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
Which of the following is true?
a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.
b. A reaction in which the entropy of the system decreases can be spontaneous only if it is endothermic.
c. A reaction in which the entropy of the system decreases can be spontaneous only if it is exothermic.
d. A reaction in which the entropy of the system increases can be spontaneous only if it is exothermic.
Answer:
a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.
Explanation:
The change in free energy (ΔG) that is, the energy available to do work, of a system for a constant-temperature process is:
[tex]ΔG = ΔH - TΔS[/tex]
When ΔG < 0 the reaction is spontaneous in the forward direction.When ΔG > 0 the reaction is nonspontaneous. The reaction isspontaneous in the opposite direction.
When ΔG = 0 the system is at equilibrium.If both ΔH and ΔS are positive, then ΔG will be negative only when the TΔS term is greater in magnitude than ΔH. This condition is met when T is large.
What type of compound is held together w/ electrostatic forces?
Answer:
Ionic compound.
Explanation:
Ionic bond:
It is the bond which is formed by the transfer of electron from one atom to the atom of another element. the electrostatic attraction is created between bonded atoms.
Both bonded atoms have very large electronegativity difference. The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.
The compound having ionic bonds generally have moderate to high boiling points and melting point because of greater electrostatic interaction. Their electrical conductivity are high and these minerals tend to dissolve in water.
For example:
Sodium chloride is ionic compound. The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion and both atoms joint together through electrostatic interaction.
Hydrogen gas produced in the laboratory by the reaction of zinc and hydrochloric acid was 23) collected over water at 25°C. The barometric pressure at the time was 742.5 mm Hg. What is the pressure of dry hydrogen gas if the vapor pressure of water at 25'C is 23.8 mm Hg?
A) 760.0 mm Hg
B)718.7 mm Hg
C)742.5 mm Hg
D) 766.3 mm Hg
E) none of the above
Answer:
Explanation:
The balanced chemical reaction is:
[tex]Zn+2HCl\rightarrow ZnCl_2+H_2[/tex]
According to Dalton's law, the total pressure is the sum of individual pressures.
[tex]p_{total}=p_1+p_2[/tex]
As the hydrogen is collected over water, the total pressure will be sum of pressure of water and pressure of dry hydrogen.
[tex]p_{total}=p_{H_2}+p_{H_2O}[/tex]
[tex]p_{total}[/tex] = 742.5 mm Hg
[tex]p_{H_2}[/tex] = ?
[tex]p_{H_2O}[/tex] = 23.8 mm Hg
Putting in the values:
[tex]742.5=p_{H_2}+23.8[/tex]
[tex]742.5-23.8=p_{H_2}[/tex]
[tex]p_{H_2}=718.7mmHg[/tex]
Thus the pressure of dry hydrogen gas is 718.7 mm Hg
A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research technician at Pharma-pill, you are told that X-281 is a monoprotic weak acid, but because of security concerns, the actual chemical formula must remain top secret. The company is interested in the drug's Ka value because only the dissociated form of the chemical is active in preventing cholesterol buildup in arteries.
To find the pKa of X-281, you prepare a 0.089 M test solution of X-281 at 25.0 ∘C. The pH of the solution is determined to be 2.40.
a. What is the pKa of X-281? Express your answer numerically.
At 25∘C, for any conjugate acid-base pair
pKa + pKb = 14.00
b. What is pKb of the conjugate base of X-281? (Assume 25 ∘C.) Express your answer numerically.
Answer:
a. pka = 3,73.
b. pkb = 10,27.
Explanation:
a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:
HX + H₂O ⇄ H₃O⁺ + X⁻
Where ka is defined as:
[tex]ka = \frac{[H_3O^+][X^-]}{[HX]}[/tex]
In equilibrium, molar concentrations are:
[HX] = 0,089M - x
[H₃O⁺] = x
[X⁻] = x
pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:
[tex][H_3O^+]} = 10^{-2,40}[/tex]
[H₃O⁺] = 0,004M
Thus:
[X⁻] = 0,004M
And:
[HX] = 0,089M - 0,004M = 0,085M
[tex]ka = \frac{[0,004][0,004]}{[0,085]}[/tex]
ka = 1,88x10⁻⁴
And pka = 3,73
b. As pka + pkb = 14,00
pkb = 14,00 - 3,73
pkb = 10,27
I hope it helps!
