Answer:
16 moles
Explanation:
16 moles of NH₃ are required to produce 12 moles of NH₄Cl
What is Stoichiometry ?Stoichiometry helps us use the balanced chemical equation to measures quantitative relationships and it is to calculate the amount of products and reactants that are given in a reaction.
What is Balanced Chemical Equation ?The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
First we have to write the balanced chemical equation.
8NH₃ + 3Cl₂ → 6NH₄Cl + N₂
Here 8 moles of NH₃ produce 3 moles of Cl₂ to form 6 moles of NH₄Cl.
So,
12 mole NH₄Cl × [tex]\frac{8\ \text{mole}\ NH_3}{6\ \text{mole}\ NH_4Cl}[/tex]
= 16 moles NH₃
Thus, we can say that 16 moles of NH₃ are required to produce 12 moles of NH₄Cl.
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Water molecules contain what type of bonds
Final answer:
Water molecules consist of polar covalent bonds within the molecule and hydrogen bonds between molecules. Polar covalent bonds are formed due to the oxygen atom sharing electrons with two hydrogen atoms and attracting them more strongly. Hydrogen bonds occur between the positive portions of hydrogen and negative oxygen atoms of adjacent water molecules.
Explanation:
Water molecules contain two types of bonds: polar covalent bonds and hydrogen bonds. Each water molecule is composed of one oxygen atom and two hydrogen atoms. The oxygen atom shares a pair of valence electrons with each hydrogen atom, forming two polar covalent bonds within the molecule. This happens because oxygen has a higher electronegativity than hydrogen, pulling the shared electrons closer and creating partial charges within the molecule. The result is that the oxygen atom acquires a partial negative charge while the hydrogen atoms have partial positive charges.
The hydrogen bonds occur between different water molecules. A hydrogen bond is a dipole interaction where the partially positive hydrogen atom of one water molecule is attracted to the electronegative oxygen atom of a nearby water molecule. While hydrogen bonds are much weaker than covalent bonds, they are strong enough in water to hold adjacent molecules together. The bridging of molecules by hydrogen bonds gives water many of its unique characteristics. Diagrams illustrating hydrogen bonds often represent them with dotted lines to indicate their relative weakness compared to solid lines representing covalent bonds.
Which of the given statements best represent what to do in the event of a spill of concentrated sulfuric acid.
a. First, rinse the affected area with copious amount of water.
b. First, rinse the affected area with copious amounts of sodium hydroxide.
c. Second, treat the area with aqueous sodium bicarbonate solution.
d. Second, treat the area with sodium thiosulfate solution.
Answer: Option A and then Option c
Explanation: Because Concentrated Sulphuric acid is a very strong acid and corrosive, when spilled on the skin, can affect the surface and internal skin. It is necessary to First rinse the affected area with copious amount of water to reduce effect of the acid and Secondly to treat the area with aqueous sodium bicarbonate solution since it will further neutralize and reduce stinging pain from the acid spill by acting as an alkali or base,
Rinsing with a copious amount of water is the first step to dilute concentrated sulfuric acid before neutralizing it with sodium bicarbonate. For a spill of 27.6 mL of 6.25 M H2SO4 solution, one would need 28.96 grams of sodium bicarbonate.
Explanation:In the event of a concentrated sulfuric acid spill, the first step is to rinse the affected area with a copious amount of water to dilute the acid. This is important because adding water to concentrated sulfuric acid can lead to a highly exothermic reaction which can cause the sulfuric acid to splatter and cause further damage or injury. Therefore, careful dilution with water is essential. Once the area is rinsed and the acid is diluted, the next step is to neutralize the acid. This is where sodium bicarbonate (NaHCO3) comes into play.
For the reaction between sodium bicarbonate and sulfuric acid:
2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) + 2CO2(g),
we can calculate the moles of sodium bicarbonate needed. If 27.6 mL of a 6.25 M H2SO4 solution were spilled, this is equivalent to 27.6 mL × 6.25 mmol/mL = 172.5 mmol of H2SO4. Since the stoichiometry of the reaction requires 2 moles of NaHCO3 for every 1 mole of H2SO4, we need 2 × 172.5 mmol = 345 mmol of NaHCO3. To convert moles to grams, we use the molar mass of NaHCO3, which is approximately 84.01 g/mol. Hence, 345 mmol × 84.01 g/mol = 28.96 g of NaHCO3 would be required to neutralize the spill.
The concentration of a mixture can be increased in which of the following ways?
Answer:
C. adding more powder solute
Explanation:
The concentration of a mixture can be increased by removing solvent.
The concentration of a mixture can be increased in several ways. If we consider a distillery wanting to achieve a higher concentration of alcohol, they need to remove solvent, which in this case is water, from their product. This can be achieved through processes such as distillation, which separates alcohol from water due to their different boiling points. Another approach is the addition of a substance that reacts with the water, effectively removing water content from the mixture.
To increase the concentration of a solution in general, one can also add more solute (the substance being dissolved) into the solution or remove solvent (the substance dissolving the solute). It is important not to confuse the process of dilution, which involves adding solvent and decreases solute concentration, with the process of concentrating a solution, which involves removing solvent and increases solute concentration. Therefore, methods such as evaporation, reverse osmosis, or chemical reaction can be employed to increase the concentration of a solute in a solution.
A cook prepare a solution for boiling by adding 12.5g of NaCl to a pot holding 0.750L of
water. At what temperature should the solution in the not boil? *
Final answer:
The boiling point of the student's solution with 12.5g of NaCl in 0.750L of water would be approximately 100.04 °C. Given the slight elevation, salt's effect on boiling temperature is negligible, and the water will practically boil at standard boiling temperature of pure water (100 °C).
Explanation:
The student has prepared a solution by adding 12.5g of NaCl to 0.750L of water and wants to know at what temperature the solution will boil. Generally, pure water boils at 100 °C at standard atmospheric pressure. However, when salt (NaCl) is added to water, the boiling point elevation can occur due to the colligative properties of the solution. The boiling point elevation (ΔTb) depends on the molality (m) of the solution and the ebullioscopic constant (kb) of the solvent, which for water is 0.512 °C/m.
Given this information, we can approximate that the boiling point of water will be slightly elevated, by about 0.04 °C, due to the addition of NaCl. However, the exact boiling point will depend on the concentration of the NaCl solution. For the student's solution, the precise boil elevation calculation would require the formula ΔTb = i * kb * m, where i is the Van't Hoff factor for NaCl (approximately equal to 2, because NaCl dissociates into two ions: Na+ and Cl-).
