Answer:
A) Degree of Freedom (DF) = 2 and this means that 2 variables are required to determine the state of the system
B) The vessel doesn't constitute an explosion hazard.
Explanation:
A) Degree of Freedom(DF)=2 + C - π
Where;
π = 2 (liquid and gas phase equilibrium)
C = 2 (air and MEK)
Thus, DF = 2 + 2 - 2 = 2
DF of 2 means that 2 variables are required to determine the state of the system
B) Using Antoine Equation;
Log(Pmek) = 6.97421 - (1209.6/(55 + 216)) = 2.51
Log(Pmek) = 2.51
Thus,Pmek = 10^(2.51) = 323.59mm of Hg or approximately 324mm of Hg.
So, fraction of mek in vapour phase = 324/1200 = 0.27 or 27%
Now this fraction is more than 11.5% in the question.
Therefore, the vessel contains no explosion hazard.
The number of degrees of freedom for the MEK and air system in the vessel at equilibrium is 2, meaning two variables can be varied independently without changing the number of phases. Without knowing the concentration of MEK in the vapor phase, we cannot determine whether the mixture is explosively hazardous.
Explanation:(a) The Gibbs phase rule is given by F = C - P + 2, where F is the number of degrees of freedom, C is the number of components and P is the number of phases. Here, we have two components (MEK and air), and two phases (liquid and vapor) existing in the vessel. So, the number of degrees of freedom F = 2 - 2 + 2 = 2. This means that two independent variables (such as temperature and pressure) can be varied independently without changing the number of phases.
(b) The mixture of MEK vapor and air in the vessel is potentially explosive if it contains between 1.8 mole% MEK and 11.5 mole% MEK. However, without knowing the exact concentration of MEK in the vapor phase, it is impossible to determine whether the mixture is explosively hazardous or not.
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During Project 3: Week 1 the goal was to synthesize an artificial kidney stone. If you were producing barium phosphate tribasic, what would be the correct stoichiometric coefficients for the reaction depicted below? Ba(NO3)2 (aq) + Na3PO4 (aq) → Ba3(PO4)2 (s) + NaNO3 (aq)
Answer: the correct stoichiometric coefficients will be
3:2= 1:6
Explanation:
3Ba(NO3)2 (aq) + 2Na3PO4 (aq) → Ba3(PO4)2 (s) + 6NaNO3 (aq)
From the equation of reaction,
3 moles of Ba(NO3)2 (aq) will require 2 moles of Na3PO4 , to give 1 mole of Ba3(PO4)2 (s) and 6moles of NaNO3 (aq)
Now use the density formula to calculate the density of each object. Round your answers to the nearest one-tenth of a gram (one decimal place). a. Bowling Ball: 8.5 g/in3 Feathers: 0.1 g/in3 b. Bowling Ball: 8.5 g/cm3 Feathers: 0.1 g/cm3 c. Bowling Ball: 0.1 lb/in3 Feathers: 8.5 lb/in3 d. Bowling Ball: 0.1 g/cm3 Feathers: 8.5 g/cm3
Question:
Imagine a six-pound bowling ball. That's generally the lightest you'll find in a bowling alley. Also imagine six pounds of feathers. Your average feathered bedroom pillow weighs about 2 pounds, so imagine three pillows in a stack. You now have six pounds of bowling ball and six pounds of feathers. If you were to take each and throw them into a swimming pool, you would expect the bowling ball to sink and the feathers to float, but why?First, let's figure out their volume.A bowling ball has a radius of 4.25 in, which converts to 10.8 cm. Using the formula to find the volume of a sphere:
Note; The result is 322 cm³(= error by calculating in 4.25 cm instead).
Next, the standard size for a bedroom pillow is 26 in x 20 in x 4 in. When we convert to centimeters, our stacked pillows are 66 cm x 51 cm x 10 cm. The volume is then 33,660 cm³. We already know the mass of each object (six pounds), but remember that mass in this formula is measured in grams. You'll first need to convert six pounds to grams.
One pound converts to 453.6 g. Round your calculation to the nearest full gram and write it down.Now use the density formula to calculate the density of each object. Round your answers to the nearest one-tenth of a gram (one decimal place).
Bowling Ball: 0.1 g/cm³Feathers: 8.5 g/cm³ Correct!
Bowling Ball: 8.5 g/cm³ Feathers: 0.1 g/cm³
Bowling Ball: 0.1 lb/in³ Feathers: 8.5 lb/in³
Bowling Ball: 8.5 g/in³ Feathers: 0.1 g/in³
Answer:
The correct answer to the question is
b. Bowling Ball: 8.5 g/cm³ Feathers: 0.1 g/cm³.
Explanation:
To solve the question, we note that both the sphere and the Bowling Ball have a mass of 6 lb which is equal to 2721.554 g
The radius of a bowling ball is 4.25 cm = 1.67 in which gives its volume as
322 cm³
The mass of the bowling ball 6 lb while
The volume of a feathered pillow = 26 in × 20 in × 4 in = 2080 in³ = 58899040.911 cm³
The density of the Bowling Ball = mass/volume = 2721.554 g/322 cm³
= 8.452 g/cm³
The density of the feathers = mass/volume
= 2721.554 g/34085 cm³ = 0.0798 g/cm³ which to one decimal place
= 0.1 g/cm³
Therefore the density of the Bowling Ball to the nearest one-tenth
= 8.5 g/cm³ and
The density of the feathers to the nearest one-tenth = = 0.1 g/cm³
The rate constant for this second‑order reaction is 0.380 M − 1 ⋅ s − 1 0.380 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶products How long, in seconds, would it take for the concentration of A A to decrease from 0.860 M 0.860 M to 0.230 M?
Answer: 8.38 seconds
Explanation:
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{a}=kt+\frac{1}{a_0}[/tex]
[tex]a_0[/tex] = initial concentartion = 0.860 M
a= concentration left after time t = 0.230 M
k = rate constant =[tex]0.380M^{-1}s^{-1}[/tex]
[tex]\frac{1}{0.860}=0.380\times t+\frac{1}{0.230 }[/tex]
[tex]t=8.38s[/tex]
Thus it will take 8.38 seconds for the concentration of A to decrease from 0.860 M to 0.230 M .