The correct answers are:
a. The pKa of X-281 is 2.40.
b. The pKb of the conjugate base of X-281 is 11.60.
a. To find the pKa of X-281, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the acid and the concentration of the acid and its conjugate base. For a monoprotic weak acid HA, the equation is:
[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]
Given that the pH of the solution is 2.40 and the concentration of X-281 (HA) is 0.089 M, we can assume that the concentration of the conjugate base A^- is negligible compared to the concentration of the acid HA because the pH is close to the pKa. This means that the ratio[tex]\([\text{A}^-] / [\text{HA}]\)[/tex]is approximately zero, and thus the log term is approximately zero. Therefore, the pH is approximately equal to the pKa of the acid:
[tex]\[ \text{pH} \approx \text{pKa} \] \[ \text{pKa} \approx 2.40 \][/tex]
b. At 25°C, the sum of pKa and pKb for any conjugate acid-base pair is 14.00, which is the pKw of water:
[tex]\[ \text{pKa} + \text{pKb} = 14.00 \][/tex]
We have already determined that the pKa of X-281 is 2.40. Therefore, we can solve for the pKb of the conjugate base of X-281:
[tex]\[ 2.40 + \text{pKb} = 14.00 \] \[ \text{pKb} = 14.00 - 2.40 \] \[ \text{pKb} = 11.60 \][/tex]
Thus, the pKb of the conjugate base of X-281 is 11.60.
A researcher is performing a gram stain on a specimen in a medical laboratory. She gets distracted by a phone call and accidentally uses water instead of alcohol to decolorize. What may happen as a result of this mistake?
Answer:
Explanation:
It wouldn't be able to separate negative gram from positive. it would not wash away some of the crystal violet stain effectively. Gram - Cells will appear to be the same hue. Water can also dilute cells or wash them off the slide
Answer:
Explanation:
Would not be able to differentiate gram - from + it would not efficiently wash some of the crystal violet stain away. gram - cells will still remain the same color. also water may dilute or wash cells off the slide
Which of the following processes is endothermic?A. a hot cup of soup (system) cools on a countertop B. the combustion of butaneC. the chemical reaction in a "hot pack" often used to treat sore musclesD. the vaporization of rubbing alcohol E. the freezing of water
Answer: Option (D) is the correct answer.
Explanation:
A reaction in which heat energy is absorbed by the molecules of a substance is known as an endothermic reaction.
Whereas when energy is released by the molecules of a substance in a reaction then it is known as an exothermic reaction.
So, when a hot cup of soup (system) cools on a countertop then energy is released into the surrounding by the soup. Hence, it is an exothermic reaction.Also, combustion reactions are always exothermic in nature.The chemical reaction in a "hot pack" often used to treat sore muscles because heat is released to the muscles. Therefore, it is an exothermic reaction.When rubbing alcohol vaporizes then heat is absorbed by the molecules of alcohol. Due to which they gain energy and escape into the surrounding. Therefore, it is an endothermic reaction.Freezing of water is an exothermic reaction because during this process energy is released by the molecules of water. Hence, it is an exothermic reaction.Thus, we can conclude that the vaporization of rubbing alcohol is endothermic.
Among the provided options, the vaporization of rubbing alcohol is the endothermic process, as it involves the absorption of heat from its surroundings.
Explanation:In the context of chemistry, an endothermic process is one in which heat is absorbed from the surroundings. This results in an overall increase in the internal energy of the system. Given the options, the vaporization of rubbing alcohol is the process that is endothermic. This is because when rubbing alcohol vaporizes, it absorbs heat from its surroundings, causing the surrounding environment to feel cooler. Other examples might be the melting of ice or the evaporation of water, both of which require an external heat source to facilitate the process.
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What is the full name of NI3
Answer:
Nitrogen triiodide is NI3.