Answer for the Student's Question:
In this educational context, the increase in boiling temperature is minimal, and the solution in the pot will boil at approximately 100.04 °C, or simply 100 °C when rounded to three significant figures. Adding salt to water used for cooking pasta causes a negligible effect on the boiling temperature and consequently on the cooking time.
g Calculate the [OH-] from the results of your titrations. Explain your calculations2. Calculate the [Ca2+]. Explain your calculations3. Calculate the Ksp for calcium hydroxide. Explain your calculations4. Temperature affects equilibrium in one direction or the other (depending on whether the system is Exo or Endothermic). Discuss5. What does the value of Ksp tell you in terms of equilibrium?6. The theoretical value for Ksp of Ca(OH)2 at 25°C is 9.0 X 10-6, compare and suggest reasons for the difference.
Answer:
Explanation:
*Since the titration is between the strong acid HCl and the strong base Ca(OH)2, the pH at the equivalent point should be 7. On interpolation, we will obtain that 9.50mL and 9.82 mL of HCl is required to completely neutralized the given Ca(OH)2 solution.
*pH at the equivalence point =7
we know that pH + pOH = 14
Hence pOH= 14-7=7
pOH= -log(OH-)
The concentration of OH-= 10-pH= 1X10-7 M
One reason for the low solubility may be the higher reaction temperature, Another reason is the common ion effect.
The titration is between the strong acid HCl and the strong base Ca(OH)2, the pH at the equivalent point should be 7.
Titration:The process of determining the quantity of a substance A by adding measured increments of substance B, the titrant, with which it reacts until exact chemical equivalence is achieved (the equivalence point).
It is possible titrations are affected by temperature. The pH range for indicators can change with temperature. strong base
Since the titration is between the strong acid HCl and the strong base Ca(OH)2, the pH at the equivalent point should be 7.
On interpolation, we will obtain that 9.50mL and 9.82 mL of HCl is required to completely neutralized the given Ca(OH)2 solution.
*pH at the equivalence point =7
we know that pH + pOH = 14
Hence pOH= 14-7=7
pOH= -log(OH-)
The concentration of [tex]OH^-= 10^{-pH}= 1*10^{-7} M[/tex].
One reason for the low solubility may be the higher reaction temperature, Another reason is the common ion effect.
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The following reactant molecules are rearranged to form the product molecules shown. The relevant bond enthalpies are given, in kJ/mol. What is the estimated enthalpy change for this reaction?
ReactionEnergy
A. 678 kJ/mol
B. 198 kJ/mol
C. –265 kJ/mol
D. –632 kJ/mol
Answer:
Correct Answer: B. 198 kJ/mol
Explanation:
In this reaction, four C–H bonds and two O–H bonds are broken, so the total energy absorbed is 4 · 413 + 2 · 463 = 1,652 + 926 = 2,578 kJ/mol. For the products, one C=O bond and three H–H bonds are formed, so the energy released is 1,072 + 3 · 436 = 1,072 + 1,308 = 2,380 kJ/mol. The total enthalpy change is equal to the energy absorbed minus the energy released, so we have 2,578 – 2,380 = 198 kJ/mol. The enthalpy change is positive, so the reaction is endothermic.
Any help would be appreciated. Confused.
Answer:
q(problem 1) = 25,050 joules; q(problem 2) = 4.52 x 10⁶ joules
Explanation:
To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...
=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:
Heat of fusion (ΔHₓ) = 80 cal/gram
Heat of vaporization (ΔHv) = 540 cal/gram
specific heat of ice [c(i)] = 0.50 cal/gram·°C
specific heat of water [c(w)] = 1.00 cal/gram·°C
specific heat of steam [c(s)] = 0.48 cal/gram·°C
Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ... Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.
q(warming ice) = m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal
q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal
q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal
q(evaporation of water) = m·ΔHv = (10g)(540cal/g) = 5400 cal
q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal
Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.
Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).
Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.
Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.
Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).
Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.
Problems containing a temperature change:
NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve. Good luck on your efforts. Doc :-)
2. How are these nutrients used by your body?
Nutrients are used by the body for energy, building materials, and maintaining physiological functions. Carbohydrates and lipids are primarily energy-yielding, while proteins provide amino acids for the growth and repair of tissues.
Nutrients are the substances obtained from food that are vital for growth and maintenance of a healthy body throughout life. Our body uses nutrients for energy, building materials, and sustaining bodily functions. Among these nutrients, carbohydrates and lipids are primarily known as the energy-yielding nutrients. Proteins, while not primarily used for energy, supply essential amino acids that serve as building blocks for the repair and growth of tissues.
Carbohydrates are broken down into glucose, which is used in the metabolic processes to create adenosine triphosphate (ATP), the energy currency of the cell. Lipids are utilized for energy storage, insulation, and cellular structure. Proteins contribute to various functions, including enzyme reactions, transport mechanisms, and cell signaling.
Micronutrients, such as vitamins and minerals, do not provide energy but are crucial in other physiological processes, including metabolism regulation, bone health, and immune system function. Water, despite not providing energy, is essential for nutrient transportation, temperature regulation, and waste excretion.
Overall, a balanced diet with the appropriate intake of macronutrients and micronutrients is vital for maintaining health and supporting the body's various functions.
Henry measures the temperatures at which three liquids change into gases. What property is Henry comparing? A. solubility B. boiling point C. melting point D. thermal conductivity
Explanation:
It is the boiling point because is the point when liquid starts to boil and changes to gases.
A steel container with a volume of 30L is filled with oxygen to a pressure of 9.00 atm at 28.0°C. What is the pressure of the temperature changes to 129.0°C
Answer:
[tex]\large \boxed{\text{12.0 atm}}[/tex]
Explanation:
The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:
At constant volume, the pressure exerted by a gas is directly proportional to its temperature.
