Answer:
a. 0.8125 moles of sucrose
b. 277.8 g of sucrose
Explanation:
Consider these relation's value:
Molarity = Mol / Volume(L)
Mol = Molarity . Volume(L)
Volume(L) = Mol / Molarity
So, Molarity = 3.25M
Volume(L) = 0.25L
Molarity . Volume(L) = Mol → 3.25mol/L . 0.25L = 0.8125 mol
Let's convert the moles to mass → 0.8125 mol . 342 g /1mol = 277.8 g of sucrose
Answer:
We have 0.8125 moles sucrose in this solution. This is 277.9 grams of sucrose
Explanation:
Step 1: Data given
Molarity of a sucose solution = 3.25 M
Molar mass of sucrose = 342 g/mol
Step 2: Calculate moles sucrose
Moles sucrose = molarity sucrose solution * volume solution
Moles sucrose = 3.25 M * 0.25 L
Moles sucrose = 0.8125 moles sucrose
Step 3: Calculate mass of sucrose
Mass sucrose = moles sucrose * molar mass sucrose
Mass sucrose = 0.8125 moles * 342 g/mol
Mass sucrose = 277.9 grams
We have 0.8125 moles sucrose in this solution. This is 277.9 grams of sucrose
How many molecules are in 23 Moles of oxygen o2
Answer:
1.38×10^25 molecules
Explanation:
Applying n= (no. of molcules)/NA
23 = N/6.02×10^23
= 1.38×10^25 molecules
When The reaction below demonstrates which characteristic of a base?
Upper C U upper S upper O subscript 4 (a q) plus 2 upper N a upper O upper H (a q) right arrow upper C u (upper O upper H) subscript 2 (s) plus upper N a subscript 2 upper S upper O subscript 4 (a q).does phenolphthalein turn pink?
It shows the neutralization property of the base and there is no pink color.
Explanation:
CuSO₄(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)
Here NaOH is the base and Copper sulfate is an salt of an acid and so it is acidic salt. When these two compounds react, we will get the precipitate of copper hydroxide and Sodium sulfate.
And this reaction seems to be a neutralization reaction, since an acidic salt is neutralized by a base.
Since NaOH is converted into salt, that is sodium sulfate (Na₂SO₄), the phenolphthalein indicator becomes colorless in this reaction. It doesn't turn pink.
At 823 °C, Kp = 490 for the equilibrium reaction CoO(s) + CO(g) ⟷ Co(s) + CO2(g) What is the value of Kc at the same temperature for the equilibrium below? Co(s) + CO2(g) ⟷ CoO(s) + CO(g)
The equilibrium constant Kc for the reaction Co(s) + [tex]CO_{2}[/tex](g) ----> CoO(s) + CO(g) at 823°C is the reciprocal of Kp for the reverse reaction, resulting in a Kc of approximately 2.04 x 10^-3.
The value of the equilibrium constant Kc for the reaction Co(s) + [tex]CO_{2}[/tex](g) --------> CoO(s) + CO(g) can be calculated using the relationship between Kp and Kc and the fact that the reaction is the reverse of the one given. For the given reaction, Kp = 490 at 823°C. Since there are the same number of moles of gas on each side of the reaction, the relationship between Kp and Kc does not require any correction for the change in moles of gas, so Kc is the reciprocal of Kp for the reverse reaction. Thus, Kc for Co(s) + [tex]CO_{2}[/tex] ------->CoO(s) + CO(g) is 1/490 or approximately 2.04 x 10^-3 at 823°C.
What solution is placed when plating metal with gold?
Answer:
The metal atoms that plate your object come from out of the electrolyte, so if you want to copper plate something you need an electrolyte made from a solution of a copper salt, while for gold plating you need a gold-based electrolyte—and so on
Explanation:
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Answer:
electrolyte which is made from a solution of copper salt
Explanation:
A proposed mechanism is: Step 1: H2(g) + ICl(g) → HI(g) + HCl(g) (slow) Step 2: HI(g) + ICl(g) → I2(g) + HCl(g) (fast) Which of the following species is a catalyst? ICl HCl This mechanism has no catalyst. HI H2
Answer: This mechanism has no catalyst
Explanation:
The proposed mechanism is :
Step 1 : [tex]H_2(g)+ICl(g)\rightarrow HI(g)+HCl(g)[/tex] (slow)
Step 2: [tex]HI(g)+ICl(g)\rightarrow I_2(g)+HCl(g)[/tex] (fast)
The combined chemical equation will be :
[tex]H_2(g)+2ICl(g)\rightarrow I_2(g)+2HCl(g)[/tex]
A catalyst is a substance which enhances the rate of chemical reaction without being consumed in the chemical reaction. Thus catalyst gets used up in first step and gets regenerated in second.
HI is formed as an intermediate as it gets formed in first step and gets used up in the second step.
Here there is no substance which is used as a catalyst.
The reaction below has a Kp value of 3.3 × 10-5. What is the value of Kc for this reaction at 700 K? Note: Type the correct answer with 1 decimal place and in scientific notation. Example: 6.5 x 10-15 should be input as 6.5E-15. Example 2: 5.4 x 10-6 should be input at 5.4E-06.
This is an incomplete question, here is a complete question.
