Answer:
-99.8 kJ
Explanation:
We are given the methodology to answer this question, which is basically Kirchhoff law . We just need to find the heats of formation for the reactants and products and perform the calculations.
The standard heat of reaction is
ΔrHº = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants
where ν are the stoichiometric coefficients in the balanced equation, and ΔfHº are the heats of formation at their standard states.
Compound ΔfHº (kJmol⁻¹)
SO₂ -296.8
O₂ 0
SO₃ -395.8
The balanced chemical equation is
SO₂(g) + ½O₂(g) → SO₃(g)
Thus
Δr, 298K Hº( kJmol⁻¹ ) = 1 x (-395.8) - 1 x (-296.8) = -99.0 kJmol⁻¹
Now the heat capacity of reaction will be be given in a similar fashion:
Cp rxn = ∑ ν x Cp of products - ∑ ν x Cp of reactants
where ν is as above the stoichiometric coefficient in the balanced chemical equation.
Cprxn ( JK⁻¹mol⁻¹) = 50.7 - ( 39.9 + 1/2 x 29.4 ) = - 3.90
= -3.90 JK⁻¹mol⁻¹
Finally Δr,500 K Hº = Δr, 298K Hº + CprxnΔT
Δr,500 K Hº = - 99 x 10³ J + (-3.90) JK⁻¹ ( 500 - 298 ) K = -99,787.8
= -99,787.8 J x 1 kJ/1000 J = -99.8 kJ
Notice thie difference is relatively small that is why in some problems it is o.k to assume the change in enthalpy is constant over a temperature range, especially if it is a small range of temperatures.
Which of the following solutions would make a good buffer system? (Check all that apply.) A. A solution that is 0.10 M NH3 and 0.10 M NH4Cl B. A solution that is 0.10 M HCN and 0.10 M NaF C. A solution that is 0.10 M HCN and 0.10 M LiCN D. A solution that is 0.10 M HF and 0.10 M NaF
Answer:
A solution that is 0.10 M HCN and 0.10 M LiCN
. A solution that is 0.10 M NH3 and 0.10 M NH4Cl
Explanation:
A buffer consists of a weak acid and its conjugate base counterpart. HCN is a weak acid and the salt LiCN contains its counterpart conjugate base which is the cyanide ion. A buffer maintains the pH by guarding against changes in acidity or alkalinity of the solution.
A solution of ammonium chloride and ammonia will also act as a basic buffer. A buffer may also contain a weak base and its conjugate acid.
Answer:
Good buffer systems are:
A) NH3 + NH4Cl
C) HCN + LiCN
D) HF + NaF
Explanation:
Buffers consist in a mixture of a weak acid with its salt or a weak alkaly with its salt. All buffer systems are conformed by:
1) Weak acid + salt
or
2) Weak alkaly + salt
It is very important these salts come from the weak acid or weak alkaly. It means, the anion of the acid must be the anion in the salt which is going to be part of the buffer system. On the other hand, the cation of the weak alkaly must be the cation of the salt which is going to form the salt in the buffer system.
Then, when we evaluate all options in this exercise, answers are the following:
A) 0.10 M NH3 and 0.10 M NH4Cl. It is a buffer because NH3 (ammonia) is a weak alkaly and NH4Cl is a salt coming from NH3.
Buffer component reactions:
Reaction weak alkaly: NH3 + H2O <-----> NH4+ + OH-
Reaction salt in water: NH4Cl ---> NH4+ + Cl-
NH4+ is the cation of the weak alkaly so it must be part of the salt in the buffer system. Then NH4Cl is a salt from NH3.
C) 0.10 M HCN and 0.10 M LiCN. It is a buffer because HCN is a weak acid and LiCN is a salt which is coming from HCN.
Buffer component reactions:
Reaction weak acid: HCN + H2O <-----> H3O+ + CN-
Reaction salt in water: LiCN --> Li+ + CN-
CN- is the anion of the acid, so it must be part of the salt in the buffer system. Then LiCN is a salt from HCN.
D) 0.10 M HF and 0.10 M NaF. It is a buffer because HF is a weak acid and NaF is a salt which is coming from HF.
Buffer component reactions:
Reaction weak acid: HF + H2O <------> H3O+ + F-
Reaction salt in water: NaF ---> Na+ + F-
F- is the anion of the weak acid (HF), so it must be part of the salt in th buffer systema. Then NaF is a salt coming from HF.
However option B, it is not a buffer, because it is a mixture of 0.10 M HCN and 0.10 M NaF. Salt is not coming from the weak acid.
Reaction weak acid: HCN + H2O <-----> H3O+ + CN- (anion of the acid is CN-)
Rection salt in water: NaF --> Na+ + F- (anion in the salt is F-, not CN-)
Anion of the acid is CN- and the anion in the salt is F- so it is not a salt coming from the weak acid. Then option B it is not a buffer system.
Snape grows tired of these conceptual questions and thinks it’s time for a problem. What is the retention factor if the distance traveled by the solvent front is 2.00 cm, and the distance traveled by the ion is 0.40 cm?
Answer:
ion travelled : 2.00cm -0.40cm
= 1.6cm
∴ Rf = 1.6/2.0
= 0.80
Explanation:
Retention factor is the ratio of distance travelled by solute divided by distance travelled by solvent.
since you given distance travelled by solvent then the distance required by solute is need in this case the ion.
A car starts at mile marker 145 on a highway and drives at 55 mi/hrmi/hr in the direction of decreasing marker numbers. What mile marker will the car reach after 2 hours
Answer:
135 mile marker will the car reach after 2 hours.