\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}
Data:
p₁ = 9.00 atm; T₁ = 28.0 °C
p₂ = ?; T₂ = 129.0 °C
Calculations:
1. Convert the temperatures to kelvins
T₁ = (28.0 + 273.15) K = 301.15
T₂ = (129.0 + 273.15) K = 402.15
2. Calculate the new pressure
[tex]\begin{array}{rcl}\dfrac{9.00}{301.15} & = & \dfrac{p_{2}}{402.15}\\\\0.02989 & = & \dfrac{p_{2}}{402.15}\\\\0.02989 \times 402.15 &=&p_{2}\\p_{2} & = & \textbf{12.0 atm}\end{array}\\\text{The new pressure is $\large \boxed{\textbf{12.0 atm}}$}[/tex]
Which of these statements are true for a neutral, aqueous solution at 25 °C? pH = 7.00 [ H + ] = [ OH − ] pOH = 7.00 Which of these statements are true for a neutral, aqueous solution regardless of temperature? [ H + ] = [ OH − ] pH = 7.00 pOH = 7.00
The statement which are true for a neutral, aqueous solution at 25 °C are:
pH = 7.00 [ H + ] = [ OH − ] pOH = 7.00The statement which is/are true for a neutral, aqueous solution regardless of temperature is;
[ H + ] = [ OH − ]We must know that at standard temperature, 25°C, a neutral, aqueous solution has it's pH = pOH = 7. Additionally, the hydrogen ion concentration, [H+] is equal to its hydroxide ion concentration, [OH-]
pH is slightly affected by change in temperature as it decreases with increase in temperature. In a neutral aqueous solution, there are always the same concentration of hydrogen ions,[H+] and hydroxide ions, [OH-] and hence, the solution is still neutral (even if its pH changes).Ultimately, the pH of a solution decreases with increase in temperature and vice versa.
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an 11 pound bowling ball has a circumference of 27 inches. calculate the density of the bowling ball and determine whether it will float or sink if placed in water. ignore the holes in the bowling ball
Answer:
The density is 0.0036023 pounds/inches³, which is 99.71 kg/ m³. The ball will float.
Explanation:
Density is mass/ volume.
The mass is 11 pounds.
Calculate the volume of a sphere by using [tex]\frac{4}{3}[/tex]πr².
V =[tex]\frac{4}{3}[/tex]π(27)²
V = 3053.63 inches³
Density = mass/volume
Density = 11 pounds/ 3053.63 inches³
Density = 0.0036023 pounds/inches³
Convert this to kg/ m³, which is 99.71 kg/ m³. Water has a density of 1000 kg/m³. Since the ball has a lower density than water, it will float
The bowling ball has a density of approximately 0.92 g/cm³ which is less than the density of water (1 g/cm³). Therefore, the bowling ball would float in water.
Explanation:To calculate the density of the bowling ball, we first need to determine its volume. The volume of a sphere can be calculated through the formula V = 4/3 * π * r³. We can find the radius 'r' from the given circumference of the ball which is 27 inches using the formula Circumference = 2 * π * r. Hence, r = Circumference/( 2 * π) = 27 inches/( 2 * π)≈ 4.29 inches.
Substituting this radius value into the volume formula, we get V ≈ 4/3 * π * (4.29)³ ≈ 331 cubic inches. The weight of the bowling ball in pounds needs to be converted into grams for density calculation in g/cm³. As 1 pound equals to 453.59g, so the weight of the bowling ball is approximately 11 * 453.59 g ≈ 4989.49 g. Then, convert volume from cubic inches to cubic cm as 1 cubic inch = 16.387 cm³, leading to a volume around 331 * 16.387 ≈ 5423 cubic cm.
The density is then calculated as density = mass/volume = 4989.49 g / 5423 cm³ ≈ 0.92 g/cm³. Since the density of water is 1 g/cm³, and the density of the bowling ball is less than this, the bowling ball would float in water.
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What is restriction enzymes and what is it’s purpose?
Answer: Endonuclease enzymes used in molecular biology that cut DNA at specified points.
Explanation:
Enzymes are specific protein types which bind to a substrate within a reaction, to increase the rate of reaction within the solution- they speed up the rate of reaction.
Restriction enzymes are bacteria-derived enzymes; these make cuts on deoxyribonucleic acid molecules or DNA. These are also called restriction endonucleases. They are utilized in molecular biology for DNA cloning and sequencing and cut DNA into smaller pieces called fragments.
Restriction enzymes make directed cuts on DNA molecules. They precisely target sites on DNA to produce mostly identical or homogenous, discrete fragments of equal sizes, producing blunt or sticky ends. In order to do this, they recognize sequences of nucleotides that correspond with a complementary sequence on the endonuclease called restriction sites.
There are several kinds that may require cofactors (chemical or metallic compounds that aid in enzyme activity) :
Type I: cleave far away from the recognition site; require ATP and SAMe S-Adenosyl-L-Methionine Type II: cleave near to the site; require Magnesium Type III: cleave near to the site; require ATP which is not hydrolysed but SAMe S-Adenosyl-L-Methionine is optionalType IV: cleavage targeted to DNA that have undergone post transcriptional modification through certain types of methylation (addition of a methyl group)Final answer:
Restriction enzymes are proteins that cut DNA at specific sequences for defensive purposes in bacteria but are also utilized in genetic engineering to facilitate DNA manipulation. They can create sticky ends for DNA recombination, and their specificity allows for applications like DNA fingerprinting and restriction mapping.
Explanation:
What are Restriction Enzymes?
Restriction enzymes, also known as restriction endonucleases, are proteins that can cut DNA at specific sequences known as recognition sites. Each enzyme has a specificity for a particular sequence, which is often palindromic, meaning the sequence reads the same in the 5' to 3' direction on both strands of the DNA double helix. Used extensively in recombinant DNA technology, restriction enzymes are crucial in genetic engineering practices.
The purpose of restriction enzymes is to defend bacteria from invading viruses, such as bacteriophages, by cleaving the foreign DNA. However, their ability to cut DNA precisely is harnessed in molecular biology for various purposes, including gene cloning, creating DNA fingerprints for forensic applications, conducting genetic research, and more.
Some restriction enzymes cut DNA to produce sticky ends, which are single-stranded overhangs that can easily bind to complementary sequences, facilitating the recombination of DNA fragments. Others create blunt ends without overhangs. To finalize the recombination process, another enzyme called DNA ligase is used to glue the DNA fragments together.
Over 800 restriction endonucleases have been discovered, and each recognizes different specific sequences. Their precise action leaves a characteristic pattern of DNA fragments, which in turn can be used for a variety of analyses, such as restriction mapping.
Question 16 A chemistry graduate student is given of a methylamine solution. Methylamine is a weak base with . What mass of should the student dissolve in the solution to turn it into a buffer with pH ? You may assume that the volume of the solution doesn't change when the is dissolved in it. Be sure your answer has a unit symbol, and round it to significant digits.