Final answer:
To find out how long it would take for the concentration of A to decrease from 0.860 M to 0.230 M in a second-order reaction, we can use the integrated rate law.
Explanation:
The reaction in question is second order and the rate constant is 0.380 M⁻¹⋅s⁻¹ at 300 ∘C. To find out how long it would take for the concentration of A to decrease from 0.860 M to 0.230 M, we can use the integrated rate law for a second-order reaction:
t = 1 / (k * [A])
Substituting the given values:
t = 1 / (0.380 M⁻¹⋅s⁻¹ * 0.860 M)
t ≈ 2.80 s
Write the overall molecular equation for the reaction of hydroiodic acid ( HI ) and potassium hydroxide. Include physical states. Enter the formula for water as H 2 O .
The molecular equation for the reaction between hydroiodic acid (HI) and potassium hydroxide (KOH) is:
HI(aq) + KOH(aq) ⇒ KI(aq) + H₂O(l).
The molecular equation represents the chemical reaction between hydroiodic acid (HI) and potassium hydroxide (KOH). In the aqueous phase (aq), hydroiodic acid dissociates into hydrogen ions (H+) and iodide ions (I-), while potassium hydroxide dissociates into potassium ions (K+) and hydroxide ions (OH-).
During the reaction, the hydrogen ions from hydroiodic acid react with the hydroxide ions from potassium hydroxide, forming water (H₂O) in liquid (l) state. Additionally, the remaining potassium ions from potassium hydroxide and iodide ions from hydroiodic acid combine to form potassium iodide (KI) in aqueous state. This balanced equation illustrates the rearrangement of ions and atoms during the chemical reaction,
HI(aq) + KOH(aq) ⇒ KI(aq) + H₂O(l).
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The reaction between hydroiodic acid and potassium hydroxide forms potassium iodide and water. The balanced molecular equation is HI(aq) + KOH(s) → KI(aq) + H₂O(l). All the physical states are indicated for each compound.
In this reaction, hydroiodic acid (HI) is added to solid potassium hydroxide (KOH).
This is an acid-base reaction, which typically results in the formation of a salt and water. The balanced molecular equation for the reaction is given below:HI(aq) + KOH(s) → KI(aq) + H₂O(l)In the equation, HI is in aqueous state (aq), KOH is in solid state (s), KI is in aqueous state (aq), and water is in liquid state (l).The reaction can be summarized as follows:
Reactants: Hydroiodic acid (HI) and Potassium hydroxide (KOH)Products: Potassium iodide (KI) and Water (H₂O)Correct question is: Write the overall molecular equation for the reaction of hydroiodic acid ( HI ) and potassium hydroxide. Include physical states. Enter the formula for water as H₂O .
The solubility product constant for MX2 is 7.2 x 10-8. How many grams of MX2 (108.75 g/mol) will dissolve in 276 ml of water at 25°C. M is the metal and X is the anion. Enter as a number to 4 decimal places.
Answer: The mass of [tex]MX_2[/tex] that will dissolve is 0.0786 grams
Explanation:
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The chemical equation for the ionization of [tex]MX_2[/tex] follows:
[tex]MX_2(aq.)\rightleftharpoons M^{2+}(aq.)+2X^-(aq.)[/tex]
s 2s
The expression of [tex]K_{sp}[/tex] for above equation follows:
[tex]K_{sp}=s\times (2s)^2[/tex]
We are given:
[tex]K_{sp}=7.2\times 10^{-8}[/tex]
Putting values in above expression, we get:
[tex]7.2\times 10^{-8}=s\times (2s)^2\\\\s=2.62\times 10^{-3}M[/tex]
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
Molar mass of [tex]MX_2[/tex] = 108.75 g/mol
Molarity of solution = [tex]2.62\times 10^{-3}mol/L[/tex]
Volume of solution = 276 mL
Putting values in above equation, we get:
[tex]2.62\times 10^{-3}mol/L=\frac{\text{Mass of }MX_2\times 1000}{108.75/mol\times 276}\\\\\text{Mass of }MX_2=\frac{2.62\times 10^{-3}\times 108.75\times 276}{1000}=0.0786g[/tex]
Hence, the mass of [tex]MX_2[/tex] that will dissolve is 0.0786 grams
The mass of MX₂ that will dissolve is 0.0786 grams.
What is Solubility Product?Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The chemical equation for the ionization of follows:
[tex]MX_2--- > M^{2+}+2X^-[/tex]
s 2s
Calculation for "s" :
Given: [tex]K_{sp}= 7.2*10^{-8}[/tex]
[tex]K_{sp}=s*(2s)^2\\\\7.2*10^{-8}=s*(2s)^2\\\\s=2.62*10^{-3}M[/tex]
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
M = n/ V
Molar mass of = 108.75 g/mol
Molarity of solution = [tex]2.62*10^{-3}mol/L[/tex]
Volume of solution = 276 mL
On substituting the values:
[tex]2.62*10^{-3} mol/L=\frac{\text{ Mass of } MX_2 * 1000}{108.75g/mol*276} \\\\\text{ Mass of } MX_2=\frac{2.62*10^{-3} mol/L*108.75g/mol*276}{1000} \\\\\text{ Mass of } MX_2=0.0786 g[/tex]
Hence, the mass of MX₂ that will dissolve is 0.0786 grams.
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Gasoline (which can be considered to be octane, C8H18) burns in oxygen to produce carbon dioxide and water. What volume (L) of oxygen at STP is necessary to react with 1.0 gal of gasoline? (The density of gasoline is 0.81 g/mL. 1 gal = 3.78 L)
To find the volume of oxygen at STP necessary to react with 1.0 gal of gasoline we perform a series of conversions: from volume of gasoline to mass using density from mass of gasoline to moles using molar mass of octane from moles of octane to moles of oxygen using the stoichiometric ratio from the balanced chemical equation and from moles of oxygen to volume using the molar volume of gas at STP. This series of conversions yields the answer.