Explanation:
Speed of the car = 55 mile/hour
Distance covered in 2 hours = d
[tex]Speed=\frac{Distance}{Time}[/tex]
[tex]55 mile/hour=\frac{d}{2 hour}[/tex]
[tex]d=55 mile/hour\times 2 hour=110 mile[/tex]
The direction of the car is in decreasing marker numbers which mienas that car had started from end where 145 mile marker was present.
So, the marker appearing after travelling 2 hours will be:
145 - 110 = 135
135 mile marker will the car reach after 2 hours.
Use a spreadsheet and construct curves for the following titrations. Calculate potentials after the addition of 10.00, 25.00, 49.00, 49.90, 50.00, 50.10, 51.00, and 60.00 mL of the reagent. Where necessary, assume that = 1.00 throughout.
Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows: C5H5N+H2O⇌C5H5NH++OH− The pKb of pyridine is 8.75. What is the pH of a 0.460 M solution of pyridine? (Assume that the temperature is 25 ∘C.) Express the pH numerically to two decimal places.
Answer:
The pH of the solution is 9.46
Explanation:
C₅H₅N + H₂O ⇌ C₅H₅NH+OH⁻
I 0.46
C - x +x +x
E 0.46 - x
-LogKb = Pkb
[tex]Kb =10^{[-PKb]} = 10^{[-8.75]} = 1.778 X 10^{-9}[/tex]
[tex]Kb = \frac{[C5H5NH][OH^-]}{[C5H5N]}[/tex]
[tex]1.778 X10^{-9}= \frac{X^2}{0.46-X} \\\\X^2 = 8.1788 X 10^{-10} - 1.778 X10^{-9}X\\\\X^2 + 1.778 X10^{-9}X -8.1788 X 10^{-10}\\\\X = 2.85977 X 10^{-5} = [OH^-][/tex]
pOH = -Log[OH⁻]
pOH = -Log [2.85977 x 10⁻⁵]
pOH = 4.54
pOH + pH = 14
pH = 14 - pOH
pH = 14 - 4.54
pH = 9.46
Therefore, the pH = 9.46
1) β-galactosidase is a unique enzyme, in that it can have multiple substrates. What are some other substrates for β-galactosidase? What are some other inhibitors for β-galactosidase?
The natural substrate of beta glycosidase enzyme is Ganglioside GMI, Lactosylceramide, lactose and glycoprotein.
Inhibitors of beta galactosidase enzyme is 1,4-dithiothreitol, beta marcaptoethanol, 4-chloromercurobenzoic acid and Acid-beta galactosidase.
Explanation:
Beta galactosidase enzyme performs the hydrolysis of beta galactosides into monosaccharides. It acts on aryl, amino, alkyl beta glycosidic linkages also.
The enzyme attacks on the bond formed between organic entity and galactose sugar.
It can act on multiple substrates. Some substrates are :
Ganglioside GMI
Lactosylceramide
Lactose : Enzyme beta galactosidase enzyme is a boon for lactose intolerant people because it breaks lactose in yoghurt, sour cream and cheese and makes it easy for consumption to such people.
glycoprotein : These have glycosidic bonds on which enzyme works to break the bonds.
The inhibitors for the β-galactosidase are :
4-dithiothreitol
beta marcaptoethanol
4-chloromercurobenzoic acid
Acid-beta galactosidase.
When inhibitors are bind to enzyme it breaks down the inhibitor and reaction does not takes place.
β-galactosidase can act on different substrates like lactose, ONPG and IPTG, demonstrating its flexibility. Its activity can be regulated by inhibitors such as glucose and PETG, affecting enzyme function either competitively or non-competitively.
Explanation:The enzyme β-galactosidase is unique as it can act on various substrates and is influenced by several inhibitors. This enzyme mainly acts on lactose but can also interact with several structurally related substrates such as o-nitrophenyl-β-D-galactoside (ONPG) and isopropyl β-D-1-thiogalactopyranoside (IPTG), exemplifying the flexibility of β-galactosidase.
The activity of β-galactosidase is subject to regulation by various inhibitors. These inhibitors could exert their effect by binding to the enzyme's active site (competitive inhibition), or noncompetitively by interacting with the enzyme's allosteric site--an alternate part where non-substrate molecules can attach. Examples of inhibitors include glucose and phenylethyl β-D-thiogalactoside (PETG).
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The pKa of an acid (one dissociable hydrogen) is -5.7. To the nearest ones, what is the Ka of this acid? Please note that we should use scientific notation and fewer significant figures, but Canvas is not configured to use scientific notation.
Answer:
The value of dissociation constant of an acid [tex]5.0\times 10^{5}[/tex].
Explanation:
The [tex]pK_a[/tex] of an acid = -5.7
The dissociation constant of the reaction = [tex]K_a[/tex]
The relation between [tex]pK_a[/tex] and [tex] K_a[/tex] is given by ;
[tex]pK_a=-\log[K_a][/tex]
[tex]-5.7=-\log[K_a][/tex]
[tex]K_a=10^{-(-5.7)}=5.012\times 10^{5}\approx 5.0\times 10^{5}[/tex]
The value of dissociation constant of an acid [tex]5.0\times 10^{5}[/tex].
Suppose you want to separate a mixture of the following compounds: salicylic acid, 4-ethylphenol, p-aminoacetophenone, and napthalene. Come up with a list of steps and chemicals needed to most efficiently isolate all four compounds as solids with the greatest purity possible. You do not need to write a formal procedure, but be sure to indicate steps needed clearly and in order.