Answer:
90 g CH₃NH₃Cl
Explanation:
It appears your question lacks the values required to solve this problem. However, an online search tells me these are the values. Be aware if your values are different your answer will also be different, but the methodology remains the same.
" A chemistry graduate student is given 450 mL of a 1.70 M methylamine solution. Methylamine is a weak base with Kb=4.4x10⁻⁴. What mass of CH₃NH₃Cl should the student dissolve in the methylamine solution to turn it into a buffer with pH = 10.40 ? You may assume that the volume of the solution doesn't change when the is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits. "
We'll use the Henderson-Hasselbach equation:
pH = pKa + log[tex]\frac{[Methylamine]}{[Salt]}[/tex]So first we use Kb to calculate Ka and then pKa:
Ka = Kw/Kb ⇒ Ka = 1x10⁻¹⁴/4.4x10⁻⁴ = 2.27x10⁻¹¹pKa = -log(Ka) = 10.64Now we can calculate the concentration of the salt, CH₃NH₃Cl:
pH = pKa + log[tex]\frac{[Methylamine]}{[Salt]}[/tex]10.40 = 10.64 + log[tex]\frac{1.70}{[Salt]}[/tex]-0.24 = log[tex]\frac{1.70}{[Salt]}[/tex][tex]10^{-0.24}[/tex]=[tex]\frac{1.70}{[Salt]}[/tex][Salt] = 2.95 MNow we use the final volume to calculate the moles of CH₃NH₃Cl and finally its mass, using its molecular weight:
450 mL ⇒ 450/1000 = 0.450 L2.95 M * 0.450 L = 1.3275 mol CH₃NH₃Cl1.3275 mol CH₃NH₃Cl * 67.45 g/mol = 89.54 g CH₃NH₃ClWhich when rounded to 2 significant digits becomes 90 g CH₃NH₃Cl
the labels of bottles that contain barium hydroxide, hydrochloric acid and sodium carbonate had fallen off. When the contents of the bottle A were mixed with B's Contents, a white precipitate formed, A with C yielded a gas and B with C gave heat. Identify A,B,C and write molecular and net ionic equations for the chemical changes observed.
Final answer:
Bottle A contains barium hydroxide, Bottle B contains hydrochloric acid, and Bottle C contains sodium carbonate. When combined, they form a white precipitate in one case, release a gas in another case, and give off heat in the last case.
Explanation:
The bottles contain:
Bottle A: Barium hydroxide (Ba(OH)2)Bottle B: Hydrochloric acid (HCl)Bottle C: Sodium carbonate (Na2CO3)The chemical changes observed are:
A + B: Formation of a white precipitate (BaSO4) - Molecular equation: Ba(OH)2(aq) + HCl(aq) → BaCl2(aq) + H2O(l) + CO2(g)A + C: Release of a gas (CO2) - Molecular equation: Ba(OH)2(aq) + Na2CO3(aq) → 2NaOH(aq) + BaCO3(s)B + C: Release of heat - Molecular equation: HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g)what molecule is responsible for chemical digestion.
Answer:
Amylase.
Explanation:
The process of digestion begin to start in mouth when food mix with saliva. An enzyme is released which is called Amylase help in digestion of carbohydrates.
For the decomposition of phosphorous pentachloride to phosphorous trichloride and chlorine at 400K the KC is 1.1x10-2. Given that 1.0g of phosphorous pentachloride is added to a 250mL reaction flask, find the percent decomposition after the system has reached equilibrium. PCl_5(g) PCl_3(g) Cl_2(g) K_C
Answer:
[tex]\% Decomposition=47.4\%[/tex]
Explanation:
Hello,
In this case, for the given decomposition of phosphorous pentachloride:
[tex]PCl_5(g)\rightleftharpoons PCl_3(g)+ Cl_2(g)[/tex]
As the equilibrium constant is [tex]1.1x10^{-2}[/tex] and the initial concentration of phosphorous pentachloride is:
[tex][PCl_5]_0=\frac{1.0gPCl_5*\frac{1molPCl_5}{208.24gPCl_5} }{250mL*\frac{1L}{1000mL} } =0.019M[/tex]
Hence, by writing the law of mass action equation:
[tex]Kc=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]
We must introduce the change [tex]x[/tex] occurring due to the reaction extent and the concentrations at equilibrium (ICE table methodology):
[tex]Kc=\frac{(x)(x)}{[PCl_5]_0-x}=\frac{x^2}{0.019-x}=1.1x10^{-2}[/tex]
Thus, solving for [tex]x[/tex] we obtain:
[tex]x=0.01M[/tex]
In such a way, the equilibrium concentration of phosphorous pentachloride results:
[tex][PCl_5]_{eq}=[PCl_5]_0-x=0.019M-0.01M\\[/tex]
[tex][PCl_5]_{eq}=0.009M[/tex]
Finally, the percent decomposition is computed by:
[tex]\% Decomposition=\frac{[PCl_5]_0}{[PCl_5]_{eq}}*100\%=\frac{0.009M}{0.019M} *100\%\\\\\% Decomposition=47.4\%[/tex]
Best regards.
It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis consists of: 70% propane C3H8; 5% butane C4H10; and 25% propene C3H6 The higher heating values of the fuels are 50.38 MJ/kg for propane, 49.56 MJ/kg for butane, and 48.95 MJ/kg for propene. a) Work out the overall combustion reaction for stoichiometric combustion of 1 mole of LPG with air, and determine the stoichiometric F/A and A/F ratios. b) What are the higher and lower heating values per unit mass of LPG?
Answer:
a)
The overall balanced combustion reaction is written as :
[tex]0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2[/tex]
[tex](F/A)_{stoichiometric} = 0.0424[/tex]
[tex](A/F)_{stoichiometric} = 23.562[/tex]
b)
the higher heating values [tex](HHV)_f[/tex] per unit mass of LPG = 49.9876 MJ/kg
the lower heating values [tex](LHV)_f[/tex] per unit mass of LPG = 46.4933 MJ/kg
Explanation:
a)
The stoichiometric equation can be expressed as :
[tex]0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2 \ + \ bH_2O \ + \ cN_2[/tex]
Now, equating the coefficient of carbon; we have:
(0.7×3)+(0.05×4)+(0.25×3) = a
a = 3.05
Also, Equating the coefficient of hydrogen : we have:
(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b
2b = 7.6
b = 3.8
Equating the coefficient of oxygen
2x = 2a + b
[tex]x = \frac{2a+b}{2} \\ \\ x = \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95[/tex]
Equating the coefficient of Nitrogen
[tex]c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612[/tex]
Therefore, The overall balanced combustion reaction can now be written as :
[tex]0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2 \ + \ 3.8H_2O \ + \ 18.612N_2[/tex]
Now; To determine the stoichiometric F/A and A/F ratios; we have:
[tex](F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\ (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424[/tex]
[tex](A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\ (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562[/tex]
b)
What are the higher and lower heating values per unit mass of LPG?
Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.
Molecular mass of the fuel [tex]M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})[/tex]
= 30.8 + 2.9 + 10.5
= 44.2 kg/mol
Mass fraction of the fuel components can now be calculated as :
[tex]m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8} = 0.7 \\ \\ \\ m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06 \\ \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6} = 0.24[/tex]
Finally; calculating the higher heating values [tex](HHV)_f[/tex] per unit mass of LPG; we have:
[tex](HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg[/tex]
calculating the lower heating values [tex](LHV)_f[/tex] per unit mass of LPG; we have:
[tex](LHV)_f = (HHV)_f - \delta H_w \\ \\ (LHV)_f = (HHV)_f - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f = 49.9876 \ MJ/kg - [\frac{3.8*18}{44.2}*2.258 \ MJ/kg] \\ \\ (LHV)_f = 46.4933 \ M/kg[/tex]
The higher heating value (HHV) per unit mass of LPG is approximately 49.91 MJ/kg, and the lower heating value (LHV) per unit mass of LPG is approximately 44.92 MJ/kg.
To solve this problem, we will first determine the overall combustion reaction for stoichiometric combustion of 1 mole of LPG with air. Then, we will calculate the stoichiometric fuel-to-air (F/A) and air-to-fuel (A/F) ratios. Finally, we will find the higher and lower heating values per unit mass of LPG.
Step 1: Determine the overall combustion reaction
First, we need to calculate the moles of each component in the LPG mixture based on the given volume percentages:
- Propane (C3H8): 70% of the mixture
- Butane (C4H10): 5% of the mixture
- Propene (C3H6): 25% of the mixture
Since the mixture is on a volume basis, we can directly use the molar percentages for the calculations. We will assume 1 mole of LPG for simplicity, which means:
- Moles of propane: 0.70 moles
- Moles of butane: 0.05 moles
- Moles of propene: 0.25 moles
Next, we write the balanced combustion reactions for each component:
Propane: C3H8 + 5O2 → 3CO2 + 4H2O
Butane: C4H10 + 6.5O2 → 4CO2 + 5H2O
Propene: C3H6 + 4.5O2 → 3CO2 + 3H2O
To find the overall reaction, we sum up the reactions for each component, taking into account their respective moles in the mixture:
Overall reaction = (0.70 * Propane reaction) + (0.05 * Butane reaction) + (0.25 * Propene reaction)
This gives us:
Overall reaction = (0.70 * (C3H8 + 5O2 → 3CO2 + 4H2O)) + (0.05 * (C4H10 + 6.5O2 → 4CO2 + 5H2O)) + (0.25 * (C3H6 + 4.5O2 → 3CO2 + 3H2O))
Combining the terms, we get:
Overall reaction = (0.70C3H8 + 0.05C4H10 + 0.25C3H6) + (3.5O2 + 0.325O2 + 1.125O2) → (2.1CO2 + 0.2CO2 + 0.75CO2) + (2.8H2O + 0.25H2O + 0.75H2O)
Simplifying the coefficients by multiplying by the least common multiple to get whole numbers, we have:
Overall reaction = (7C3H8 + C4H10 + 2.5C3H6) + (35O2 + 3.25O2 + 11.25O2) → (21CO2 + 2CO2 + 7.5CO2) + (28H2O + 2.5H2O + 7.5H2O)
Further simplifying, we get:
Overall reaction = (7C3H8 + C4H10 + 2.5C3H6) + 50O2 → (29CO2 + 2.5CO2) + (35H2O + 2.5H2O)
Finally, the overall balanced reaction is:
Overall reaction = (7C3H8 + C4H10 + 2.5C3H6) + 50O2 → 31.5CO2 + 37.5H2O
Step 2: Calculate the stoichiometric F/A and A/F ratios
The stoichiometric F/A ratio is the ratio of the mass of fuel to the mass of air required for complete combustion. The stoichiometric A/F ratio is the inverse of the F/A ratio.
First, we need to calculate the molar mass of the LPG mixture:
Molar mass of LPG = (0.70 * Molar mass of C3H8) + (0.05 * Molar mass of C4H10) + (0.25 * Molar mass of C3H6)
Molar mass of C3H8 = 3 * 12.01 (for C) + 8 * 1.008 (for H) = 44.092 g/mol Molar mass of C4H10 = 4 * 12.01 + 10 * 1.008 = 58.124 g/mol Molar mass of C3H6 = 3 * 12.01 + 6 * 1.008 = 42.08 g/mol Molar mass of LPG = (0.70 * 44.092) + (0.05 * 58.124) + (0.25 * 42.08) ≈ 47.49 g/molNext, we calculate the mass of air required for the combustion of 1 mole of LPG. The stoichiometric combustion of hydrocarbons with air can be represented as
CxHy + (x + y/4)O2 → xCO2 + (y/2)H2OFor 1 mole of LPG, the mass of oxygen required is:
Mass of O2 = (7 + 1 + 2.5) * 32 (molar mass of O2) = 10.5 * 32 = 336 gThe mass of air required is the mass of O2 multiplied by the ratio of the molar mass of air to the molar mass of O2 (approximately 28.97/32):
Mass of air = 336 * (28.97/32) ≈ 309.7 gNow, we can calculate the F/A and A/F ratios:
F/A = Mass of LPG / Mass of air = 47.49 / 309.7 ≈ 0.1533 A/F = Mass of air / Mass of LPG = 309.7 / 47.49 ≈ 6.52Step 3: Calculate the higher and lower heating values per unit mass of LPG
The higher heating value (HHV) of the LPG mixture is calculated by taking the weighted average of the HHVs of its components:
HHV of LPG = (0.70 * HHV of C3H8) + (0.05 * HHV of C4H10) + (0.25 * HHV of C3H6) HHV of C3H8 = 50.38 MJ/kg HHV of C4H10 = 49.56 MJ/kg HHV of C3H6 = 48.95 MJ/kg HHV of LPG = (0.70 * 50.38) + (0.05 * 49.56) + (0.25 * 48.95) ≈ 49.91 MJ/kgThe lower heating value (LHV) is typically about 10% less than the HHV due to the energy used to vaporize the water produced during combustion. We can estimate the LHV as follows:
LHV of LPG ≈ HHV of LPG - 0.10 * HHV of LPG LHV of LPG ≈ 49.91 - 0.10 * 49.91 ≈ 44.92 MJ/kgFinal
The overall combustion reaction for stoichiometric combustion of 1 mole of LPG with air is:
[tex]\[ (7C3H8 + C4H10 + 2.5C3H6) + 50O2 → 31.5CO2 + 37.5H2O \][/tex]The stoichiometric F/A and A/F ratios are approximately 0.1533 and 6.52, respectively.