Explanation:The question you asked pertains to how gasoline, or octane (C8H18), reacts with oxygen to produce carbon dioxide and water, and how much oxygen volume at STP is necessary for this reaction with 1.0 gal of gasoline. To solve this task, we first convert gasoline volume to grams using its given density. Knowing that 1 gal of gasoline equals 3.78 L and the density of gasoline is 0.81 g/mL, we multiply these values to get the total mass of gasoline. Then, we consider the balanced chemical equation of the combustion of octane: 2C8H18 + 25O2 -> 16CO2 + 18H2O. This equation tells us that 25 moles of oxygen react with 2 moles of octane. Using octane's molar mass, we convert the mass of gasoline to moles. Once we have the moles of octane, we use the stoichiometric ratio from the balanced equation to find the moles of oxygen necessary for the reaction. Finally using the molar volume of a gas at STP which is approximately 22.4 L, we convert the moles of oxygen to volume. This lengthy chemical calculation processes yields the volume of oxygen necessary to react with 1 gal of gasoline.
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Place the following substances in order of increasing boiling point. Ne Cl2 O2
a. O2 < Cl2 < Ne
b. Cl2 < Ne < O2
c. Cl2 < O2 < Ne
d. Ne < O2 < Cl2
e. Ne < Cl2 < O2
Answer: = D
Explanation:
The atomic mass increases from Ne to O2 to Cl2 hence the boiling point also increases, therefore
Ne < O2 < Cl2
Hydroxyapatite, Ca 10 ( PO 4 ) 6 ( OH ) 2 Ca10(PO4)6(OH)2 , has a solubility constant of Ksp = 2.34 × 10 − 59 2.34×10−59 , and dissociates according to Ca 10 ( PO 4 ) 6 ( OH ) 2 ( s ) − ⇀ ↽ − 10 Ca 2 + ( aq ) + 6 PO 3 − 4 ( aq ) + 2 OH − ( aq ) Ca10(PO4)6(OH)2(s)↽−−⇀10Ca2+(aq)+6PO43−(aq)+2OH−(aq) Solid hydroxyapatite is dissolved in water to form a saturated solution. What is the concentration of Ca 2 + Ca2+ in this solution if [ OH − ] [OH−] is fixed at 2.50 × 10 − 6 M 2.50×10−6 M ?
Answer:
1.315x10⁻³M = [Ca²⁺]
Explanation:
Based in the reaction:
Ca₁₀(PO₄)₆(OH)₂(s) ⇄ 10Ca²⁺(aq) + 6PO₄³⁻(aq) + 2OH⁻(aq)
Solubility product, ksp, is defined as:
ksp = [Ca²⁺]¹⁰ [PO₄³⁻]⁶ [OH⁻]²
From 1 mole of hydroxyapatite are produced 10 moles of Ca²⁺ and 6 moles of PO₄³⁻. That means moles of PO₄³⁻ are:
6/10 Ca²⁺ = PO₄³⁻
Replacing in ksp formula:
ksp = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [OH⁻]²
As [OH⁻] is 2.50x10⁻⁶M and ksp is 2.34x10⁻⁵⁹:
2.34x10⁻⁵⁹ = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [2.50x10⁻⁶]²
3.744x10⁻⁴⁸ = 0.046656[Ca²⁺]¹⁶
1.315x10⁻³M = [Ca²⁺]
I hope it helps!
Net Ionic Equations for mixing Strong Acids with Strong Bases Consider a reaction between hydrochloric acid and potassium hydroxide.
a) In an aqueous solution of hydrochloric acid, check all of the major species found in solution (ignore the trace hydronium ions and hydroxide ions that would come from the autoionization of water). HCI (1) CIo (aq) CI (aq) OH (ag)H20 () H3o (aq)
b) In an aqueous solution of potassium hydroxide, check all of the major species found in solution (ignore the trace hydronium ons and hydroxide ions that would come from the autoionization of water). H20 () K+ (aq) OH (aq) Hyo (aq) KOH (5)
c) When the acid and base react together, they will neutralize each other to form water and a salt. Give the chemical formula for the salt formed. chemPad Help Greek acid and potassium hydroxide. Write it out in this order. Remember that, by convention, a net ionic equation has a single reaction arrow
d) When you cancel out the spectator ions, what is the net ionic equation that remains for the reaction between hydrochloric Help chemPad Oreek ▼
Answer:
sorry but I am just answering the questions because I need points
Explanation:
thank you
e sure to answer all parts. Calculate the pH of the following aqueous solutions at 25°C: (a) 9.5 × 10−8 M NaOH (b) 6.3 × 10−2 M LiOH (c) 6.3 × 10−2 M Ba(OH)2
Final answer:
To calculate the pH of the given aqueous solutions at 25°C, we can use the formula pH = -log[H3O+]. For NaOH, LiOH, and Ba(OH)2, the concentration of hydroxide ions is equal to the concentration of the base, which allows us to find the pH.
Explanation:
To calculate the pH of an aqueous solution, we need to determine the concentration of the hydronium ions (H3O+). Since NaOH, LiOH, and Ba(OH)2 are strong bases that ionize completely, we can assume that the concentration of the hydroxide ions (OH-) is equal to the concentration of the base. To find the pH, we can use the formula:
pH = -log[H3O+]
For example:
For 9.5 × 10^(-8) M NaOH, the concentration of hydroxide ions is 9.5 × 10^(-8) M. Taking the negative logarithm, the pH is approximately 7.02.
For 6.3 × 10^(-2) M LiOH, the concentration of hydroxide ions is 6.3 × 10^(-2) M. Taking the negative logarithm, the pH is approximately 11.20.
For 6.3 × 10^(-2) M Ba(OH)2, the concentration of hydroxide ions is twice the concentration of the base, so it is 2 * 6.3 × 10^(-2) M = 1.26 × 10^(-1) M. Taking the negative logarithm, the pH is approximately 1.90.
Consider the second-order reaction:
2HI(g)→H2(g)+I2(g)
Use the simulation to find the initial concentration [HI]0 and the rate constant k for the reaction. What will be the concentration of HI after t = 4.53×1010 s ([HI]t) for a reaction starting under the condition in the simulation?