Answer:
The procedure you will use in this exercise exploits the difference in acidity and solubility just described.
(a) you will dissolve your unknown in ethyl acetate (an organic solvent). All of the possible compounds are soluble in ethyl acetate.
(b) you will extract with sodium bicarbonate to remove any carboxylic acid that is present.
(c) you will extract with sodium hydroxide to remove any phenol that is present.
(d) you will acidify both of the resulting aqueous solutions to cause any compounds that were extracted to precipitate.
onsider the following reaction at equilibrium: CO(g) + Cl2(g)=======COCl2(g) Predict whether the reaction will shift left,shift right, or remain the unchanged upon each of the following disturbances.a) COCl2 is added to thereaction mixtureb) Cl2 is added to thereaction mixturec) COCl2 is removed fromthe reaction mixture
Explanation:
CO(g) + Cl2(g) ⇄ COCl2(g)
This question is based on Le Chatelier's principle.
Le Chatelier's principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.
a) COCl2 is added to the reaction mixture
COCL2 is a product in the reaction. If we add additional product to a system, the equilibrium will shift to the left, in order to produce more reactants. The reaction would shift to the left.
b) Cl2 is added to the reaction mixture
if we add reactants to the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.
c) COCl2 is removed from the reaction mixture
if we remove products from the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.
Chemical reactions at equilibrium, such as CO(g) + Cl2(g) ⇌ COCl2(g), shift in response to changes to re-establish equilibrium. If a product or reactant is added, the reaction will respectively shift towards reactants or products. Similarly, if a product or reactant is removed, the reaction will respectively shift towards products or reactants.
Explanation:In chemistry, chemical reactions at equilibrium respond to disturbances according to Le Châtelier's principle: the system shifts in a way that counters the disturbance and re-establishes equilibrium. Now, let's consider the reaction: CO(g) + Cl2(g) ⇌ COCl2(g).
(a) When COCl2 is added, it will increase the product's concentration, thus, the reaction will shift to the left (towards the reactants) to re-establish equilibrium. (b) When Cl2 is added, it will increase the reactant's concentration. The reaction will shift to the right (towards the products) to counter this and re-establish equilibrium. (c) When COCl2 is removed, it decreases the product's concentration. To counteract this, the reaction shifts to the right (towards the products) to re-establish equilibrium.Learn more about Chemical equilibrium here:
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Draw the structure of a compound with the molecular formula C9H10O2 that exhibits the following spectral data. (a) IR: 3005 cm-1, 1676 cm-1, 1603 cm-1 (b) 1H NMR: 2.6 ppm (singlet, I
Answer:
The answer is 3-Phenylpropanoic acid (see attached structure)
Explanation:
From spectral data:
3005 cm-1 ⇒ carboxylic acid (broad band)
1670 cm-1 ⇒ C=C
1603 cm-1 ⇒ Aromatic C-C bond
H NMR frequency at 2.6 ppm, singlet, ⇒ OH with no surrounding protons, possible deshielding (clearer investigation of spectrum would be expedient).
Hence, our C9H10O2 compound has an aromatic ring and carboxylic acid group attached to it.
The compound is probably a molecule with an aromatic ring (benzene) with an attached ester group (COOCH3) that satisfies all the given spectral data.
Explanation:The spectral data given corresponds to a compound with molecular formula C9H10O2. From the IR data, the bands at 3005 cm-1 suggests C-H sp2 bond (alkene or aromatic), at 1676 cm-1 indicates carbonyl group (C=O), and 1603 cm-1 suggests a carbon-carbon double bond (C=C) which might be in an aromatic ring. The 1H NMR data: 2.6 ppm singlet signifies the protons of a methyl group (CH3) attached to an electronegative atom like a carbonyl carbon.
Based on these data, a probable structure for the compound could be a molecule, which is an aromatic ring (benzene) with an attached ester group (COOCH3). That gives the right molecular formula, the required carbonyl, alkene and sp2 hybridized C-H bonds for the IR, and the singlet in the NMR for the methyl group (CH3) of the ester.
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Propose a mechanism to account for the formation of a cyclic acetal from 4-hydroxypentanal and one equivalent of methanol. If the carbonyl oxygen of 4-hydroxypentanal is enriched with oxygen-18, do you predict that the oxygen label appears in the cyclic acetal or in the water
Answer:
If carbonyl oxygen of 4-hydroxypentanal is enriched with [tex]O^{18}[/tex], then the oxygen label appears in the water .
Explanation:
In the first step, -OH group at C-4 gives intramolecular nucleophilic addition reaction at carbonyl center to produce a cyclic hemiacetal.Then, one equivalent of methanol gives nucleophilic substitiution reaction by substituting -OH group in cyclic hemiacetal to produce cyclic acetal.If carbonyl oxygen of 4-hydroxypentanal is enriched with [tex]O^{18}[/tex], then the oxygen label appears in the water produced at the end of reaction.Full reaction mechanism has been shown below.Estimate the following: a) The volume occupied by 18 kg of ethylene at 55°C and 35 bar. b) The mass of ethylene contained in a 0.25-m3 cylinder at 50°C and 115 bar.
The estimated volume occupied by 18 kg of ethylene at 55℃ and 35 bar is approximately 476.6 liters. The mass of ethylene contained in a [tex]0.25m^3[/tex] cylinder at 50℃ and 115 bar is approximately 3.03 kg.