The higher heating value (HHV) per unit mass of LPG is approximately 49.91 MJ/kg, and the lower heating value (LHV) per unit mass of LPG is approximately 44.92 MJ/kg.
If the solubility of KCl in 100 mL of H₂O is 34 g at 20 °C and 43 g at 50 °C, label each of the following solutions as unsaturated, saturated, or supersaturated. If more solid is added then can dissolve in the solvent, assume undissolved solid remains at the bottom of the flask.
a) adding 30 g to 100 mL of H₂O at 20 °C
b) adding 65 g to 100 mL of H₂O at 50 °C
c) adding 42 g to 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution with no precipitate
Answer:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Explanation:
A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.
An unsaturated solution contains less solute than it has the capacity to dissolve.
A supersaturated solution, contains more solute than is present in a saturated solution. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at 20 °C is 34 g, and at 50 °C is 43 g we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated
b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution with no precipitate ⇒ unsaturated (if it were saturated it would have had precipitate)
The labels for the solution is as follows:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Different types of solutions:A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than it has the capacity to dissolve. A supersaturated solution, contains more solute than is present in a saturated solution. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at 20 °C is 34 g, and at 50 °C is 43 g we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C is an unsaturated solution.
b) 65 g in 100 mL of H₂O at 50 °C is an supersaturated solution.
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution with no precipitate is an unsaturated solution (if it were saturated it would have had precipitate)
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Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a certain temperature, a chemist finds that a 5.4 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition: Calculate the value of the equilibrium constant K_c for this reaction. Round your answer to 2 significant digits.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The equilibrium constant is [tex]K_c= 2.8*10^{-4}[/tex]
Explanation:
From the question we are told that
The chemical reaction equation is
[tex]Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}[/tex]
The voume of the misture is [tex]V_m = 5.4L[/tex]
The molar mass of [tex]Fe_{2} O_{3}_{(s)}[/tex] is a constant with value of [tex]M_{Fe_{2} O_{3}_{(s)} } = 160g/mol[/tex]
The molar mass of [tex]H_{2}_{(g)}[/tex] is a constant with value of [tex]H_2 = 2g/mol[/tex]
The molar mass of [tex]H_{2}O[/tex] is a constant with value of [tex]H_2O = 18g/mol[/tex]
Generally the number of moles is mathematically given as
[tex]No \ of \ moles \ = \frac{mass}{molar\ mass}[/tex]
For [tex]Fe_{2} O_{3}_{(s)}[/tex]
[tex]No \ of\ moles = \frac{3.54}{160}[/tex]
[tex]= 0.022125 \ mols[/tex]
For [tex]H_{2}[/tex]
[tex]No \ of\ moles = \frac{3.63}{2}[/tex]
[tex]= 1.815 \ mols[/tex]
For [tex]H_{2}O[/tex]
[tex]No \ of\ moles = \frac{2.13}{18}[/tex]
[tex]= 0.12 \ mols[/tex]
Generally the concentration of a compound is mathematicallyrepresented as
[tex]Concentration = \frac{No \ of \ moles }{Volume }[/tex]
For [tex]Fe_{2} O_{3}_{(s)}[/tex]
[tex]Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}[/tex]
[tex]= 4.10*10^{-3}M[/tex]
For [tex]H_{2}[/tex]
[tex]Concentration[H_2] = \frac{1.815}{5.4}[/tex]
[tex]= 0.336M[/tex]
For [tex]H_{2}O[/tex]
[tex]Concentration [H_2O] = \frac{0.12}{5.4}[/tex]
[tex]= 0.022M[/tex]
The equilibrium constant is mathematically represented as
[tex]K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}[/tex]
Considering [tex]H_2O \ for \ product[/tex]
And [tex]H_2 \ for \ reactant[/tex]
At equilibrium the
[tex]K_c = \frac{0.022}{0.336}[/tex]
[tex]K_c= 2.8*10^{-4}[/tex]
A solution is prepared by dissolving 0.23 mol of acetic acid and 0.27 mol of sodium acetate in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of HCl to this buffer solution causes the pH to drop slightly. The pH does not decrease drastically because the HCl reacts with the __________ present in the buffer solution.
Answer:
Acetate ion
Explanation:
CH₃COOH ↔ CH₃COO⁻ + H⁺
HCl -> H⁺ + Cl⁻
The H+ ions from added HCl reacts with acetate ions and more amount of acetic acid will be formed due to common ion effect.
The pH of the buffer solution does not decrease drastically because the added HCl reacts with the sodium acetate present in the buffer, maintaining the buffer's capacity to neutralize added protons.
Sodium acetate acts as a conjugate base in the buffer system, reacting with added HCl to form acetic acid and water. This reaction prevents the pH from decreasing drastically because the added protons from HCl are consumed in the formation of acetic acid, thereby maintaining the buffer capacity.
A buffer system works to maintain a relatively constant pH upon the addition of acids or bases, demonstrating its capacity to minimize changes in the hydrogen ion concentration of a solution.
a cell contains 180 m^3 of gas at 7400 Pa and a machine. The machine is turned on remotely and expands the box. During this process, the machine also gives off 260 kJ of heat to the gas, and the internal energy is determined to be -69kJ. What is the final volume of the cell? Assume pressure stays constant.
Please Show All Work
Answer:
[tex]224.5 m^3[/tex]
Explanation:
By using the first law of thermodynamics, we can find the work done by the gas:
[tex]\Delta U=Q-W[/tex]
where in this problem:
[tex]\Delta U=-69 kJ[/tex] is the change in internal energy of the gas
[tex]Q=+260 kJ[/tex] is the heat absorbed by the gas
W is the work done by the gas (positive if done by the gas, negative otherwise)
Therefore, solving for W,
[tex]W=Q-\Delta U=+260-(-69)=+329 kJ = +3.29\cdot 10^5 J[/tex]
So, the gas has done positive work: it means it is expanding.