Explanation:
Consider the following reduction potentials:
2H+ + 2e– → H2 E° = 0.00 V Ni2+ + 2e– → Ni E° = -0.25 V Cd2+ + 2e– → Cd E° = -0.40 V
Based on this information which pairs of substances will react spontaneously under standard state conditions?
Answer:
H⁺/H₂//Ni/Ni⁺²; H⁺/H₂//Cd/Cd⁺²; Ni⁺²/Ni//Cd/Cd⁺²
Explanation:
The redox reactions are spontaneously when the E° of the substance that is reduced is higher than the E° of the substance that is oxidized. The substance is reduced when gaining electrons and is oxidized when loses electrons.
So, the reactions that occur spontaneously are:
2H⁺ + Ni → H₂ + Ni⁺² (H⁺ is reduced and Ni is oxidized)
2H⁺ + Cd → H₂ + Cd⁺² (H⁺ is reduced and Cd is oxidized)
Ni⁺² + Cd → Ni + Cd⁺² (Ni⁺² is reduced and Cd is oxidized)
Pairs: H⁺/H₂//Ni/Ni⁺²; H⁺/H₂//Cd/Cd⁺²; Ni⁺²/Ni//Cd/Cd⁺²
How much carbon dioxide in kilograms is produced upon the complete combustion of 29.4 L of propane (approximate contents of one 5-gal tank)? Assume that the density of the liquid propane in the tank is 0.621 g / mL.
Answer:
61kg
Explanation:
Like all other hydrocarbons, the combustion of propane will yield water and carbon iv oxide.
Let’s write a complete and balanced chemical equation for this.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
Now, we first convert the information into a mass.
We know that mass = density * volume
Let’s convert the volume here into ml since density is in mL. It must be remembered that 1000ml = 1L. Hence, 29.4L = 29,400ml
The mass is thus = 0.6921 * 29,400 = 20,347.74g
Now from the balanced equation, we can see that one mole of propane produced 3 moles of carbon iv oxide. This is the theoretical relation.
Let’s now calculate the actual number of moles. The number of moles of propane produced is the mass of propane divided by the molar mass of propane. The molar mass of propane is 3(12) + 8(1) = 36 + 8 = 44g/mol
The number of moles = 20,347.74/44 = 462.44
We simply multiply this number by 3 to get the actual number of moles of carbon dioxide produced. This is equals 1387.35 moles
Since we know this, we now calculate the mass of carbon iv oxide produced. The mass is equals the number of moles multiplied by the molar mass. The molar mass of carbon iv oxide is 44g/mol.
The mass produced is thus 1387.35 * 44 = 61,043g
To get the kilogram equivalent, we simply divide by 1000 = 61,043/1000 = 61kg
Draw the structure of the starting material needed to make 2-methylhept-3-yne using sodium amide in liquid ammonia
Answer:
3,4-dibromo 2-methylheptane
Explanation:
Alkynes can be synthesized by using sodium amide in liquid ammonia, when the starting material is a vicinal dibromoalkane. This means an alkane in which there's a Br group in two carbons next to one another.
In the attached picture you can see the reaction.
Answer:
see picture
Explanation:
Although the previous answer may be correct, the question is incomplete, the completed question is:
Draw the structure of the starting material needed to make 2-methylhept-3-yne using sodium amide in liquid ammonia followed by 1-bromopropane.
Now, the previous answer is also incorrect, because the starting material with the two bromine in carbon 3 and 4, the first reaction taking place is the substracting of the hydrogen from carbon 5, and not carbon 3 because it's bulkier than carbon 5. So the majority product formed will be a double bonded product and not a triple bond. Therefore, this answer is incorrect.
Now, that we have completed the question the Starting material would have to be the following:
CH3 - CH(CH3) - C ≡ CH
When this reactant, is in presence of NaNH2, the NH2 substract the Hydrogen from carbon 1, and then, this will attack as nucleophyle to the carbon 1 of the bromopropane and formed the desired product.
The mechanism is as following in the picture.
Dalton's experiments advanced our understanding of the atom withoutA.carefully observing and measuring events.B.discovering the internal structure of the atom.C.examining the ratios of elements involved in chemical reactions.D.testing a hypothesis.
Answer:
B: Discovering the internal structure of the atom.