To estimate the volume occupied by 18 kg of ethylene at 55℃ and 35 bar, we can use the ideal gas law expressed as PV = nRT. First, we need to calculate the number of moles (n) using the molar mass of ethylene ([tex]C_2H_4[/tex]) which is approximately 28 g/mol. Therefore, n = 18000 g / 28 g/mol = 642.86 moles. The gas constant (R) is 0.08314 L*bar/mol*K. Converting the temperature to Kelvin, T = 55 + 273 = 328 K. Now, we can solve for V:
V = (nRT) / P = (642.86 mol * 0.08314 L*bar/mol*K * 328 K) / 35 bar = 476.6 L
To determine the mass of ethylene in a 0.25-[tex]m^3[/tex] cylinder at 50℃ and 115 bar, we apply the ideal gas law again. We first convert the volume to liters (V = 0.25 [tex]m^3[/tex] * 1000 L/[tex]m^3[/tex] = 250 L), and temperature to Kelvin (T = 50 + 273 = 323 K). The pressure is already in bar. Now, we calculate the number of moles, solving for n:
n = PV / RT = (115 bar * 250 L) / (0.08314 L*bar/mol*K * 323 K) = 108.14 moles
The mass of ethylene is then m = n * molar mass = 108.14 moles * 28 g/mol = 3027.92 g or approximately 3.03 kg of ethylene.
The properties of elements are different than the compound that the elements form. Is this statement True or False
Answer:
true
Explanation:
The decay constant for 14C is .00012 In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14C found in living trees. Write a formula for the age of the charcoal (hence of the associated paintings). Show your work to find this formula.
The question is incomplete, here is the complete question:
The decay constant for 14-C is [tex]0.00012yr^{-1}[/tex] In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14-C found in living trees. Write a formula for the age of the charcoal (hence of the associated paintings). Show your work to find this formula.
Answer: The formula for the age of the charcoal is [tex]t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]
Explanation:
Carbon-14 isotope is a radioisotope and its decay process follows first order kinetics.
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]0.00012yr^{-1}=1.2\times 10^{-4}yr^{-1}[/tex]
t = time taken for decay process = ? yr
[tex][A_o][/tex] = initial amount of the sample = 100 grams
[A] = amount left after decay process = (100 - 20) = 80 grams
Putting values in above equation, we get:
[tex]1.2\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{20}\\\\t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]
Hence, the formula for the age of the charcoal is [tex]t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]
Consider the following reaction: Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)If you react an excess of Pb(NO3)2with 26.3 g of NaCl, and you isolate 52.1 g of PbCl2, what is your percent yield?
Answer:
[tex]\large \boxed{84.7 \, \%}[/tex]
Explanation:
Mᵣ: 58.44 278.11
Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃
m/g: 26.3
1. Moles of NaCl
[tex]\text{Moles of NaCl} = \text{26.3 g NaCl} \times \dfrac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.4505 mol NaCl}[/tex]
(b) Moles of PbCl₂
[tex]\text{Moles of PbCl${_2}$} = \text{0.4505 mol NaCl} \times \dfrac{\text{1 mol PbCl${_2}$}}{\text{2 mol NaCl}} = \text{0.2253 mol PbCl${_2}$}[/tex]
(c) Theoretical yield of PbCl₂
[tex]\text{Mass of PbCl${_2}$} = \text{0.2253 mol PbCl${_2}$} \times \dfrac{\text{278.11 g PbCl${_2}$}}{\text{1 mol PbCl${_2}$}} = \text{61.52 g PbCl${_2}$}[/tex]
(d) Percent yield
[tex]\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \,\% = \dfrac{\text{52.1 g}}{\text{61.52 g}} \times 100 \, \% = \mathbf{84.7 \,\%}\\\\\text{The percent yield is $\large \boxed{\mathbf{84.7 \, \% }}$}[/tex]
Final answer:
In a reaction where lead (II) nitrate reacts with sodium chloride to form lead (II) chloride, given 26.3 g of NaCl and 52.1 g of PbCl2 produced, the percent yield is calculated to be 83.27%.
Explanation:
The student is performing a reaction where lead (II) nitrate reacts with sodium chloride to produce lead (II) chloride and sodium nitrate. To calculate the percent yield, we use the actual mass of PbCl2 obtained from the experiment (52.1 g), and compare it with the theoretical mass that should have been produced if the reaction were 100% efficient.
First, you need to calculate the moles of NaCl:
Molecular weight of NaCl = 22.99 (Na) + 35.45 (Cl) = 58.44 g/molMoles of NaCl = mass (g) / molar mass (g/mol) = 26.3 g / 58.44 g/mol = 0.45 molesBased on the stoichiometry of the balanced equation, 2 moles of NaCl will produce 1 mole of PbCl2 (2:1 ratio). Therefore, 0.45 moles of NaCl should theoretically produce 0.225 moles of PbCl2.
Now, calculate the theoretical yield:
Molecular weight of PbCl2 = 207.2 (Pb) + 2*35.45 (Cl) = 278.1 g/molTheoretical yield (g) = moles * molar mass = 0.225 moles * 278.1 g/mol = 62.57 g of PbCl2To find the percent yield:
Percent yield = (actual yield / theoretical yield) * 100% = (52.1 g / 62.57 g) * 100% ≈ 83.27%The percent yield of the reaction is therefore 83.27%.
5. At 20°C, the water autoionization constant, Kw, is 6.8 ´ 10–15. What is the H3O+ concentration in neutral water at this temperature? A. 6.8 × 10–7 M B. 3.4 × 10–15 M C. 6.8 × 10–15 M D. 8.2 × 10–8 M E. 1.0 × 10–7 M
Explanation:
Let us assume that the concentration of [[tex]OH^{-}[/tex] and [tex]H^{+}[/tex] is equal to x. Then expression for [tex]K_{w}[/tex] for the given reaction is as follows.