Then we can rewrite the work done by the gas as
[tex]W=p(V_f-V_i)[/tex]
where:
[tex]p=7400 Pa[/tex] is the pressure of the gas
[tex]V_i=180 m^3[/tex] is the initial volume of the gas
[tex]V_f[/tex] is the final volume
And solving for Vf, we find
[tex]V_f=V_i+\frac{W}{p}=180+\frac{3.29\cdot 10^5}{7400}=224.5 m^3[/tex]
Final answer:
To determine the final volume of the gas, the first law of thermodynamics was used, considering the process as isobaric due to constant pressure. The final volume was calculated to be approximately 45045.9 m³ after applying the formula and accounting for the work done and heat supplied.
Explanation:
To find the final volume of the gas, we can use the first law of thermodynamics, which is given by ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. Since the question states the pressure remains constant (isobaric process), the work done by the gas can be found by W = pΔV, where p is the pressure and ΔV is the change in volume.
From the given data, we have:
Initial internal energy change (ΔU) = -69 kJ (since the value is negative, this indicates that the gas loses energy)Heat added to the system (Q) = 260 kJInitial volume (V₁) = 180 m³Pressure (p) = 7400 PaUsing ΔU = Q - W and substituting W with pΔV, we get:
-69 kJ = 260 kJ - (7400 Pa × ΔV)
Convert kJ to J by multiplying by 1000 (since 1 kJ = 1000 J):
-69000 J = 260000 J - (7400 Pa × ΔV)
We can then solve for ΔV:
ΔV = (260000 J + 69000 J) / 7400 Pa
ΔV = 44865.9 m³ (rounded to one decimal place)
To find the final volume (V₂), we add the change in volume to the initial volume:
V₂ = V₁ + ΔV
V₂ = 180 m³ + 44865.9 m³
V₂ = 45045.9 m³
So, the final volume of the gas after the process is approximately 45045.9 m³.
We separate U-235 from U-238 by fluorinating a sample of uranium to form UF6 (which is a gas) and then taking advantage of the different rates of effusion and diffusion for compounds containing the two isotopes. Calculate the ratio of effusion rates for 238UF6 and 235UF6. The atomic mass of U-235 is 235.054 amu and that of U-238 is 238.051 amu.
Answer:
The ratio of effusion rates for [tex]^{238}UF_6[/tex] and [tex]^{235}UF_6[/tex] is 0.995734.
Explanation:
Effusion rate of the [tex]^{235}UF_6 [/tex]gas = [tex]R[/tex]
Effusion rate of the [tex]^{238}UF_6 [/tex]gas = [tex]r[/tex]
Molar mass of [tex]^{235}UF_6=235.054+6\times 19.00 amu=349.054 amu[/tex]
Molar mass of [tex]^{238}UF_6=238.051+6\times 19.00 amu=352.051 amu[/tex]
Graham's law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:
[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
[tex]\frac{r}{R}=\sqrt{\frac{349.054 amu}{352.051 amu}}=0.995734[/tex]
The ratio of effusion rates for [tex]^{238}UF_6[/tex] and [tex]^{235}UF_6[/tex] is 0.995734.
Final answer:
The ratio of effusion rates for 238UF6 and 235UF6 can be calculated using Graham's law, which states that the ratio of effusion rates is equal to the square root of the ratio of the molar masses of the gases.
Explanation:
The ratio of effusion rates for 238UF6 and 235UF6 can be calculated using Graham's law, which states that the ratio of effusion rates is equal to the square root of the ratio of the molar masses of the gases.
So, the ratio of effusion rates for 238UF6 and 235UF6 is:
(352.04 g/mol) / (349.03 g/mol) = 1.008
This means that the effusion rate of 238UF6 is approximately 1.008 times higher than the effusion rate of 235UF6.
The procedure for testing your unknown solution in this week's lab is identical to the procedure which you conducted in Week 1. The only difference is, of course, your Unknown Solution may or may not contain all of the ions which you tested for in Week 1. With that being said, please consider the following scenario: You enter the lab and obtain an Unknown Solution from the Stockroom. You begin testing the solution through the steps outlined in the flowchart on p. 9 of the Exp 22 document. You first add HCl, and centrifuge your mixture. You observe the formation of a white precipitate in the bottom of the test tube. After pouring off the supernatant liquid, you add hot water to the white precipitate. Upon addition of the hot water, you still have some white precipitate in the bottom of the test tube. You add ammonia, NH3, to the test tube and observe the formation of a gray-black precipitate. Which of the following is the best conclusion to draw at this point
A. The unknown solution definitely has Ag+ present.
B. The unknown solution could have Agt present, or Hg22+ present, or BOTH.
C. The unknown solution definitely has Hg22+ present.
D.The Unknown Solution definitely has Pb2+ present.
Answer:
The correct answer is option C. The unknown solution definitely has Hg22+ present.
Explanation:
In the analysis of group 1 metal cation, the unknown solution is treated with sufficient quantity of 6 M HCl solution and if group 1 metal cations are present then white precipitate of Agcl, PbCl2 or Hg2Cl2 is formed. The precipitate of PbCl2 is soluble in hot water but the other two remains insoluble after treating with hot water. Precipitate of AgCl disappears upon treatment of NH3 solution but Hg2Cl2 becomes black in the reaction with NH3. The black Colour appears due to the formation of metallic Hg.
Balanced chemical equation of the reation is -
Hg2Cl2 + 2NH3 ---------> HgNH2Cl (white ppt.) + Hg (black ppt.) + NH4Cl
Therefore, from the given information the conclusion which can be drawn is that the unknown solution definitely has Hg22+ present.
Answer:
C- the unknown solution definitely has Hg22+ present.
Explanation:,
• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL volumetric flask and diluting to 25.00 mL prior to removing the 10.00 mL sample used above. What was the molar concentration of phosphoric acid in the original stock solution?
Answer:
a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) 0.0035 mole
c) 0.166 M
Explanation:
Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.
The equation of the reaction is expressed as:
[tex]H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O[/tex]
1 mole 3 mole
The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows
[tex]H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O[/tex]
10 ml 17.50 ml
(x) M 0.200 M
Molarity = [tex]\frac{0.2*17.5}{1000}[/tex]
= 0.0035 mole
c) What was the molar concentration of phosphoric acid in the original stock solution?