Explanation:
Dalton’s experiments with gases marked the beginning of the modern era of chemistry. The hypotheses about the nature of matter on which Dalton’s atomic theory is based can be
summarized as follows:
Elements are composed of extremely small particles called atoms. All atoms of a given element are identical, having the same size, mass, and chemical properties. The atoms of one element are different from the atoms of all other elements. Compounds are composed of atoms of more than one element. In any compound, the ratio of the numbers of atoms of any two of the elements present is either an integer or a simple fraction. A chemical reaction involves only the separation, combination, or rearrangement of atoms; it does not result in their creation or destruction.Dalton made no attempt to describe the structure or composition of atom —he had no idea what an atom is really like. But he did realize that the different properties shown by elements such as hydrogen and oxygen can be explained by assuming that hydrogen atoms are not the same as oxygen atoms.
To determine the moles of ammonium chloride produced, we need to find the limiting reactant. H2O is present in an unlimited amount, so it cannot be the limiting reactant. Let’s then determine the amount of each remaining reactant present (in moles). NaCl is already given in moles (0.250 mol). NH3 and CO2 are not. They are given in liters. How many moles are present in 3.08 L of NH3 and 2.09 L of CO2?
Answer:
0.094 mols of CO2 are present in 2.09 L.
0.132 mols of NH3 are present in 3.08 L
The moles of ammonium chloride produced are 0.094
Explanation:
First of all, think the reaction
NH3 + 2H2O + NaCl + CO2 → NH4Cl + NaOH + H2CO3
To get the moles with volume, you must need density
Density = mass / volume
Density CO2 = 0,001976 g/ml
Density NH3 = 0,00073 g/ml
Volume of CO2 = 2.09L
2.09 L . 1000 = 2090 mL
0,001976 g/ml = mass / 2090 mL
4.13 g = mass
Molar weight = 44 g/m
mass / molar weight = 4.13 g / 44 g/m = 0.094 moles
Volume of NH3 = 3.08 L
3.08 L . 1000 = 3080 mL
0,00073 g/ml = mass/ 3080 mL
2.25 g = mass
Molar weight = 17 g/m
mass / molar weight = mols → 2.25 g / 17g/m = 0.132 mols
In my reagents, the least amount I have is CO2. This is my limiting reactant. Take account the reaction.
1 mol of NH3 reacts with 1 mol of CO2, so as I have 0.132 moles of NH3 I need 0.132 moles of CO2; I only have 0.094.
Relation between CO2 and NH4Cl is 1:1 so, 1 mol of CO2 is needed to produce 1 mol e chloride, so 0.094 mols are needed to produce the same amount of chloride.
Which of the following were found to destroy ozone in the upper atmosphere?A. sulfur oxides.B. CFCs.C. nitrogen oxides.D. carbon dioxide.
Answer:
sulfur oxides
Explanation:
One difference between mixtures and pure substances is that
Answer:
The mixture is made up of different atoms and pure substance is made up of same type of atom.
The main difference is that mixture can be separated into its component by physical mean while pure substances can not be separated by physical process
Explanation:
Mixture:
The properties of the mixture are not same and contains the properties of all those component present in it.it is a combination of one or more Pure substances and can be separated by simple physical methods.it have varying boiling and melting pointExamples are:
mixture of salt and sandSalt water is mixture of water and NaCl and can be separated by physical mean.Alloys: its a mixture of different metalAir: mixture of gasesPure Substance:
Pure substances are those made of same type of atoms all elements and compounds are pure substances.
it can not be separated by simple physical meanit have very constant and consistent melting and boiling pointExamples are:
Water : contain only water moleculeAll elements: all elements are pure substance made of same atomsAll compounds: can not be separated by physical mean.In which sample of water do the molecules have the highest average kinetic energy 1)20. ML at 100.°C 2)40. ML at 80. °C 3)60. ML at 60.° C 4)80. Ml at 40.° C
Answer:
1)20. ML at 100.°C
Explanation:
Average kinetic energy is not related with volume. so increase in temperature will also increase kinetic energy. so this increase means the highest temperature will be a right answer.