Given from simulation:
Rate Law: k[HI]^2
k= 6.4 x 10^-9 l/(mol x s) at 500K
Initial Rate= 1.6 x 10^-7 mol/(l x s)
Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after [tex]4.53\times 10^{10} s[/tex] is 0.00345 mol/L.
Explanation:
[tex]2HI(g)\rightarrow H_2(g)+I_2(g) [/tex]
Rate Law: [tex]k[HI]^2 [/tex]
Rate constant of the reaction = k = [tex]6.4\times 10^{-9} L/mol s[/tex]
Order of the reaction = 2
Initial rate of reaction = [tex]R=1.6\times 10^{-7} Mol/L s[/tex]
Initial concentration of HI =[tex][A_o][/tex]
[tex]1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2[/tex]
[tex][A_o]=5 mol/L[/tex]
Final concentration of HI after t = [A]
t = [tex]4.53\times 10^{10} s[/tex]
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{[A]}=kt+\frac{1}{[A_o]}[/tex]
[tex]\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}[/tex]
[tex][A]=0.00345 mol/L[/tex]
The concentration of HI after [tex]4.53\times 10^{10} s[/tex] is 0.00345 mol/L.
Use the pull-down menus to specify the state of each reactant and product. Ag2S 2Ag + S2- Is silver sulfide considered soluble or not soluble ? ... ... A. Soluble ... B. Not soluble Based upon this, the equilibrium constant for this reaction will be: ... ... A. Greater than 1 ... B. Less than 1 Submit AnswerRetry Entire Group
Explanation:
For the given reaction equation, we will write the state of each specie as follows.
[tex]Ag_{2}S(s) \rightarrow 2Ag^{+}(aq) + S^{2-}(aq)[/tex]
Since, silver sulfide ([tex]Ag_{2}S[/tex]) will remain in solid state. Therefore, it acts as a precipitate, that is, insoluble solid. Hence, it is insoluble.
And, expression for the equilibrium constant of this reaction is as follows.
[tex]K_{eq} = \frac{[Ag^{+}]^{2}[S^{2-}]}{[AgS]^{2}}[/tex]
For solids, it is considered to be equal to 1. Hence, the equilibrium constant expression will be as follows.
[tex]K_{eq} = [Ag^{+}]^{2}[S^{2-}][/tex]
Therefore, we can conclude that its equilibrium constant for this reaction will be greater than 1.
Consider the reaction: N2(g) + 2O2(g)2NO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.90 moles of N2(g) react at standard conditions. S°system = J/K Submit Answer
Explanation:
It is known that standard entropies for [tex]N_{2}(g)[/tex] is 191.6 J/mol K, [tex]O_{2}(g)[/tex] = 205 J/mol K, and [tex]NO_{2}(g)[/tex] is 239.7 J/mol K at 298 K.
Therefore, we will calculate the value of [tex]\Delta S^{o}_{reaction}[/tex] from standard absolute entropies as follows.
[tex]\Delta S^{o}_{reaction} = \sum \Delta S^{o}_{products} - \sum \Delta S^{o}_{reactants}[/tex]
= 2 mole of [tex]NO_{2}(g)[/tex] - 1 mole of [tex]N_{2}(g)[/tex] + 2 mole of [tex]O_{2}(g)[/tex]
= [tex]2 \times 239.7 J/mol K - 1 \times 191.6 J/mol K + 2 \times 205 J/mol K[/tex]
= -122.2 J/K
The entropy change for 1.90 moles of [tex]N_{2}(g)[/tex] reacting is as follows.
[tex]\Delta S^{o}_{system}[/tex] = 1.90 moles of [tex]N_{2}(g) \times 122.2 J/K/ 1 \text{mol of} N_{2}(g)[/tex]
= 232.18 J/K
Thus, we can conclude that the entropy change for the given system is 232.18 J/K.
A sample of impure NaHCO3 with an initial mass of 0.654 g yielded a solid residue (consisting of Na2CO3 and other solids) with a final mass of 0.456 g. Determine the mass percent of NaHCO3 in the sample.
Answer:
81.7 %
Explanation:
Equation of the decomposition of NaHCO₃
2 NaHCO₃ → Na₂CO₃ + CO₂ + H₂O
2 mole of NaHCO₃ yielded 1 mole of CO₂ and 1 mole of H₂O
molar mass of CO₂ = 44 g
molar mass of H₂O = 18 g
1 mole of CO₂ : 1 mole H₂O = ( mass of CO₂ / molar mass of CO₂) : ( mass of H₂O / molar mass of H₂O
also mass of CO₂ + mass of H₂O = ( 0.654 - 0.456 ) g = 0.198 g
1 = ( mass of CO₂/ 44g) : ( mass of H₂O / 18g)
44 mass of H₂O = 18 mass of CO₂
1 mass of CO₂ = 44 / 18 mass of H₂O
substitute into equation 1
mass of CO₂ + mass of H₂O = 0.198 g
2.44 mass of H₂O + mass of H₂O = 0.198 g
3.44 mass of H₂O = 0.198 g
mass of H₂O = 0.198 g / 3.44 = 0.0576 g
mass of CO₂ = 0.198 g - 0.0576 g = 0.140 g
2 mole of NaHCO₃ yielded 1 mole of CO₂
168 g of NaHCO₃ yielded 44 g of CO₂
unknown mass of NaHCO₃ yielded 0.140 g CO₂
unknown mass of NaHCO₃ = 168 g × 0.140 g / 44 g = 0.535 g
mass percent of NaHCO₃ = 0.535 g / 0.654 g = 81.7 %
Answer:
Explanation:
Mass of impure NaHCO3 = 0.654g
Mass of residue= 0.456g
Mass loss of NaHCO3= 0.654-0.456= 0.198g
Balanced reaction equation:
2NaHCO3(s)-------> Na2CO3(s) + H2O(g)+CO2(g)
Note CO2 and water vapour produced in the decomposition combines to give H2CO3
84g of NaHCO3 yields 62g of H2CO3
Xg of NaHCO3 yields 0.198g of H2CO3
Therefore X= 84 × 0.198/ 62
=0.268g
Mass%= 0.268/0.654 × 100
=41% NaHCO3
A rectangular block floats in pure water with 0.5 in, above the surface and 1.5 in below the surface. When placed in an aqueous solution, the block of material floats with 1 in. below the surface. Estimate the specific gravities of the block and the solution. (Suggestion: Call the horizontal cross-sectional area of the block A. A should cancel in your calculations.)