[tex]K_{w} = [OH^{-}][H^{+}][/tex]
[tex]K_{w} = x^{2}[/tex]
[tex]6.8 \times 10^{-15} = x^{2}[/tex]
Now, we will take square root on both the sides as follows.
[tex]\sqrt{6.8 \times 10^{-15}} = \sqrt{x^{2}}[/tex]
[tex][H^{+}] = 8.2 \times 10^{-8}[/tex] M
Thus, we can conclude that the [tex]H_{3}O^{+}[/tex] concentration in neutral water at this temperature is [tex]8.2 \times 10^{-8}[/tex] M.
Answer: The concentration of [tex]H_3O^+[/tex] in neutral water is [tex]8.2\times 10^{-8}M[/tex]
Explanation:
The chemical equation for the ionization of water follows:
[tex]2H_2O\rightleftharpoons H_3O^++OH^-[/tex]
The expression of [tex]K_w[/tex] for above equation, we get:
[tex]K_w=[H_3O^+]\times [OH^-][/tex]
We are given:
[tex]K_w=6.8\times 10^{-15}[/tex]
[tex][H^+]=[OH^-]=x[/tex]
Putting values in above equation, we get:
[tex]6.8\times 10^{-15}=x\times x\\\\x=8.2\times 10^{-8}M[/tex]
Hence, the concentration of [tex]H_3O^+[/tex] in neutral water is [tex]8.2\times 10^{-8}M[/tex]
The Br⊘⊘nsted-Lowry model focuses on the transfer of _______ in an acid-base reaction. The Brnsted-Lowry model focuses on the transfer of _______ in an acid-base reaction. electrons neutrons OH−OH− orbitals H+H+
Answer:
The Brönsted-Lowry model focuses on the transfer of H⁺ in an acid-base reaction.
Explanation:
According to the theory of Brönsted-Lowry , an acid is a chemical substance capable of releasing hydrogen ions, while a base is that chemical substance capable of accepting hydrogen ions. That is, acids are substances capable of yielding protons (hydrogen H + ions) and substance bases capable of accepting them.
On the other hand, the conjugate base of a Brønsted-Lowry acid is the species that forms after an acid donated a proton. The conjugate acid of a Brønsted-Lowry base is the species that forms when a base accepts a proton.
Thus, the acid-base reaction is one in which the acid transfers a proton to a base (proton transfer H⁺).
This theory, unlike another theory like Arrhenius, does not require the presence of water as a means of reaction for the transfer of H⁺.
The Brönsted-Lowry model focuses on the transfer of H⁺ in an acid-base reaction.
When making a solution of sodium hydroxide and water, a student weighed out an certain amount of sodium hydroxide pellets and dissolved them in an certain amount of water. However, the sodium hydroxide concentration of the resulting solution was lower than the concentration that the student thought they made. What was the problem?
Hygroscopic nature of NaOH is the main reason for the lower concentration of the sodium hydroxide in the solution.
Explanation:
It is a well known fact that Sodium hydroxide pellets are hygroscopic in nature, which clearly means that the sodium hydroxide pellets absorbs moisture from the air, so that it becomes deliquescent. When the NaOH crystals are weighed, the crystals absorb moisture from the surroundings, and so the weight of the crystals may change, so the concentration of the solution was lower than the required one.So it is not possible to prepare NaOH solution under normal room temperature. So, while preparing the solution of NaOH we have to be more careful.The main reason for the lower concentration of the sodium hydroxide in the solution is Hygroscopic nature of NaOH.
Nature of Sodium hydroxide:It's obviously true that Sodium hydroxide pellets are hygroscopic in nature, which plainly implies that the sodium hydroxide pellets retains dampness from the air, so it becomes deliquescent. Whenever the NaOH precious stones are gauged, the gems assimilate dampness from the environmental factors, thus the heaviness of the gems might change, so the centralization of the arrangement was lower than the expected one. So it is absurd to expect to get ready NaOH arrangement under ordinary room temperature. In this way, while setting up the arrangement of NaOH we must be more cautious.
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Your friend, a business major, overhears you talking about chemical energy and food and decides to develop a new super food for busy students that just don’t have time to cook but want to eat a well-balanced diet for a reasonable price. If your friend wants to design a meal bar that contains a high amount of potential energy that the body could use, how would you recommend they rank the possible atom interactions they could incorporate into the chemical formula? Rank from highest chemical energy to lowest.
To design a high energy meal bar, ingredients should be chosen based on their potential chemical energy, with high-energy phosphates, lipids, proteins, and carbohydrates ranked from highest to lowest energy contents.
Ranking Atom Interactions by Chemical Energy
When designing a meal bar with high amounts of potential energy for busy students, your friend should consider the chemical structure of the ingredients. Chemical energy is stored within the bonds of atoms in food molecules. This chemical potential energy is what our bodies metabolize to generate energy, measured in Calories (calorimeter units). Atom interactions in food molecules can be broadly ranked from highest to lowest potential chemical energy as follows:
High-energy phosphates (such as in ATP)
Lipid molecules/fats (like triglycerides)
Protein (amino acid chains)
Carbohydrates (like glucose)
It is important to note that while lipids generally contain more energy per gram than proteins or carbohydrates, the balance and quality of nutrients also play a crucial role in the healthfulness of a meal bar. Therefore, your friend should focus on including a mix of macronutrients that release energy over time and sustain a student's activity levels efficiently.
