefore, the percent chlorine is 100% - (25% + 75%) = 0%.
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Which tags do not depend on a silicon microchip and use plastic or conductive polymers instead of silicon-based microchips allowing them to be washed or exposed to water without damaging the chip?
Answer: Chipless RFID tags
Which process is a chemical reaction? Which process is a chemical reaction? Limescale is formed on the kettle walls Dew condensing on grass during the night Gasoline evaporating from a gasoline tank Lake freezes in winter
Answer:
Limescale formed on kettle walls
Explanation:
A chemical reaction is one which is associated with a chemical change. While the other two examples are mere change in physical state, the formation of limescale on kettle is a chemical change. It is called the furring of kettles.
These limescales are formed when Calcium bicarbonate decomposes into calcium carbonate. It is this calcium carbonate that causes the furring of kettles.
It is one of the consequence of using temporary hard water. Temporarily hard water contains soluble magnesium bicarbonate and calcium bicarbonate. Now the heating of this water causes the decomposition of the calcium bicarbonate into calcium carbonate which forms these scales on the body of the kettle.
Calcium bicarbonate decomposes into calcium carbonate according to the following equation;
CaH(CO3)2 (aq) ---------> CaCO3 (s) + H2O (l) + CO2 (g)
If the volume of wet gas collected over water is 85.0 mL at 20°C and 760 mm Hg , what is the volume of dry gas at STP conditions? (The vapor pressure of water at 20°C is 17.5 mm Hg.) Express your answer with the appropriate units.
Answer: 77.4 mL
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is:
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of dry gas = (760 - 17.5) mmHg= 742.5 mm Hg
[tex]P_2[/tex] = final pressure of dry gas at STP = 760 mm Hg
[tex]V_1[/tex] = initial volume of dry gas = 85.0 mL
[tex]V_2[/tex] = final volume of dry gas at STP = ?
[tex]T_1[/tex] = initial temperature of dry gas = [tex]20^oC=273+20=293K[/tex]
[tex]T_2[/tex] = final temperature of dry gas at STP = [tex]0^oC=273+0=273K[/tex]
Now put all the given values in the above equation, we get the final volume of wet gas at STP
[tex]\frac{742.5mmHg\times 85.0ml}{293K}=\frac{760mmHg\times V_2}{273K}[/tex]
[tex]V_2=77.4mL[/tex]
Volume of dry gas at STP is 77.4 mL.
How many moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl?
2 Al(s) + 6 HCl(aq) → 2 AlCl3 (aq) + 3 H2(g)
Answer:
0.121 moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl
Explanation:
This is the reaction:
2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂(g)
To make 3 moles of H₂, we need 2 moles of Al.
By conditions given, we will find out how many moles of H₂ do we have.
Let's use the Ideal Gas Law
P. V = n . R . T
1.11 atm . 4.04L = n . 0.082 L.atm/mol.K . 300K
(1.11 atm . 4.04L) / (0.082 mol.K/L.atm . 300K) = n
0.182 mol = n
So the rule of three will be:
If 3 moles of H₂ came from 2 moles of Al
0.182 moles of H₂ will come from x
(0.182 .2) / 3 = 0.121 moles
How is the atomic mass of an element calculated from isotope data
Answer:
[tex]aam=\frac{\Sigma m_{i} \times ab_{i} }{100}[/tex]
Explanation:
When an atom has 2 or more isotopes, the average atomic mass (aam) depends on the mass of each isotope (mi) and the percentual abundance in nature of each isotope (abi). The average atomic mass can be calculated using the following expression:
[tex]aam=\frac{\Sigma m_{i} \times ab_{i} }{100}[/tex]
in the reaction a + b -> c the following information applies. at the beginning of the reaction there was 1.0 mol of chemical a and 0.5 mol of chemical c was recovered. what are the limiting reagent, the theoretical yield, and the percent yield for this reaction
For the reaction, calculate how many grams of the product form when 14.4 g of Br2 completely reacts. Assume that there is more than enough of the other reactant. 2K(s)+Br2(l)→2KBr(s)
Answer:
The answer to your question is 21.45 g of KBr
Explanation:
Chemical reaction
2K + Br₂ ⇒ 2KBr
14.4 ?