Explanation:
We will calculate the specific gravity of the block as follows.
Specific gravity of block = [tex]\frac{\text{block vol below}}{\text{total block vol } \times \text{specific gravity of water}}[/tex]
= [tex]\frac{1.5}{(1.5 + 0.5) \times 1}[/tex]
= 0.75
And, the specific gravity of the solution is as follows.
Specific gravity of solution = [tex]\frac{\text{total block vol}}{\text{block vol below} \times \text{specific gravity of block}}[/tex]
= [tex]\frac{2}{1 \times 0.75}[/tex]
= 1.5
As the given block is rectangular, and its volume is directly proportional to its height and A is not needed in these calculations.
Also, we can note here that the term specific gravity is no longer approved and has been replaced by the term relative density.
The specific gravity of the block is 0.75 and the specific gravity of the solution is also 0.75. This is calculated using the principles of fluid dynamics and buoyancy, specifically Archimedes' principle of equal weight displacement.
Explanation:When analysing fluid dynamics and buoyancy, the density of an object and the fluid it's submerged in is key. According to Archimedes' principle, the weight of the block equals the weight of the water displaced when it is fully submerged. Therefore, mass of the block is the total volume of the block (2 in) times density of the block, and mass of the water displaced is (1.5 in) times the density of the water. Thus, density of the block would be (0.75/1) times density of the water, making the specific gravity of the block 0.75. In the aqueous solution, the block floats with 1 in below the surface, meaning specific gravity in relation to the solution equals 1 (since 1 in/1 in = 1), therefore, the specific gravity of the aqueous solution equals 0.75, the specific gravity of the block.
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A pair of students determining the molarity of their unknown HCl solution calculates the concentration to be 0.0961 M on their first trial and 0.104 M on their second trial. Do they need to run a third trial
Answer:
Yes.
Explanation:
It should be noted that the meaning of molarity is the ratio of moles of solute per liter of solution.
It should be understood that when determining or finding the molarity of an unknown compound ,the process should be performed or carried out at least 3 times. This is done to remove any form of doubt.
The first calculated value for the concentration of the compound will be regarded as rough value, while the second and the third will be regarded as the first and second values respectively.
In this case, the third value for the concentration of HCl will be calculated to for confirmation of other value, that is to be finally sure of its concentration.
Answer:
Yes.
Explanation:
For a typical experiment, you should plan to repeat it at least three times (more is better).
The value gotten in the first trial is not close to the value gotten from the second trial.
The solution to these problems is to do repeated trials. Repeated trials are when you do a measurement multiple times - at least three, commonly five, but the more the better. When you measure something once, the chance that the number you get is accurate is much lower. But if you measure it several times, you can take an average of those numbers and get a result that is much closer to the truth.
Repeating a trial gives a RELIABLE result.
Mercury and oxygen react to form mercury(II) oxide, like this: 2 Hg(l)+02(g)--HgO(s) At a certain temperature, a chemist finds that a 6.9 L reaction vessel containing a mixture of mercury, oxygen, and mercury(II) oxide at equilibrium has the following composition: compound amount Hg 16.9 g O 10.9 g HgO 23.8 g Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.
Answer:
Kc = 20
Explanation:
We have the equilibrium:
2 Hg (l) + O₂ ( g) ⇄ HgO (s)
Kc = 1/ [O₂]
The key here is to remember that pure solids and liquids do not enter into the calculation for the equilibrium expression, and Hg is a pure liquid and HgO is a solid
So what we need to do to solve this question is to calculate the concentration of oxygen at equilibrium. We are given its mass, and the volume so we are equipped to calculate the concentration of oxygen as follows:
[O₂] = # moles O₂ / V
# moles O₂ = mass / molar mass = 10.9 g / 32g/mol = 0.34 mol
[O₂] = 0.34 mol / 6.9 L = 0.049 M
⇒ Kc = 1 / 0.049 = 20 ( rounded to 2 significant figures )
The spectator ions in the reaction between aqueous perchloric acid and aqueous barium hydroxide are ________. OH- and ClO4- H , OH- , ClO4-, and Ba2 H and OH- H and Ba2 ClO4- and Ba2
Answer:
ClO4-, and Ba2+.
Explanation:
A spectator ion is an ion that exists as a reactant and a product in a chemical equation.
Equation of the reaction.
2HClO4 + Ba(OH)2 --> Ba(ClO4)2 + 2H2O
ClO4- + OH- --> 2ClO4- + H2O
The spectator ions in the reaction between aqueous perchloric acid and aqueous barium hydroxide are ClO₄⁻ and Ba⁺². These ions do not participate in the reaction and remain unchanged in the solution after the reaction.
The spectator ions in the reaction between aqueous perchloric acid and aqueous barium hydroxide are ions that do not participate in the chemical reaction and do not change during the course of the reaction. Aqueous solutions of acids and bases typically dissociate into their constituent ions in water. For perchloric acid (HClO₄), it dissociates into H⁺ and ClO⁴⁻ ions, while barium hydroxide (Ba(OH)₂) dissociates into Ba²⁺ and OH⁻ ions.
In a neutralization reaction, the H+ ions from the acid combine with the OH⁻ ions from the base to form water (H₂O), meaning they are not spectator ions. Therefore, the spectator ions would be the ions that remain unchanged, which are ClO⁴⁻ and Ba²⁺. The complete ionic equation for this reaction would show all the ions separated, but the net ionic equation would exclude these spectators, focusing only on the ions that participate directly in forming the product, water.