Final answer:
The water sample with the highest average kinetic energy is the 20 mL of water at 100.0°C, as average kinetic energy is directly proportional to temperature.
Explanation:
The molecules with the highest average kinetic energy are in the sample of water at the highest temperature. Kinetic energy is directly proportional to the absolute temperature of the substance, meaning that as temperature increases, kinetic energy also increases. Therefore, ignoring the volume of the water samples, the rank from highest kinetic energy to lowest based on temperature would be:
20 mL of water at 100.0°C40 mL of water at 80.0°C60 mL of water at 60.0°C80 mL of water at 40.0°CThe sample at 100.0°C holds the highest average kinetic energy because it has the highest temperature.
Redox reactions can be written as two half-reactions, focusing on the gain or loss of electrons by one of the chemical substances. One half-reaction shows the oxidation while the other shows the reduction. When the two half-reactions are combined, the overall reaction is obtained.The half-reactions can aid in the balancing of redox equations because the number of each element must be balanced as well as the number of electrons gained and lost. What substance is added to balance the hydrogen in a half-reaction?
A. H₂O₂
B. H₂O
C. H⁺
D. H₂
Answer:
C) H⁺
Explanation:
When we are balancing the reaction in an acid medium, hydrogen is balanced using the H⁺ species. This is most likely the intended answer of your question.
When the reaction takes place not in an acid medium, but in an alkaline one, then hydrogen is present as the OH⁻ species. However this option is not given in your question.
Thus the answer is option C).
Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ ⋅ mol − 1 calculate the average molar bond enthalpy of the carbon‑bromine bond in a CBr 4 molecule.
Answer:
283.725 kJ ⋅ mol − 1
Explanation:
C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1
[tex]\frac{1}{2}[/tex]Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1
C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1
4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1
eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1
so,
average bond enthalpy is [tex]\frac{1134.9}{4}[/tex] = 283.725 kJ ⋅ mol − 1
The average molar bond enthalpy of the carbon‑bromine bond in a CBr 4 molecule is 283.725 kJ ⋅ mol − 1
Calculation of average molar bond:C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1
Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1
C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1
4*eqn(2) + eqn(3)
⇒ 2Br2(g) + C(s)
⇒ 4 Br(g) + C(g) , Δ H ∘
= 1164.3 kJ ⋅ mol − 1
eqn(1) - eqn(4)
⇒ 4 Br(g) + C(g)
⇒ CBr4(g) , Δ H ∘
= -1134.9 kJ ⋅ mol − 1
so,
average bond enthalpy is = 283.725 kJ ⋅ mol − 1
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The atomic number of an element is the same as the number of its
Answer: The atomic number of an element is the same as the number of its protons.
Explanation: For example, the atomic number of oxygen is 8. So, this would mean that the number of protons would be 8.
Hope this helps^
A 1.50-kilogram ball is attached to the end of a 0.520-meter string and swung in a circle. The velocity of the ball is 9.78 m/s. What is the tension in the string?
Answer:
F centripetal force (tension) = 275.9 N
Explanation:
Given data:
Mass = 1.50 kg
Radius = 0.520 m
Velocity of ball = 9.78 m/s
Tension = ?
Solution:
F centripetal force (tension) = m.v² / R
F centripetal force (tension) = 1.50 kg . (9.78 m/s)² / 0.520 m
F centripetal force (tension) = 1.50 kg . 95.65 m²/s² / 0.520 m
F centripetal force (tension) = 143.5 kg. m²/s² / 0.520 m
F centripetal force (tension) = 275.9 N
A quantity of N2 gas originally held at 5.23 atm pressure in a 1.20 −L container at 26 ∘C is transferred to a 14.5 −L container at 20 ∘C. A quantity of O2 gas originally at 5.21 atm and 26 ∘C in a 5.10 −L container is transferred to this same container.
To find the final pressure of the gas mixture after transferring the gases, we can use the ideal gas law equation. The final pressure is calculated based on the total number of moles of gas and the final volume of the container. The final pressure of the gas mixture is 0.229 atm.