Process
1.- Calculate the molecular mass of bromine and potassium bromide
Bromine = 2 x 79.9 = 159.8g
Potassium bromide = 2(79.9 + 39.1) = 238 g
2.- Solve it using proportions
159.8 g of Bromine ------------ 238 g of potassium bromide
14.4 g of Bromine ------------ x
x = (14.4 x 238) / 159.8
x = 3427.2 / 159.8
x = 21.45g of KBr
A solution of phosphoric acid was made by dissolving 10.8 g of H3PO4 in 133.00 mL of water. The resulting volume was 137 mL. Calculate the density, mole fraction, molarity, and molality of the solution.
Answer:
Density is: 1.05 g/ml
Mole fraction solute: 0.015
Mole fraction solvent: 0.095
Molarity: 0.80 M
Molality: 0.82 m
Explanation:
A typical excersise of solution.
It is more confortable to make a table for this.
| masss | volume | mol
solute | | |
solvent | | |
solution | | |
Let's complete, what we have.
| masss | volume | mol
solute | 10.8g | |
solvent | | 133 mL |
solution | | 137 mL |
We can first, know how many moles are 10.8 g
Molar Mass H3PO4 = 97.99 g/mol
Mass / Molar mass = mol
10.8 g / 97.99 g/m = 0.110 mol
Density of water is 1 g/ml (it is a very knowly value)
From this data, we can know water mass, solvent.
Density = mass / volume
1 g/ml = mass / 133 mL
Mass = 133 g
We can also have the moles, by the molar mass of water 18 g/m
133 g / 18 g/m = 7.39 mol
| masss | volume | mol
solute | 10.8g | | 0.110 mol
solvent | 133g | 133 mL | 7.39 mol
solution | 143.8g | 137 mL | 7.50 mol
Mass of solution will be solute mass + solvent mass
Moles of solution will be solute moles + solvent moles
Now we can calculate everything.
Molarity means mol of solute in 1 L of solution. (mol/L)
We have to convert 137 mL in L (/1000)
0.137L so → 0.110 m / 0.137L = 0.80 M
Molality means mol of solute in 1kg of solvent.
We have to convert 133g in kg (/1000)
0.133 kg so → 0.110 m/0.133 kg = 0.82 m
Density is mass / volume
Solution density will be solution mass / solution volume
143.8 g/137 mL = 1.05 g/m
Molar fraction is : solute moles / total moles or solvent moles/total moles.
You can also (x 100%) to have a percent of them.
Remember sum of molar fraction = 1
Molar fraction of solute = 0.110 mol / 7.50mol = 0.015
Molar fraction of solvent = 7.39 mol / 7.50 mol = 0.985
Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and 791 mmHg. Calculate the percent by mass of calcium carbonate in the sample.
Final answer:
To calculate the percent by mass of calcium carbonate in the sample, we need to determine the mass of calcium carbonate present in the 1.14 L of carbon dioxide produced. We can use the ideal gas law to calculate the moles of carbon dioxide, and then use stoichiometry to convert from moles of CO2 to moles of CaCO3. Finally, we can calculate the percent by mass of CaCO3 in the sample by dividing the mass of CaCO3 by the mass of the sample and multiplying by 100.
Explanation:
To calculate the percent by mass of calcium carbonate in the sample, we need to determine the mass of calcium carbonate present in the 1.14 L of carbon dioxide produced. We can use the ideal gas law to calculate the moles of carbon dioxide, and then use stoichiometry to convert from moles of CO2 to moles of CaCO3. Finally, we can calculate the percent by mass of CaCO3 in the sample by dividing the mass of CaCO3 by the mass of the sample and multiplying by 100.