A student used water as the solvent and encountered some problems. Comment on the effect, if any, each of the following situations could have had on the experimental results...a)The unknown, a white powder, failed to dissolve.b)The student returned to the laboratory instructor for a different solid unknown. This unknown dissolved, but bubbles were seen escaping from the solution almost immediately after the addition of the solid.c)As the student was setting up the apparatus to measure the freezing point of the unknown solution, the thermometer assembly rolled off the lab bench, and the thermometer broke. The student observed a new thermometer and performed the experiment as instructed.
Answer:
a) If the white powder didn't dissolve completely, that could have a great effect on the experiment. The concentration of the solution cant get to the point where it needs to get for the rest of the experiment which can skew results.
b) If a solution forms bubbles immediately after an addition of a solid, that could simply mean that gas is formed in that reaction. That has no negative effect on an experiment since that's what is supposed to happen during the reaction.
c) Using a different measuring device can really effect a students calculations later on during the experiment. Different devices are calibrated differently, which can skew the results or calculations later on. Its best to stay consistent with what you are measuring with during an experiment.
Explanation:
If the unknown does not dissolve, it affects the results. Bubbles during dissolution indicate a chemical reaction. Replacing the thermometer does not greatly affect the results.
Explanation:a) If the unknown white powder failed to dissolve in water, it means that it is insoluble in water. This could affect the experimental results as the solubility of the unknown substance is an important factor in determining its properties.
b) The bubbling observed when the different solid unknown was added to water indicates a chemical reaction. This could affect the experimental results as the reaction may lead to the formation of a new compound, changing the composition of the solution.
c) The breaking of the thermometer and the replacement with a new one should not significantly affect the experimental results. As long as the new thermometer is calibrated and accurate, it can still be used to measure the freezing point of the unknown solution.
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Check the box next to each molecule on the right that has the shape of the model molecule on the left: model molecules (check all that apply X 5 ? | O CH20 CNH, You can drag the slider to rotate the model molecule. + 1 O Brf 4 CH2Cl2 Note for advanced students: the length of bonds and size of atoms in the model is not necessarily realistic. The model is only meant to show you the general geometry and 3D shape of the molecule.
Answer:
[tex]NH^{+} _{4}[/tex] ammonium ion.
Explanation:
Check the box next to each molecule on the right that has the shape of the model molecule on the left: model molecules (check all that apply X 5 ? | O CH20 CNH, You can drag the slider to rotate the model molecule. + 1 O Brf 4 CH2Cl2 Note for advanced students: the length of bonds and size of atoms in the model is not necessarily realistic. The model is only meant to show you the general geometry and 3D shape of the molecule.
when you look at the diagram from the source page
one can conclude that the diagram is tetrahedral and the angle between the molecules is 109.5 deg
There is one central atom bonded to four atoms in a tetrahedral molecule .it has no lone electron pairs.m
NH4+ is the answer
Other molecules that are tetrahedral in shape are methane ion and phosphate ion.
Answer:
CH3O- , BrF4- and NH4+ have tetrahedral geometry on the basis of their electron domain geometry..
Explanation:
The object on the picture as shown on the fig below describes a compound with a total of 4 pair electron domain with it's electron domain typically described as tetrahedral.
The task is to sort out which of those in the options fort into the category.
Although NH3 and NH4+ ion both have the SP3 hybridization their electron pair geometry differs. In the NH3 molecule one lone pair and three bond pairs are present. While Distortion is caused by repulsion between lone pair and bond pair the geometry of NH3 causes it to become pyramidal in NH4+ irrespective of possessing the sp3 hybridization.
Its resulting trigonal pyramidal geometry is thus described as tetrahedral. It consequently has 3 bonding domains and 1 nonbinding domain.
CH3O- is also tetrahedral with an idealized bond angle of 109.5°.
BrF4- It has 4 bond pair present hence the tetrahedral geometry. The presence of two lone pair makes it square planar described sometimes as AE2X4. It has 6 electron regions.
C2Cl4 has a linear geometry it has one triple bond and two single bonds this giving hints that its coordinate and steric number is 2 and its bond angle is 180°.
So, CH3O- , BrF4- and NH4+ have tetrahedral geometry
When individuals are looking for jobs but are unable to find work, they are said to be
.
Answer:
Unemployed
Explanation:
(of a person) without a paid job but available to work.
"I was unemployed for three years" - ----- Example
Similar:
jobless
out of work
out of a job
not working
between jobs .
Consider the following mechanism for the decomposition of NO2Cl to NO2 and Cl2: (1) NO2Cl ⇌ NO2 + Cl (2) NO2Cl + Cl → NO2 + Cl2 (a) Use steady-state approximation to express the rate of Cl2 production. Select the single best answer
Final answer:
The rate of Cl2 production can be determined using the steady-state approximation, which states that the rate of the forward reaction in the first step is equal to the rate of the reverse reaction. Therefore, the rate of Cl2 production is given by the rate of the forward reaction of step 1. This can be expressed as k1 [NO2Cl] [Cl].
Explanation:
The rate of Cl2 production can be determined using the steady-state approximation. According to the given mechanism, the first step is in equilibrium, so the forward and reverse reaction rates are equal. This allows us to express the rate of the forward reaction of step 1 as:
rate of forward reaction of step 1 = k1 [NO2Cl] [Cl]
Since the reverse reaction of step 1 is negligible, the overall rate of Cl2 production is equal to the rate of the forward reaction of step 1. Therefore, the rate of Cl2 production is given by:
rate of Cl2 production = k1 [NO2Cl] [Cl]
Final answer:
To express the rate of Cl2 production using the steady-state approximation, one must balance the rate of chlorine atom creation with its consumption and solve for its concentration in terms of the reactants and rate constants. This value is then used in the rate equation for Cl2 production.
Explanation:
The question pertains to the mechanism of the decomposition of nitryl chloride (NO2Cl) to nitrogen dioxide (NO2) and chlorine gas (Cl2). Since the steady-state approximation is used, we consider the intermediate species such as chlorine atoms (Cl) to be in a steady state, meaning their concentrations do not change over time. The provided steps do not correspond directly to the decomposition of NO2Cl, but they outline a similar mechanism which can be analyzed in the same manner using steady-state approximation.