To create a meal bar with a high amount of potential chemical energy, the macronutrients should be ranked as follows: lipids (fats) with the most chemical energy, followed by proteins, and then carbohydrates.
When designing a superfood meal bar with a high amount of potential chemical energy, we need to consider the atom interactions that store the most energy. The rank of atom interactions from highest to lowest chemical energy typically aligns with the macronutrient composition, as these are the main carriers of chemical energy in food.
Macronutrients Ranked by Chemical Energy:
Lipids (fats) - They have the highest amount of energy per gram due to the presence of long chains of hydrocarbons which, when metabolized, release large amounts of chemical energy.Proteins - Proteins are made up of amino acids linked by peptide bonds, and while they have a considerable energy content, it is lower than that of lipids.Carbohydrates - These include sugars, starches, and fibers. Simple carbohydrates can quickly release energy, while complex carbohydrates provide a more sustained energy release, although both have less energy per gram than lipids and proteins.These rankings are based on the caloric content obtained through oxidation processes like metabolism. In designing a high-energy food bar, your friend should focus on incorporating healthy fats, complete proteins, and complex carbohydrates while considering the overall nutritional balance.
A(g) + 2B(g) → C(g) + D(g)
If you initially start with 1.00 atm of both A and B and find that at equilibrium 0.211 atm of C is present, what is the value of Kp for the reaction at the temperature the reaction was run?
Answer:
0.169
Explanation:
Let's consider the following reaction.
A(g) + 2B(g) ⇄ C(g) + D(g)
We can find the pressures at equilibrium using an ICE chart.
A(g) + 2 B(g) ⇄ C(g) + D(g)
I 1.00 1.00 0 0
C -x -2x +x +x
E 1.00-x 1.00-2x x x
The pressure at equilibrium of C is 0.211 atm, so x = 0.211.
The pressures at equilibrium are:
pA = 1.00-x = 1.00-0.211 = 0.789 atm
pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm
pC = x = 0.211 atm
pD = x = 0.211 atm
The pressure equilibrium constant (Kp) is:
Kp = pC × pD / pA × pB²
Kp = 0.211 × 0.211 / 0.789 × 0.578²
Kp = 0.169
A student fills her burette with NaOH to the 2.5 mL mark. She titrated her sample of the NaOH until she reaches the endpoint (i.e. all the acid has been neutralized by the NaOH). The volume marking on her burette - at the endpoint - is 52.5 mL. How many mL of NaOH did she use to reach the endpoint
Answer:
She used 50 mL of NaOH to reach the endpoint.
Explanation:
Assuming the burette is filled to the point marked 3.30 ml, You would record the initial point as 3.30 ml:
If at the end of the titration the level of the NaOH is at 20.30 mL; Subtract the initial reading from the final burette reading to get how many mL was used to reach an end point.
That is 20.3 - 3.3 = 17.00 mL
Therefore, the titration would have required 17.00 mL.
Remember that you should read the number that is at the bottom of the meniscus and at an eye level in order to avoid error.
Initial mark = 2.5 mL
Final mark = 52.5 mL.
volume used = 52.5 - 2.5
= 50 mL
How many joules of heat are required to heat 110 g of aluminum from 52.0 oC to 91.5 oC?
Answer:
We need 3910.5 joules of energy
Explanation:
Step 1: Data given
Mass of aluminium = 110 grams
Initial temperature = 52.0 °C
Final temperature = 91.5 °C
Specific heat of aluminium = 0.900 J/g°C
Step 2: Calculate energy required
Q = m*c*ΔT
⇒with Q = the energy required = TO BE DETERMINED
⇒with m = the mass of aluminium = 110 grams
⇒with c = the specific heat of aluminium = 0.900 J/g°C
⇒with ΔT = the change in temperature = T2 - T1 = 91.5 °C - 52.0 °C = 39.5 °C
Q = 110 grams * 0.900 J/g°C * 39.5
Q = 3910.5 J
We need 3910.5 joules of energy
what is the density of a substance that has a mass of 2.0 g, and when placed in a graduated cylinder the volume rose from 70 mL to 75 mL
Answer:
0.4g/mL
Explanation:
The following data were obtained from the question:
Mass = 2g
Volume = 75 — 70 = 5mL
Density =?
Density = Mass /volume
Density = 2/5
Density = 0.4g/mL
The density of the substance is 0.4g/mL
The density of a substance that has a mass of 2.0 g, and when placed in a graduated cylinder the volume rose from 70 mL to 75 mL is 0.4 g/mL.
To find the density of the substance, we need to calculate the volume it displaces and then divide its mass by that volume.
The volume displaced by the substance can be found by subtracting the initial volume in the graduated cylinder from the final volume after the substance is added. The initial volume is 70 mL and the final volume is 75 mL. Therefore, the volume displaced by the substance is:
Volume displaced = Final volume - Initial volume
Volume displaced = 75 mL - 70 mL
Volume displaced = 5 mL
Now, we have the mass of the substance, which is 2.0 g, and the volume it displaces, which is 5 mL. The density [tex]\rho[/tex] is calculated by dividing the mass (m) by the volume (V):
[tex]\rho = \frac{m}{V}[/tex]
[tex]\rho = \frac{2.0 \text{ g}}{5 \text{ mL}}[/tex]
[tex]\rho = 0.4 g/mL[/tex]
A mixture of 0.577 M H_2O , 0.314 M Cl_2O , and 0.666 M HClO are enclosed in a vessel at 25°C .