First, let's calculate the moles of CO2 using the ideal gas law:
mass of CO2 = (volume of CO2) x (molar mass of CO2) = (1.14 L) x (22.4 g/mol) = 25.536 g
Next, let's calculate the moles of CaCO3 using stoichiometry. From the balanced equation CaCO3 + 2HCl → CaCl2 + H2O + CO2, we know that 1 mole of CaCO3 produces 1 mole of CO2. Therefore, moles of CaCO3 = moles of CO2 = 25.536 g / (molar mass of CaCO3) = 25.536 g / 100.09 g/mol = 0.255 mol
Finally, let's calculate the percent by mass of CaCO3:
percent by mass of CaCO3 = (mass of CaCO3 / mass of the sample) x 100 = (0.255 mol x (100.09 g/mol)) / 5.28 g x 100 = 4.85%
The principal source of the difference in the normal boiling points of ICI (97°C; molecular mass 162 amu) and Br2(59°C; molecular mass 160 amu) is:a. ICI has greater strength of hydrogen bonding than Br2 b. ICI has stronger dipole-dipole interactions than Br2c. ICI has stronger dispersion forces than Br2 d. The I-CI bond in ICl is stronger than the Br-Br bond in Br2 e. ICI has a greater molecular mass than Br2
Answer:
Option b
Explanation:
The difference in the boiling points of ICl and Br2 is mainly due to the dipole-dipole interactions present between iodine and chlorine in the ICl molecule because as there is difference in electronegativity between iodine and chlorine these type of forces arise
As the dipole-dipole interactions are stronger, the stronger will be the boiling point of that compound because the forces between the molecules increases and as a result the boiling point of the compound increases
In case of Br2 as both are bromine atoms there will be no difference in electronegativity and therefore these type of interactions are not present
Molecular mass can also be a explanation for difference in normal boiling points because more molecular mass means more will be the vander waals forces but as dipole interactions are stronger than vander waals forces the major factor will be due to dipole interactions
The principal source of difference in the normal boiling points of ICI and Br2 is that ICI has stronger dipole-dipole interactions. Although both have similar London dispersion forces due to similar molecular masses, ICI being a polar molecule exhibits stronger dipole-dipole attractions, requiring more energy to overcome and hence has a higher boiling point.
Explanation:The principal source of difference in the normal boiling points of ICI (97°C; molecular mass 162 amu) and Br2(59°C; molecular mass 160 amu) is that ICI has stronger dipole-dipole interactions than Br2. Both ICI and Br2 have similar masses and therefore experience similar London dispersion forces. However, compared to Br2 which is nonpolar, ICI is a polar molecule and thus also exhibits dipole-dipole attractions. These dipole-dipole attractions in ICI are stronger and require more energy to overcome, resulting in a higher boiling point for ICI.
Larger and heavier atoms and molecules like ICI exhibit stronger dispersion forces, leading to higher melting and boiling points. However, in the case of ICI and Br2, the effect of polar dipole-dipole attraction in ICI is more significant. This is apparent when we compare substances with similar molecular masses - the substance with the polar molecules has a higher boiling point, due to the stronger dipole-dipole attractions.
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The process where bacteria takes nitrogen out of the air and makes it usable for other organisms is called
Answer:
The process is called Nitrogen fixation
Explanation:
The nitrogen fixation is a process carried out by some prokaryotic microorganisms (bacteria), specifically those have the presence of the nitrogenase enzyme. The bacteria absorb the atmospheric nitrogen (N2) from the roots of plants, and the nitrogenase enzyme, with the help of two proteins that act as electron donors and acceptors (nitrogenase complex) reduce the nitrogen to ammonia (NH3), then the ammonia is ionized to NH4+ (ammonium). Followed by that, the ammonia is oxidated to nitrates and nitrites, which are finally absorbed again by plants.
Enter your answer in the provided box.S(rhombic) + O2(g) → SO2(g) ΔHo rxn= −296.06 kJ/molS(monoclinic) + O2(g) → SO2(g) ΔHo rxn= −296.36 kJ/molcalculate the enthalpy change for the transformationS(rhombic) → S(monoclinic)(Monoclinic and rhombic are different allotropic forms of elemental sulfur.)_______kJ/mol
Answer: [tex]\Delta H^0=+0.3kJ/mol[/tex].
Explanation:
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
[tex]S_{rhombic}+O_2(g)\rightarrow SO_2(g)[/tex] [tex]\Delta H^0_1=-296.06kJ[/tex] (1)
[tex]S_{monoclinic}+O_2(g)\rightarrow SO_2(g)/tex] [tex]\Delta H^0_2=-296.36kJ[/tex] (2)
The final reaction is:
[tex]S_{rhombic}\rightarrow S_{monoclinic}[/tex] [tex]\Delta H^0_3=?[/tex] (3)
By subtracting (1) and (2)
[tex]\Delta H^0_3=\Delta H^0_1-\Delta H^0_2=-296.06kJ-(-296.36kJ)=0.3kJ[/tex]
Hence the enthalpy change for the transformation S(rhombic) → S(monoclinic) is 0.3kJ
The enthalpy change for the transformation from rhombic sulfur to monoclinic sulfur is -0.30 kJ/mol.