If we follow a similar approach for NO2Cl decomposition, we would set the rate of creation of Cl equal to the rate of its consumption. This would lead to a system of equations that can then be solved to express the rate of Cl2 production in terms of the concentrations of the reactants and the rate constants of the elementary steps.
The rate of reaction (2) can be determined using the rate law, which is determined experimentally. However, since we are not given any experimental data or rate law, we cannot provide a specific answer.
To summarize, to find the rate of Cl2 production in the decomposition of NO2Cl, we need the rate law for reaction (2). Without the rate law, we cannot determine the specific rate of Cl2 production.
Compare the mass of the Mg ribbon with the mass of the magnesium oxide. Notice that the the mass of the magnesium oxide is greater than the mass of the Mg. How do you account for this apparent increase in mass?
The mass of magnesium oxide is greater than that of the Magnesium ribbon because the Magnesium ribbon reacts with oxygen in the air to form magnesium oxide. The extra mass is from the oxygen atoms.
Explanation:When comparing the mass of the Mg ribbon and the magnesium oxide, you may notice that the mass of the magnesium oxide seems to be greater. This apparent increase in mass can be attributed to a chemical reaction. In this case, when the Mg ribbon is exposed to air, it reacts with the oxygen present in the air to create magnesium oxide (MgO).
The increase in mass is due to the additional weight of the oxygen atoms that are now part of the compound. Moreover, smaller pieces of magnesium metal will react more rapidly than larger pieces because there is more reactive surface available. The increase in mass is thus directly related to the magnesium oxide formation with the interaction of magnesium and oxygen.
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The increase in mass of Magnesium Oxide compared to Magnesium ribbon is due to the incorporation of Oxygen during the oxidation process which forms Magnesium Oxide.
Explanation:
In comparing the mass of a Magnesium (Mg) ribbon and that of Magnesium Oxide (MgO), it's observed that the mass of the MgO is greater. This apparent increase in mass can be attributed to the chemical reaction that takes place when Magnesium reacts with Oxygen in the air to form Magnesium Oxide.
Magnesium + Oxygen -> Magnesium Oxide
Magnesium (Mg) atoms combine with Oxygen (O) molecules from the air. As a result, the atoms bond to form Magnesium Oxide (MgO) which results in an increase in mass since the mass of the Oxygen is now being taken into consideration.
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Which type of lipid is found in the body and is converted into vitamins, hormones, and bile salts?
A.fats
B.waxes
C.sterols
D.phospholipids
Answer:
C.sterols
Explanation:
Sterols or steroid alcohols are type of lipids. In plants they are present as phytosterols and in animals as zoosterols. They maintain the fluidity of cell membrane, act as signalling molecules and also form the skin oils in animals.
Cholesterol is an important zoosterol. It is a fatty waxy substance. It is present in cell membrane. It is also a precursor for vitamin D. It is precursor for steroid hormones like cortisol and aldosterone. When it is non esterified, it gets converted to bile.
Answer:
Option-C
Explanation:
Steroids are the hydrophobic molecules that can be structurally characterised by the fused rings and -OH group. The steroids since are hydrophobic therefore are kept with the phospholipids.
A sterol which is present as a component of the lipid layer is known as the cholesterol. The cholesterol acts as a precursor to a variety of steroid hormone-like estrogens, glucocorticoids, progestagens and many others. The cholesterol also acts as precursor to vitamin D and bile acids in the body.
Thus, Option-C is correct.
Determine the change in entropy for 2.7 moles of an ideal gas originally placed in a container with a volume of 4.0 L when the container was expanded to a final volume of 6.0 L at constant temperature.
Answer:
The value of entropy change for the process [tex]dS = 0.009 \frac{KJ}{K}[/tex]
Explanation:
Mass of the ideal gas = 0.0027 kilo mol
Initial volume [tex]V_{1}[/tex] = 4 L
Final volume [tex]V_{2}[/tex] = 6 L
Gas constant for this ideal gas ( R ) = [tex]R_{u} M[/tex]
Where [tex]R_{u}[/tex] = Universal gas constant = 8.314 [tex]\frac{KJ}{Kmol K}[/tex]
⇒ Gas constant R = 8.314 × 0.0027 = 0.0224 [tex]\frac{KJ}{K}[/tex]
Entropy change at constant temperature is given by,
[tex]dS = R log _{e} \frac{V_{2}}{V_{1}}[/tex]
Put all the values in above formula we get,
[tex]dS = 0.0224 log _{e} [\frac{6}{4}][/tex]
[tex]dS = 0.009 \frac{KJ}{K}[/tex]
This is the value of entropy change for the process.
With respect to NAD+ and NADP+, which electron carrier is generally preferred for anabolic reactions? Group of answer choices Just NAD+ Just NADP+ Either of these Neither of these
Answer:
Just NADP+.
Explanation:
Two main type of biochemical reaction are anabolic reaction and catabolic reaction. Anabolic reactions joins the small molecules to form the large products.
The electron carrier may be defined as the molecule that has the ability to transport the electron from one molecule to the other molecule. NADP+ acts as electron carrier in the synthesis of molecule especially in the cholesterol metabolism. NAD+ acts as electron carrier in the catabolic reactions.
Thus, the correct answer is option (2).
Write the dissolution reaction for iron(III) nitrate in water. Use the pull-down menus to specify the state of each reactant and product. + Is iron(III) nitrate considered soluble or not soluble ? ... ... A. Soluble ... B. Not soluble Based upon this, the equilibrium constant for this reaction will be: ... ... A. Greater than 1 ... B. Less than 1
Explanation:
As Iron (III) nitrate is an ionic compound so when it is dissolved in water then it will dissociate to give nitrate ions and ferric ions. Also, we know that like dissolves like and here iron (III) nitrate being an ionic compound will readily dissolve in water (polar solvent).
The chemical equation for this reaction is as follows.
[tex]Fe(NO_{3})_{3}(s) + H_{2}O(l) \rightarrow Fe^{3+}(aq) + 3NO^{-}_{3}(aq)[/tex]
Hence, iron(III) nitrate considered soluble in water.