Explanation:To find the final pressure of the gas mixture after the transfer, we can use the ideal gas law equation, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
We can calculate the number of moles of N2 gas using the information given:
n = PV/RT = (5.23 atm)(1.20 L)/(0.0821 atm·L/mol·K)(26 + 273 K) = 0.287 mol
Similarly, we can calculate the number of moles of O2 gas:
n = PV/RT = (5.21 atm)(5.10 L)/(0.0821 atm·L/mol·K)(26 + 273 K) = 1.05 mol
To find the final pressure, we need to find the total number of moles of gas and the final volume of the container:
Total moles of gas = moles of N2 + moles of O2 = 0.287 mol + 1.05 mol = 1.34 mol
Final volume of the container = sum of the initial volumes = 1.20 L + 14.5 L = 15.7 L
Now we can use the ideal gas law again to calculate the final pressure:
P = nRT/V = (1.34 mol)(0.0821 atm·L/mol·K)(20 + 273 K)/(15.7 L) = 0.229 atm
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An aqueous mixture containing starch (a colloid), Nacl, glucose, and albumin (a colloid) is placed in a dialyzing bag and in distilled water. Which of the following correctly describes the location of the indicated substance after dialysis?
a) starch outside
b) water inside only
c) albumin inside
d) albumin inside and outside
e) starch inside and outside
Answer:
c) albumin inside
Explanation:
Let us see look closer at each answer and see why it is wrong/correct.
a) Starch cannot be outside the bag because the bag isn't permeable to starch. The starch stays inside
b) There cannot be water inside only because it would mean starch and albumin diffused out and they cannot pass through the bag
c) This is correct. The bag isn't permeable to albumin so it stays inside
d) Albumin cannot be outside the bag because the bag isn't permeable to albumin. The albumin stays inside
e) Starch cannot be outside the bag because the bag isn't permeable to starch. The starch stays inside
After dialysis, albumin - a colloid, will remain inside the dialyzing bag because the dialysis membrane is designed to let only small molecules and ions pass through, not large colloidal particles. Starch, another colloid, will remain inside for the same reason. Water will reach equilibrium, moving in and out of the membrane.
Explanation:The question relates to the process of dialysis, which is used to separate different substances in a mixture based upon their size. An aquatic mixture containing both colloids like starch and albumin and small molecules like NaCl (sodium chloride) and glucose is placed in a dialyzing bag to see where the substances will be after the process is complete. A characteristic of dialysis is that the dialyzing membrane allows small molecules and ions to pass through, while retaining larger colloidal particles.
Referring to the given options, the correct descriptions after dialysis are as follows:
Therefore, the accurate statement concerning the location of the substances post-dialysis is
c) albumin inside
A .115 L sample of dry air has a pressure of 1.0895633 atm at 377 K. What is the volume of the sample if the temperature is increased to 483 K when the pressure is kept constant?
Answer:
V= 0.147 L
Explanation:
This is simply the application of combined gas law twice, to find the unknowns.
Combined gas law states that: [tex]PV = nRT[/tex]
P= pressure of air
V= volume of air
n= moles of air
R= Universal gas constant ( 0.08205 L atm mol⁻¹ K⁻¹)
T= Absolute temperature in kelvin.
[tex]n=\frac{P * 0.115}{R * 377}[/tex]
Now, applying the same gas law at 483K and substituting for n
[tex]V = \frac{P * 0.115}{R * 377} * \frac{R* 483}{P}[/tex]
V= [tex]\frac{0.115 * 483}{377}[/tex]
V= 0.147 L
When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3,
a yellow precipitate forms and Ag+ becomes negligibly small. Which
of the following is the correct listing of the ions remaining in solution
in order of increasing concentration?