The enthalpy change for the transformation S(rhombic) → S(monoclinic), we will use the enthalpy changes provided for the reactions involving sulfur and oxygen:
S(rhombic) + O₂(g) → SO₂(g) ΔHo rxn = −296.06 kJ/molS(monoclinic) + O₂(g) → SO₂(g) ΔHo rxn = −296.36 kJ/molGiven these values, the enthalpy change for the transformation from rhombic sulfur to monoclinic sulfur can be found as follows:
ΔH (transformation) = ΔHo(S(monoclinic) → SO₂) - ΔHo(S(rhombic) → SO₂)
Using the provided enthalpy changes:
ΔH (transformation) = -296.36 kJ/mol - (-296.06 kJ/mol)
ΔH (transformation) = -296.36 kJ/mol + 296.06 kJ/mol
ΔH (transformation) = -0.30 kJ/mol
The enthalpy change for the transformation S(rhombic) → S(monoclinic) is -0.30 kJ/mol.
Newton's 2nd Law says the acceleration of an object depends on its ______ and the amount of _______ acting on it.
Friction and inertia
Mass and force
Force and friction
Inertia and Gravity
Answer:
Newton's 2nd Law says the acceleration of an object depends on its mass and the amount of net force acting on it.
Explanation:
Definition of acceleration:
The acceleration is rate of change of velocity of an object with respect to time.
Formula:
a = Δv/Δt
a = acceleration
Δv = change in velocity
Δt = change in time
Units:
The unit of acceleration is m.s⁻².
Acceleration can also be determine through following formula,
F = m × a
a = F/m
This is the newton's second law:
"The acceleration of an object depends on its mass and the amount of net force acting on it"
The acceleration is depend directly on the force while inversely on the mass.
Answer:
Mass and Force
Explanation:
The flavor of anise is due to anethole, a compound with the molecular forumal C10H12O. Combustion of one mole of anethole produces 5541 kJ of thermal energy. If 0.950 g of anethole is combusted in a bomb calorimeter whose heat capacity (Ccalorimeter) is 7.854 kJ/°C, what is the change in temperature of the calorimeter?
Final answer:
The change in temperature of the calorimeter upon combusting 0.950 grams of anethole is approximately 24.90°C. This is calculated using the energy released in the combustion and the calorimeter's heat capacity.
Explanation:
The change in temperature of the calorimeter can be calculated using the relationship between the heat capacity of the calorimeter and the amount of heat released during combustion of the substance. Given the combustion of anethole releases 5541 kJ per mole, we first need to determine the amount of energy released during the combustion of 0.950 grams of anethole. Since the molecular weight of anethole (C10H12O) is approximately 148.2 g/mol, we can calculate q (the heat released) for the given mass, and then use this to determine the change in temperature (ΔT).
To calculate the energy released:
q = (0.950 g / 148.2 g/mol) * 5541 kJ/molq = 0.0353 mol * 5541 kJ/mol = 195.59 kJThen, using the calorimeter's heat capacity:
ΔT = q / CcalorimeterΔT = 195.59 kJ / 7.854 kJ/°C ≈ 24.90°CHence, the change in temperature of the calorimeter is approximately 24.90°C.
Assuming constant pressure, rank these reactions from most energy released by the system to most energy absorbed by the system, based on the following descriptions:
A. Surroundings get colder and the system decreases in volume.
B. Surroundings get hotter and the system expands in volume.
C.Surroundings get hotter and the system decreases in volume.
D. Surroundings get hotter and the system does not change in volume.
Also assume that the magnitude of the volume and temperature changes are similar among the reactions.
Rank from most energy released to most energy absorbed. To rank items as equivalent, overlap them.
Answer:
B > D > C > A
Explanation:
For the first law of the thermodynamics, the total energy variation in a process is:
ΔU = Q - W
Where Q is the heat, and W the work. If the system loses heat, Q < 0, if it absorbs heat, Q>0. If work is done in the system (volume decreases), W < 0, if the system does the work (volume increases), W > 0.