Now, equilibrium constant expression for this reaction is as follows.
[tex]K_{eq} = [Fe^{3+}][(NO_{3})_{3}]^{3}[/tex]
Therefore, the equilibrium constant for this reaction will be greater than 1.
A student has 540.0 mL of a 0.1035 M aqueous solution of Na2CrO4 to use in an experiment. She accidentally leaves the container uncovered and comes back the next week to find only a solid residue. The mass of the residue is 19.12 g. Determine the chemical formula of this residue.
Answer:
The chemical formula of this residue: [tex]Na_2CrO_4.10H_2O[/tex]
Explanation:
Mass of residue = 19.12
Let the formula of the residue be [tex]Na_2CrO_4.xH_2O[/tex]
Moles of residue: n
[tex]n=\frac{19.12}{162 g/mol+x\times 18 g/mol}[/tex]
Moles of sodium chromate = n'
Molarity of sodium chromate = 0.1035 M
Volume of sodium chromate solution = 540.0 mL = 0.540 L
1 mL = 0.001 L
[tex]Concentration=\frac{moles}{Volume(L)}[/tex]
[tex]0.1035 M=\frac{n}{0.540 L}[/tex]
[tex]n'=0.1035 M\times 0.540 L=0.05589 mol[/tex]
[tex]Na_2CrO_4+xH_2O\rightarrow Na_2CrO_4.xH_2O[/tex]
According to reaction,1 mole of [tex]Na_2CrO_4[/tex] gives 1 mole of[tex]Na_2CrO_4.xH_2O[/tex]
So, n = n'
[tex]\frac{19.12}{162 g/mol+x\times 18 g/mol}=0.05589 mol[/tex]
x = 10
The chemical formula of this residue: [tex]Na_2CrO_4.10H_2O[/tex]
10. What is the function of mitochondria?
Mitochondria is the site of ATP synthesis in eukaryotic cell during aerobic respiration. It regulates the biochemical processes of the cell and helps in generation of co enzymes as sulphur and iron.
Explanation:
Mitochondria is a cell membrane bound organelles present in cytoplasm of eukaryotic cells.
The major roles of mitochondria in eukaryotic cell is to synthesize ATP by cellular respiration. It also regulates the metabolism or biochemical processes of the cell.
Mitochondria also aids in generating the iron and sulphur which acts as a coenzyme in several biochemical reactions.
The electron transport chain takes place in inner mitochondrial membrane.
There are 2 aerobic phases of cellular respiration those are Kreb's cycle and electron transport chain. The machinery for Kreb's Cycle is in matrix. The ETC is present as embedded in the inner membrane of matrix.
The cellular compartment of mitochondria performs following functions:
The space in intermembrane has electron transport chain and holds ATP synthase (enzyme required for
The cristae is a finger like projection which increases surface area for increase ATP synthesis.
In matrix of the mitochondria the Electron transport takes place.
In Kreb's cycle mitochondria aids in production of NADH and GTP. Also, synthesis of phospholipid takes place in mitochondria.
A certain mass of carbon reacts with 13.6 g of oxygen to form carbon monoxide. ________ grams of oxygen would react with that same mass of carbon to form carbon dioxide, according to the law of multiple proportions.
Answer: The mass of oxygen that will react with same amount of carbon as in carbon monoxide is 23.31 grams.
Explanation:
Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: [tex]Cu_2O\text{ and }CuO[/tex]
We are given:
Mass of oxygen in CO = 13.6 grams
For carbon monoxide (CO):Applying unitary method:
16 grams of oxygen are reacting with 12 grams of carbon
So, 13.6 grams of oxygen will be reacting with = [tex]\frac{12}{16}\times 13.6=10.2g[/tex] of carbon
For carbon dioxide [tex](CO_2)[/tex] :Applying unitary method:
12 grams of carbon are reacting with 32 grams of oxygen
So, 10.2 grams of carbon will be reacting with = [tex]\frac{32}{12}\times 10.2=23.31g[/tex] of oxygen
Hence, the mass of oxygen that will react with same amount of carbon as in carbon monoxide is 23.31 grams.
In the given question, 27.2 grams of oxygen will react with 10.2 grams of carbon to give 39.1 grams of carbon dioxide.
Calculations based on the law of multiple proportions:The mass of oxygen given in the question is 13.6 grams. The molecular mass of oxygen or O2 is 32 g/mol.
The reactions taking place in the given case are:
2C + O₂ ⇔ 2CO --------- (i)C + O₂ ⇔ CO₂ ---------- (ii)Now the number of moles of O₂ will be,
[tex]Moles of O2= \frac{Weight}{Molecular mass}[/tex]
[tex]Moles = \frac{13.6}{32} \\Moles = 0.425 moles[/tex]
Based on the reaction, it can be seen that with 1 mole of O₂, 2 moles of C react to produce 2 moles of CO.
Now for 0.425 moles of oxygen,
The moles of C, that is, 0.425 × 2 = 0.850 moles will react 0.425 moles of O₂ to produce 0.850 moles of CO
Now the number of moles of C is 0.850 moles, and the molecular mass of C is 12 g/mol.
The mass of C will be,
Mass = Number of moles × Molecular mass
Mass = 0.850 × 12 = 10.2 grams
Now in the second reaction, that is, in the formation of CO₂, 1 mole of C will react with 1 mole of O₂ to produce 1 mole of CO₂.
Therefore, 0.850 moles of C will react with 0.850 moles of O₂ to give 0.850 moles of CO₂.
Now the molar mass of CO₂ is 46 g/mol, the mass of CO₂ will be,
Mass = 0.850 moles × 46 = 39.1 grams
The moles of O₂ is 0.850 moles, the molar mass of O₂ is 32 g/mol. Now the mass of O₂ is,
Mass = Number of moles × Molar mass
Mass = 0.850 × 32 g/mol
Mass = 27.2 grams
Thus, 27.2 grams of oxygen will react with 10.2 grams of carbon to produce 39.1 grams of CO₂.
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