(A) PO43- < NO3- < Na+
(B) PO43- < Na+ < NO3-
(C) NO3- < PO43- < Na+
(D) Na+ < NO3- < PO43-
(E) Na+ < PO43- < NO3-
Answer:
[tex]PO_4^{-3} < NO_3^- < Na^+[/tex]
Explanation:
Solutions.
a) 100 ml 1 M of Na3PO4
b) 100 ml 1 M of AgNO3
c) Mixture: 200 ml 0.5 M of Na3PO4 and 0.5 M of AgNO3
Reaction:
[tex] 3 Ag^+ + PO_4^{-3} \longrightarrow Ag_3PO_4[/tex]
So if silver ion is consumed almost completely:
[tex][Ag^+]=0 M[/tex]
[tex][PO_4^{-3}]=0.5 M-0.5 M \frac{1 mol PO4}{3 mol Ag}=0.33 M[/tex]
[tex][Na^+]=0.5 M * \frac{3 mol Na}{mol Na_3PO_4}=1.5 M[/tex]
[tex][NO_3^-]=0.5 M[/tex]
In increasing order of concentration:
[tex]PO_4^{-3} < NO_3^- < Na^+[/tex]
The correct order of ions remaining in solution after mixing Na3PO4 with AgNO3, by increasing concentration, is phosphate (PO43-), nitrate (NO3-), and sodium (Na+), which is option (A).
Explanation:When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3, a reaction occurs where silver phosphate (Ag3PO4), a yellow precipitate, forms due to its low solubility and Ag+ ions become almost absent in the solution. The remaining ions in the solution will be Na+, NO3-, and a very small amount of PO43-. Since all of the Ag+ reacts to form the precipitate, we can assume that there are three times as many Na+ ions compared to PO43- ions originally, because each formula unit of Na3PO4 produces three sodium ions. Therefore, this leaves Na+ with the highest concentration in solution. NO3- concentration remains unchanged because it is a spectator ion and does not participate in the reaction.
The correct order of ions in solution by increasing concentration is PO43- < NO3- < Na+, which corresponds to option (A).
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An apparatus consists of a 4.0 dm^3 flask containing nitrogen gas at 25°c and 803 kPa, it is joined by a valve to a 10.0dm^3 flask containing argon gas 25°c and 47.2KPa. The valve is opened and the gases mix , what is the partial pressure of each gas and calculate the total pressure of the gas mixture
Answer:
so total pressure is = 261.92 kPa
partial pressure of Ar gas= 33.50 Kpa
partial pressure of N2 gas = 227 kPa
Explanation:
PV = nRT
n = PV / RT
as we know that 1dm3 = 1 Liter
smaller flask before mixing
ideal gas constant
n = (803 kPa) x (4L) / ((8.3144621 L kPa/K mol) x (25 + 273 K))
n = (3212 kPa L) / ((8.3144621 L kPa/K mol) x (298 K))
n = (3212 kPa L) /2477.7
n= 1.29 mol
In the larger flask:
n = (47.2 kPa) x (10 L) / ((8.3144621 L kPa/K mol) x (25 + 273 K)) =
n = (472 kpaL/ 2477.7
n= 0.19 mol
PV = nRT
P = nRT / V
After mixing:
P = (1.29 mol + 0.19 mol) x (8.3144621 L kPa/K mol) x (25+ 273 K) / (4 L + 10 L)
= (1.48) x (2477.7 / (14 L)
= 261.92 kPa total pressure
so total pressure is = 261.92 kPa
a)
(261 kPa) x (0.19 mol Ar) / (1.29 + 0.19mol) =
= 49.59/1.48 = 33.50 Kpa for Arg
b)
(261 kPa total) - (33.50 kPa Ar) = 227 kPa for N2
partial pressure of Ar gas= 33.50 Kpa
partial pressure of N2 gas = 227 kPa
The charge of B is +3 and the charge of O is -2. What is its formula?
A) B3O2
B) B2O3
C) O2B3
According to Newton�s first law of motion, if there is no net force (unbalanced forces) acting on an object that is moving at a constant 30 mph, the object will:
Continue to move at 30 mph
Eventually come to a stop
Change its inertia
Accelerate
Answer:
Continue to move at 30 mphExplanation:
Newton's first law of motion :
"Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces exerted on it."
From the question we know that the net forces on the object were zero or that the there were no unbalanced forces on it.
Therefore we can assume that the object is moving along a straight line.
And the object was moving at a constant speed of 30 mph.
So it is clear from the Newton's first law that the object will remain in the state of motion as it was earlier.
That is the object will remain in motion at constant speed of 30 mph.
Which of the following is not true for an exothermic reaction?