A. If the surroundings get colder, the system is absorbing heat, so Q>0, and the system decreases in volume so W < 0 :
ΔU = +Q - (-W) = +Q + W (absorbs a higher energy)
B. If the surroundings ger hotter, the system is losing heat, so Q<0, and the system expands, so W>0:
ΔU = -Q -W (loses higher energy)
C. Surroundings get hotter, Q<0, and the system decreases in volume, W<0
ΔU = - Q + W = 0 (magnitude of heat and work is similar)
D. Surroundings get hotter, Q<0, and the system is not changing in volume, W = 0.
ΔU = -Q (loses energy)
For the most released (more negative) for the most absorbed (most positive):
B > D > C > A
Final answer:
Reaction B is the most exothermic as it releases energy and the surroundings get hotter. Reaction D is less exothermic as the surroundings get warmer without volume change. Reaction A is most endothermic, absorbing energy, indicated by cooler surroundings.
Explanation:
Ranking the reactions from most energy released to most energy absorbed:
Exothermic reaction: Surroundings get hotter and the system expands in volume (B). This reaction releases the most energy, indicated by the heat given off to the surroundings.Surroundings get hotter but volume doesn't change (D). This is still an exothermic reaction but less energy is released compared to B due to no work done on expansion.Surroundings get hotter and the system decreases in volume (C). This describes an endothermic process where energy is absorbed, but because the surroundings get warmer, it's indicative of a net release of energy after accounting for the work done on the system (compression).Endothermic reaction: Surroundings get colder and the system decreases in volume (A). This reaction absorbs the most energy, indicated by the surroundings getting colder.In summary, the reactions involving the surroundings getting warmer are generally exothermic, while those involving the surroundings getting colder are endothermic. The volume change provides additional clues about energy changes; expansion suggests work is done by the system (releasing energy), while a decrease in volume suggests work is done on the system (absorbing energy).
Calculate the percent ionization of 0.120 M lactic acid in a solution containing 8.5×10−3 M sodium lactate.
Express the percent ionization to two significant digits.
To calculate the percent ionization in this scenario, information regarding the ionization constant of lactic acid and the pH is required. The calculation would typically use the Henderson-Hasselbalch equation, but without additional data, we cannot provide a specific answer.
Explanation:The question involves the calculation of the percent ionization of a weak acid in the presence of its conjugate base. To calculate this, one would need the initial concentration of the acid, the acid ionization constant (Ka), and the concentration of the conjugate base. Generally, percent ionization is given by the ratio of the concentration of ionized acid to the initial concentration of the acid, multiplied by 100%. However, since the pH or hydronium ion concentration is not provided and the solution is a buffer system (acid with its conjugate base), the Henderson-Hasselbalch equation would typically be used to find the pH first, then the percent ionization could be calculated.
As specific values are not provided for the ionization constant of lactic acid or the pH of the solution, an accurate calculation cannot be completed without them. If these values were provided, the calculation would involve using the Henderson-Hasselbalch equation to solve for pH, determining the hydronium ion concentration, and subsequently finding the percent ionization.
When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2 Al ( s ) + 6 HCl ( aq ) ⟶ 2 AlCl 3 ( aq ) + 3 H 2 ( g ) What volume of H 2 ( g ) is produced when 3.60 g Al ( s ) reacts at STP?
When 3.60 g of aluminum reacts with hydrochloric acid at STP, approximately 4.48 liters of hydrogen gas is produced. This is calculated using stoichiometry and the conditions of STP, which state that one mole of any gas occupies 22.4 liters.
Explanation:To calculate the volume of hydrogen gas produced when 3.60 g of Al reacts with hydrochloric acid, stoichiometry and the concept of Standard Temperature and Pressure (STP) can be used. Firstly, we need the molar mass of Al, which is approximately 26.98 g/mol. The number of moles of Al in 3.60 g can be calculated by dividing the mass by the molar mass. This gives us approximately 0.133 mol of Al. From the balanced chemical equation, we know that every 2 moles of Al produces 3 moles of H2, so 0.133 mol of Al would produce approximately 0.200 mol of H2.
As per the conditions of STP (Standard Temperature and Pressure), one mole of any gas occupies a volume of 22.4 liters. Therefore, 0.200 mol of H2 would occupy a volume of 0.200 * 22.4 L, which approximates to 4.48 L. So, when 3.60 g of aluminum reacts with hydrochloric acid under the conditions of STP, roughly 4.48 liters of hydrogen gas would be produced.