By stoichiometry, converting moles of NaOH to H₃PO₄; we have
= [tex]0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}[/tex]
= 0.00166 mole of H₃PO₄
Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:
Molar Concentration = [tex]\frac{mole \ \ of \ soulte }{ Volume \ of \ solution }[/tex]
Molar Concentration = [tex]\frac{0.00166 \ mole \ of \ H_3PO_4 }{10}*1000[/tex]
Molar Concentration = 0.166 M
∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M
Calculate the pH of a weak acid which dissociates as AH ⇌ A- + H+ ( like CH3COOH ⇌ CH3COO- + H+ or HNO2 ⇌ NO2- + H+ ) knowing that the initial concentrations are [AH]0 0.97 M and the acid dissociation constant, Ka , is 7.23e-7
Answer:
The pH is 3.08
Explanation:
Step 1:
The equation for the reaction.
AH ⇌ A- + H+
Step 2:
Data obtained from the question.
Initial concentration of AH, [AH] = 0.97 M
Dissociation constant, Ka = 7.23x10^-7
Step 3:
Determination of the concentration of A-, [A-], the concentration of H+, [H+] and the concentration of AH, [AH] after the reaction. This is illustrated below:
Before the reaction:
[AH] = 0.97 M
[A-] = 0
[H+] = 0
During the reaction:
[AH] = -y
[A-] = +y
[H+] = +y
After the reaction
[AH] = 0.97 - y
[A-] = y
[H+] = y
Step 4:
Obtaining the value of y. This is illustrated below:
Ka = [A-] [H+] / [AH]
Ka = 7.23x10^-7
[AH] = 0.97
[A-] = y
[H+] = y
Ka = [A-] [H+] / [AH]
7.23x10^-7 = y x y / 0.97
Cross multiply to express in linear form
7.23x10^-7 x 0.97 = y^2
7.0131x10^-7 = y^2
Take the square root of both side
y = √(7.0131x10^-7)
y = 8.37x10^-4
Therefore [H+] = y = 8.37x10^-4 M
Step 5:
Determination of the pH.
pH = - log [H+]
[H+] = 8.37x10^-4 M
pH = - log [H+]
pH = - log 8.37x10^-4
pH = 3.08
what structures would normally be found in a plant cell but not in an animal cell
Answer:
Cell walls. Also plant cells are rectangular shaped while animal cells are usually circular.
Explanation:
A chemist dissolves of pure sodium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (The temperature of the solution is .) Be sure your answer has the correct number of significant digits.
The pH of the solution can be calculated using the equation -log[H+]. Given that 40 grams of NaOH is dissolved in enough water to make up 60 grams of solution, the concentration of NaOH is 67 M. Taking the negative logarithm of 67 M gives the pOH, and the pH can be calculated using the equation pH = 14 - pOH.
Explanation:The pH of a solution can be calculated using the equation -log[H+]. Since sodium hydroxide (NaOH) is a strong base, it completely dissociates in water to yield Na+ and OH- ions. The concentration of OH- ions can be obtained from the molar amount of NaOH and the volume of the solution. Given that 40 grams of NaOH is dissolved in enough water to make up 60 grams of solution, the concentration of NaOH is calculated as follows:
Concentration of NaOH (M) = (40 g NaOH) / (40 g/mol NaOH) / (0.06 L solution) = 67 M
Since one NaOH molecule dissociates to one OH- ion, the concentration of OH- ions is also 67 M. Taking the negative logarithm of 67 M gives the pOH:
pOH = -log(67) ≈ -1.826
Finally, the pH can be calculated using the equation:
pH = 14 - pOH ≈ 14 - (-1.826) ≈ 15.826
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To calculate the pH of a sodium hydroxide solution, first determine the molarity, then calculate the pOH, and finally find the pH by subtracting the pOH from 14. The pH of the solution is approximately 13.88.
1. Determine the concentration of sodium hydroxide (NaOH) in the solution.
Let's assume 15.0 grams of NaOH is dissolved in 500.0 mL (0.500 L) of solution.
Calculate the moles of NaOH:The molar mass of NaOH is approximately 40.00 g/mol.Number of moles of NaOH = mass / molar mass = 15.0 g / 40.00 g/mol = 0.375 molCalculate the molarity (concentration) of the solution:Molarity (M) = moles of solute / liters of solution = 0.375 mol / 0.500 L = 0.750 MCalculate the pOH of the NaOH solution:Since NaOH is a strong base, it completely dissociates in water:NaOH → Na⁺ + OH⁻[OH⁻] = 0.750 MpOH = −log[OH⁻] = −log(0.750) ≈ 0.125Relate pOH to pH:pH + pOH = 14pH = 14 − pOH = 14 − 0.125 = 13.875Thus, the pH of the solution is approximately 13.88, when rounded to the correct number of significant digits.
Z.O. returns to the pediatric intensive care unit after surgery. He is arousable but cannot answer questions. His pupils are equal and reactive to light.
He has a head dressing covering the entire scalp with
small amount of serosanguineous drainage. His IV is intact and infusing as ordered through a new central
venous line. His breath sounds are equal and clear, and spO2
is 98% on room air. you get him settled in his
bed and leave the room.
You check the postoperative orders, which are listed in the chart. Which orders are
appropriate, and which would you question? State your rationale.
Answer:
Explanation:
Attached is the solution
Labels of many food products have expiration dates, at which point they are typically removed from the supermarket shelves. A particular natural yogurt degrades (reacts) with a half-life of 45 days. The manufacturer of the yogurt wants unsold product pulled from the shelves when it degrades to no more than 80% (0.8) of its original quality. Assume the degradation process is first order. What should be the "best if used before" date on the container with respect to the date the yogurt was packaged?
Answer:
Explanation:
Since it first order, we use order rate equation
In ( [tex]\frac{A1}{A0}[/tex]) = -kt where A1 is the final quality = 0.8 (80%), A0 is the initial quality = 1 ( 100%)
also, t half life = [tex]\frac{In2}{k}[/tex] where k is rate constant
k = [tex]\frac{In 2}{45 days}[/tex] = 0.0154
In ( [tex]\frac{0.8}{1}[/tex]) = - 0.0154 t
-0.223 / -0.0154 = t
t = 14.49 approx 14.5 days from the date the yogurt was packaged