A. the products have a higher enthalpy than reactants
B. the temperature of the surroundings rise
C. the enthalpy change for the reaction is negative heat flows from the rxn system to the surroundings
Answer:
The false statement regarding an exothermic reaction is: A. the products have a higher enthalpy than reactants
Explanation:
An exothermic reaction is a type of chemical reaction that involves the release of energy from the system to the surroundings. Thus increasing the temperature of the surroundings.
In this reaction, the enthalpy or energy of the reactants is greater than the enthalpy or energy of the products. ([tex]\Delta H_{f} (Products) < \Delta H_{f} (reactants)[/tex])
As the enthalpy change of a reaction: [tex]\Delta H_{r} = \sum \Delta H_{f} (Products) - \sum \Delta H_{f} (reactants)[/tex]
Therefore, the enthalpy change for an exothermic reaction is negative ([tex]\Delta H_{r} < 0[/tex])
The pH of 0.50 M benzoic acid is 2.24. Calculate the change in pH when 2.64 g of C6H5COONa is added to 38 mL of 0.50 M benzoic acid, C6H5COOH. Ignore any changes in volume. The Ka value for C6H5COOH is 6.5 x 10-5.
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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Barium can be analyzed by precipitating it as BaS04 and weighing the precipitate. When a 0.713-g sample of a barium compound was treated with excess H2S04, 0.5331 g of BaSO4 formed. What is the percentage of barium in the compound?
Molecular mass of BaSO4 233.39 g/mol
Molecular mass of Ba = 137.327 g/mol
Answer:
31.37%
Explanation:
For this case, you should consider the following reaction:
Ba⁺²₍aq₎ + H₂SO₄ ₍aq₎ → BaSO₄ ₍s₎ + H₂O
For which you obtain the precipitate of BaSO₄
In order to obtain the mas of Barium on the precipitate, you may use the following formula:
gBa= M₍BaSO₄₎x(M₍Ba₎/M₍BaSO₄₎)
Where:
gBa= mass of Barium
M₍BaSO₄ ₎= mass of BaSO₄ from the precipitate
M₍Ba₎= mass of Barium from the original sample
M₍BaSO₄₎= mass of BaSO₄ from the precipitate
gBa= (0.5331)x(137.327/233.39)= 0.3136 g
Then we ontain the percentage of Barium multiplying by 100:
% Ba on the original sample= 31.36%
Final answer:
The percentage of barium in the compound is calculated by finding the moles of BaSO₄, converting it to moles of Ba, and then finding the mass of Ba before expressing it as a percentage of the initial sample mass. The resulting percentage of barium is approximately 43.99%.
Explanation:
To determine the percentage of barium in the compound, we must calculate the moles of BaSO₄ and then use the molar mass of barium to find the mass of barium in the initial sample. We can calculate the moles of BaSO₄ using the following steps:
First, use the mass of the BaSO₄ precipitate formed (0.5331 g) and the molar mass of BaSO₄ (233.39 g/mol) to find the moles of BaSO₄.Moles of BaSO₄ = mass of precipitate / molar mass of BaSO₄
= 0.5331 g / 233.39 g/mol
= 0.002284 mol (rounded to six decimal places)
Since the molar ratio of Ba to BaSO₄ is 1:1, the moles of Ba will be the same as the moles of BaSO₄.Next, multiply the moles of Ba by the molar mass of Ba to find the mass of Ba in the sample.Mass of Ba = moles of Ba × molar mass of Ba= 0.002284 mol × 137.327 g/mol
= 0.3137 g (rounded to four decimal places)
Finally, we calculate the percentage of barium in the sample using the initial sample mass (0.713 g).
Percentage of barium = (mass of Ba / initial sample mass) × 100%
= (0.3137 g / 0.713 g) × 100%
= 43.99%
Therefore, the percentage of barium in the compound is approximately 43.99%.
On Earth, the white colored paper represents areas that are A) covered with snow and reflect sunlight. B) heavily populated and use the sunlight. Eliminate C) covered with water and reflect sunlight. D) covered with concrete and absorb sunlight.
In an ecosystem model of earth , white colored paper represents areas that are covered with snow and reflect sunlight.
What is an ecosystem?Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.
Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.
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