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A 17.55 mL solution of potassium nitrate (KNO3) was diluted to 125.0 mL, and 25.00 mL of this solution was then diluted to 1.000 × 103 mL. The concentration of the final solution is 0.00383 M. Calculate the concentration of the original solution.
Answer:
[tex]M_2 = 1.094 M[/tex]
Explanation:
Given data:
V1 = 25.0 ml
M2 = 0.00383 M
V2 = 1000 ml
we knwo that [tex]M_1 V_1 = M_2 V_2[/tex]
∴ [tex]M_1 = \frac{M_2 V_ 2}{V_1}[/tex]
[tex] = \frac{0.00383 M \times 1000}{25.0}[/tex]
= 0.1532 M
0.1532 M is concentration in 25 ml but taken from 125 ml solution.
∴ The concentration in 125.0 ml solution is = 0.1532 M
M1 = 0.1532 M
V1 = 125.0ml and V2 = 17.5.0ml
[tex]M_2 = \frac{M_1 V_1}{V_2}[/tex]
[tex] = \frac{0.1532 M \times 125.0}{17.5}[/tex]
[tex]M_2 = 1.094 M[/tex]
Determine the bond order from the molecular electron configurations. ( σ 1 s ) 2 ( σ 1 s * ) 2 ( σ 2 s ) 2 ( σ 2 s * ) 2 ( σ 2 p ) 2 ( π 2 p ) 4 ( π 2 p * ) 2 bond order: ( σ 1 s ) 2 ( σ 1 s * ) 2 ( σ 2 s ) 2 bond order:
Explanation:
The bond order is defined as number of electron pairs present in a bond of the two atoms.
The formula of bond order is given by:
= [tex]\frac{1}{2}\times (\text{Number of bonding electrons}-\text{Number of anti-bonding electrons})[/tex]
1) [tex](\sigma 1 s )^2 ( \sigma 1 s*)^2 (\sigma 2s )^2 ( \sigma 2 s*)^2 ( \sigma 2 p )^2 ( \pi2 p )^4 (\pi 2 p *)^2 [/tex]
Number of bonding electrons = 10
Number of anti-bonding electrons = 6
The bond order : [tex]\frac{1}{2}\times (10-6)=2[/tex]
2) [tex]( \sigma 1 s )^2 ( \sigma 1 s * )^2 ( \sigma 2 s )^2[/tex]
Number of bonding electrons = 4
Number of anti-bonding electrons = 2
The bond order : [tex]\frac{1}{2}\times (4-2)=1[/tex]
The bond order can be determined from the molecular electron configurations by counting the number of bonding and antibonding electrons and calculating the bond order as (number of bonding electrons - number of antibonding electrons)/2.
Explanation:The bond order can be determined from the molecular electron configurations. In this case, the electron configuration given is ( σ 1 s ) 2 ( σ 1 s * ) 2 ( σ 2 s ) 2 ( σ 2 s * ) 2 ( σ 2 p ) 2 ( π 2 p ) 4 ( π 2 p * ) 2. We need to count the number of bonding and antibonding electrons to calculate the bond order. The number of bonding electrons is the sum of the electrons in all the bonding molecular orbitals. In contrast, the number of antibonding electrons is the sum of the electrons in all the antibonding molecular orbitals. The bond order is then calculated as (number of bonding electrons - number of antibonding electrons)/2. In this case, the bond order is (2+2+2+4-2)/2 = 4/2 = 2, indicating a double bond.
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A hypothetical covalent molecule, X–Y, has a dipole moment of 1.93 1.93 D and a bond length of 109 pm. 109 pm. Calculate the partial charge on a pole of this molecule in terms of e , e, where e e is the charge on an electron.
Answer:
q= 110.5 ke
Explanation:
Dipole moment is the product of the separation of the ends of a dipole and the magnitude of the charges.
μ = q * d
μ= Dipole moment (1.93 D)
q= partial charge on each pole
d= separation between the poles(109 pm).
e= electronic charge ( 1.60217662 × 10⁻¹⁹ coulombs)
So,
q= [tex]\frac{1.93}{109 * 10^{-12} }[/tex] coulombs
q = [tex]\frac{1.93}{109 * 10^{-12} * 1.60217662 * 10^{-19} }[/tex] e
q = 1.105 * 10⁵ e
q= 110.5 ke