H_2O(g) + Cl_2O(g) <-------> 2 HOCl (g) Kc = 0.0900 at 25°C
1. Calculate the equilibrium concentrations of each gas at 25°C .
Answer:
Equilibrium Concentration of H₂O(g) = 0.803
Equilibrium Concentration of Cl₂O(g) = 0.540
Equilibrium Concentration of HOCl (g) = 0.214
Explanation:
Given;
H₂O(g) + Cl₂O(g) <-------> 2HOCl (g)
I 0.577 0.314 0.666
C - x -x +2x
E 0.577 - x 0.314 - x 0.666 +2x
[tex]K_c = \frac{[HOCL]^2}{[H_2O][CL_2O]} \\\\0.09 = \frac{[0.666+2x]^2}{[0.577-x][0.314-x]}\\\\0.09(0.1812 -0.891x+x^2) = (0.666+2x)(0.666+2x)\\\\0.0163-0.0802x+0.09x^2 = 0.4436+2.664x+4x^2\\\\3.91x^2+2.7742x+0.4273 =0\\\\x = -0.226, or -0.483[/tex]
Equilibrium Concentration of H₂O(g) = 0.577 - (- 0.226) = 0.803
Equilibrium Concentration of Cl₂O(g) = 0.314 - (- 0.226) = 0.540
Equilibrium Concentration of HOCl (g) = 0.666 +2(- 0.226) = 0.214
Thus, from the result it can be seen that at equilibrium, the reactants are favored.
Which compartment has the higher osmotic pressure? Which compartment has the higher osmotic pressure? 1%% (m/vm/v) starch solution 10%% (m/vm/v) starch solution
Answer : The compartment that has the higher osmotic pressure is, 10 % (m/v) starch solution.
Explanation :
Formula used for osmotic pressure :
[tex]\pi=\frac{nRT}{V}\\\\\pi=\frac{wRT}{MV}[/tex]
where,
= osmotic pressure
V = volume of solution
R = solution constant = 0.0821 L.atm/mole.K
T= temperature of solution = [tex]25^oC=273+25=298K [/tex]
M = molar mass of solute
w = mass of solute
Now we have to determine the osmotic pressure for the following solution.
For 1 % (m/v) starch solution :
1 % (m/v) starch solution means that 1 grams of starch present in 100 mL or 0.1 L of solution.
Molar mass of starch = 692.7 g/mol
[tex]\pi=\frac{(1g)\times (0.0821Latm/moleK)\times (298K)}{(692.7g/mol)\times (0.1L)}[/tex]
[tex]\pi=0.353atm[/tex]
For 10 % (m/v) starch solution :
10 % (m/v) starch solution means that 10 grams of starch present in 100 mL or 0.1 L of solution.
Molar mass of starch = 692.7 g/mol
[tex]\pi=\frac{(10g)\times (0.0821Latm/moleK)\times (298K)}{(692.7g/mol)\times (0.1L)}[/tex]
[tex]\pi=3.53atm[/tex]
From this we conclude that, 10 % (m/v) starch solution has the higher osmotic pressure as compared to 1 % (m/v) starch solution.
Hence, the compartment that has the higher osmotic pressure is, 10 % (m/v) starch solution.
The 10% (m/v) starch solution will have a higher osmotic pressure compared to the 1% (m/v) starch solution because osmotic pressure increases with solute concentration.
Explanation:The osmotic pressure is a property that depends on the solute concentration in a solution. In comparing a 1% (m/v) starch solution to a 10% (m/v) starch solution, the solution with the higher solute concentration is the one with higher osmotic pressure. Therefore, the 10% (m/v) starch solution will have a higher osmotic pressure because it has a greater concentration of starch molecules.
The osmotic pressure of a solution can be calculated using the formula Π = MRT, where Π is the osmotic pressure, M is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin. As solute concentration increases, so does the osmotic pressure, provided that temperature and the gas constant remain the same. Consequently, hypertonic solutions will have higher osmotic pressure than hypotonic solutions.
Explain why a fossil record is not 100% accurate.
The charges and sizes of the ions in an ionic compound affect the strength of the electrostatic interaction between the ions and thus the strength of the lattice energy of the ionic compound. Arrange the compounds according to the magnitudes of their lattice energies based on the relative ion charges and sizes.a. SrO b. CsIc. RbFd. SrF2
Answer:
the correct answer:
C, B, D, A.
Explanation:
The chemical compounds named are compounds that have strong chemical bonds, forming cubic structures and crystals with very high boiling and melting points to less solid structures.
some are oxides, others salts, others are even used to emit radiation.
Consider the mechanism. Step 1: A + B − ⇀ ↽ − C equilibrium Step 2: C + A ⟶ D slow Overall: 2 A + B ⟶ D Determine the rate law for the overall reaction, where the overall rate constant is represented as
The rate law for the overall reaction derived from a two-step reaction mechanism is determined by the slowest (rate-determining) step. In this case, the rate law would be rate = k * K_eq * [A]^2[B], where k is the rate constant for the slow step, K_eq is the equilibrium constant for the fast step, and [A] and [B] represent the concentrations of A and B, respectively.
Explanation:In the context of chemistry, one step in a multistep reaction mechanism is often significantly slower than the others. This slow step is known as the rate-determining step or the rate-limiting step. The reaction cannot proceed faster than this slowest step.
In your specific example, the given mechanism consists of two steps. The rate law for each step is generally expressed in terms of the concentration of the reactants involved in that step.
For Step 1 (A + B ⇄ C), assuming that the reaction reaches an equilibrium, the concentrations of A, B, and C would remain constant over time, and won't affect the overall rate. Therefore, we would ignore this step when deriving the rate law for the overall reaction.
On the other hand, Step 2 (C + A ⟶ D) is the slow step, and thus determines the rate of the overall reaction. The rate law for this step would be rate = k * [C][A]. But since C is also a product of Step 1, we need to express C in terms of A and B. From equilibrium of Step 1, we know [C] = K_eq*[A][B] (where K_eq is equilibrium constant).
Substituting this in rate law of step 2, we get rate = k * K_eq * [A]^2[B] for the overall reaction.
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The rate law for the overall reaction 2A + B ⟶ D, considering Step 2 is the rate-determining step, is calculated by using the equilibrium from Step 1 to express the concentration of intermediate C in terms of A and B. After substitution and simplification, the rate law for the overall reaction is rate = k [A]²[B], indicating second-order dependence on A and first-order dependence on B.
Explanation:To determine the rate law for the overall reaction, we must look at the mechanism provided. Since Step 2, which involves the conversion of C and A to D, is the rate-determining step, the rate law for the overall reaction will reflect this slowest step. However, since C is an intermediate that we cannot measure directly, we must use the equilibrium established in Step 1 to express the concentration of C in terms of the concentrations of A and B.
Assuming Step 1 is at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, and thus:
Rate of forward reaction (k₁ [A][B]) = Rate of reverse reaction (k₁₁ [C])
We can rearrange this to solve for [C]:
[C] = k₁/k₁₁ [A][B]
Now, since the rate-determining step is Step 2, we write the rate law based on this step:
rate = k₂ [C][A]
Substituting in the expression for [C] gives us:
rate = k₂ (k₁/k₁₁ [A][B])[A]
rate = (k₂ * k₁/k₁₁) [A]²[B]
Thus, after simplifying and combining the rate constants into a single overall rate constant (k), we have:
rate = k [A]²[B]
This shows that the reaction is second order with respect to A and first order with respect to B.
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Which of the following is TRUE? Group of answer choices None of the above is true. The equivalence point is where the amount of acid equals the amount of base during any acid-base titration. An indicator is not pH sensitive. A titration curve is a plot of pH vs. the [base]/[acid] ratio. At the equivalence point, the pH is always 7.
Answer:
TRUE: The equivalence point is where the amount of acid equals the amount of base during any acid-base titration.
Explanation:
The point on the titration curve where the number of base equivalents added equals the number of acid equivalents is the equivalence point or neutralization point.
Chemical indicators are substances that change color thanks to a chemical change, depending on the pH of the medium, and thus indicate the end point or point of equivalence of an acid-base volumetry.
A titration curve occurs by representing the measured pH as a function of the added volume of titrant, where the rapid change in pH for a given volume is observed. The inflection point of this curve is called the equivalence point and its volume indicates the volume of titrant consumed to fully react with the analyte.
In some cases, there are multiple equivalence points that are multiples of the first equivalence point, as in the valuation of a diprotic acid, which indicates that its pH value will not always be 7.
5. Phosphoric acid (H3PO4) is a triprotic acid with three ionizable protons. Write a balance equation for the neutralization of phosphoric acid with NaOH. How many milliliters of 0.120 M NaOH would be required to completely neutralize 35.0 ml of 0.0440 M H3PO4
Answer:
1. H3PO4 + 3NaOH —> Na3PO4 + 3H2O
2. 38.5mL
Explanation:
1. We'll begin by writing a balanced equation for the reaction. This is illustrated below:
H3PO4 + 3NaOH —> Na3PO4 + 3H2O
2. H3PO4 + 3NaOH —> Na3PO4 + 3H2O
From the equation above, the following data were obtained:
nA (mole of the acid) = 1
nB (mole of the base) = 3
Data obtained from the question include:
Vb (volume of base) =?
Mb (Molarity of base) = 0.120 M
Va (volume of acid) = 35.0 mL
Ma (Molarity of acid) = 0.0440 M
Using the formula MaVa/MbVb = nA/nB, the volume of the base (i.e NaOH) can be obtained as follow:
MaVa/MbVb = nA/nB
0.0440 x 35/ 0.120 x Vb = 1/3
Cross multiply to express in linear form as shown below:
0.120 x Vb = 0.0440 x 35 x 3
Divide both side by 0.120
Vb = (0.0440 x 35 x 3) /0.120
Vb = 38.5mL
Therefore, 38.5mL of 0.120 M NaOH is needed for the complete neutralization.
Answer:
We need 38.5 mL of NaOH to neutralize the H3PO4 solution
Explanation:
Step 1: Data given
Molarity of NaOH = 0.120 M
Volume of H3PO4 = 35.0 mL = 0.035 L
Molarity of H3PO4 = 0.0440 M
Step 2: The balanced equation
H3PO4 + 3NaOH —> Na3PO4 + 3H2O
Step 3: Calculate the volume of NaOH
b*Ca*Va = a *Cb*Vb
⇒with b = the coefficient of NaOH = 3
⇒with Ca = the concentration of H3PO4 = 0.0440 M
⇒with Va = the volume of H3PO4 = 35.0 mL = 0.0350 L
⇒with a = the coefficient of H3PO4 = 1
⇒with Cb = the concentration of NaOH = 0.120 M
⇒with Vb = the volume of NaOH = TO BE DETERMINED
3*0.0440 * 0.0350 = 0.120 * Vb
Vb = 0.0385 L = 38.5 mL
We need 38.5 mL of NaOH to neutralize the H3PO4 solution