The reaction below has a Kp value of 3.3 × 10⁻⁵. What is the value of Kc for this reaction at 700 K?
[tex]2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)[/tex]
Answer : The value of [tex]K_p[/tex] is, [tex]5.79E^{-7}[/tex]
Explanation :
The relation between [tex]K_c[/tex] and [tex]K_p[/tex] is:
[tex]K_p=K_c\times (RT)^{\Delta n}[/tex]
where,
[tex]K_p[/tex] = equilibrium constant at constant pressure = [tex]3.3\times 10^{-5}[/tex]
R = gas constant = 0.0821 L.atm/mol.K
T = temperature = 700 K
[tex]\Delta n[/tex] = change in number of gaseous moles = Product moles - Reactant moles = (2+1) - (2) = 3 - 1 = 1 mol
[tex]K_c[/tex] = equilibrium constant
Now put all the given values in the above expression, we get:
[tex]K_p=K_c\times (RT)^{\Delta n}[/tex]
[tex]3.3\times 10^{-5}=K_c(0.0821\times 700)^{1}[/tex]
[tex]K_c=5.7\times 10^{-7}=5.7E^{-7}[/tex]
Therefore, the value of [tex]K_p[/tex] is, [tex]5.7E^{-7}[/tex]
The value of the equilibrium constant of the reaction (Kc) is [tex]5.74 \times 10^{-7}[/tex]
The given parameters;
equilibrium constant at constant pressure, [tex]K_p[/tex] = 3.3 x 10⁻⁵.gas constant, R = 0.0821 L.atm/mol.Ktemperature, T = 700 KThe change in the number of gaseous moles is calculated as follows;
[tex]\Delta n = product \ - reactant\\\\\Delta n = (2+1) - 2 = 1 \ mole[/tex]
The value of the equilibrium constant (Kc) is calculated as follows;
[tex]K_p = K_c \times (RT)^{\Delta n}\\\\3.3 \times 10^{-5} = K_c \times (0.0821 \times 700)^1\\\\3.3 \times 10^{-5} = 57.47K_c\\\\K_c = \frac{3.3 \times 10^{-5} }{57.47} \\\\K_c = 5.74 \times 10^{-7} \[/tex]
Thus, the value of the equilibrium constant of the reaction (Kc) is [tex]5.74 \times 10^{-7}[/tex]
"Your question is not complete, it seems to be missing the following information;"
The reaction below has a Kp value of 3.3 × 10⁻⁵. What is the value of Kc for this reaction at 700 K?
[tex]2SO_3(g) \ ---> \ 2SO_2(g) \ + \ O_2[/tex]
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Which of the following is one way that nuclear reactions differ from chemical
reactions?
Answer:
Nuclear reaction takes place at the nucleus whereas chemical reaction involves valence electrons
Explanation:
Can 1750 mL of water dissolve 4.6 moles of copper sulfate CuSO4
Answer:
No
Explanation:
We first find the solubility of the copper sulphate salt
Number of moles= 4.6 moles
Volume =1.75 L
Molarity= number of moles/ volume = 4.6/1.75= 2.6 molL-1
Hence, only 2.6 moles of Copper II sulphate dissolves in 1.75L of water. Hence 4.6 moles of copper II Sulphate dies not dissolve in 1.75 L of water.
list the 5 main factors that can affect the rates of a chemical reaction
Answer:
1.surface area of a solid reactant.2.concentration or pressure of a reactant.
3temperature.
4nature of the reactants.
5presence/absence of a catalyst.
Explanation:
The five main factors that can affect rates of a chemical reaction are concentration of reactants, temperature, pressure, presence of a catalyst, and surface area of solid reactants.
Explanation:The rate of a chemical reaction can be influenced by several factors, including:
Concentration of reactants: A higher concentration typically leads to an increased reaction rate because there are more molecules available to react.Temperature: Increasing the temperature often increases the reaction rate. This happens because higher temperatures give molecules more kinetic energy, so they collide more frequently and more forcefully.Pressure: Increasing the pressure of a gas reaction will increase the reaction rate because it effectively increases the concentration of the reactants.Presence of a catalyst: Catalysts lower the energy barrier for a reaction, making it easier for the reaction to proceed and thus increasing the reaction rate.Surface area of solid reactants: A larger surface area allows for more collisions between molecules, leading to faster reactions.Learn more about Chemical Reaction Rate Factors here:https://brainly.com/question/34204565
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This type of intermolecular force occurs when two molecules attract due to their electron density shifting to create opposite short-lived partial positives and negatives?
Answer:
London dispersion forces
Explanation:
The London dispersion force is the weakest kind of intermolecular force. The London dispersion force is a temporary attractive force that occurs when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. This force is sometimes called an induced dipole-induced dipole attraction.
These London dispersion forces are mostly seen in the halogens (e.g., Cl2 and I2), the noble gases (e.g., Xe and Ar), and in many non-polar molecules, such as carbon dioxide and propane. London dispersion forces are part of the van der Waals forces, and are very weak intermolecular attractions.
In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y 4 − Y4− ) and metal chelate (abbreviated MY n − 4 MYn−4 ) can buffer the free metal ion concentration at values near the dissociation constant of the metal chelate, just as a weak acid and a salt can buffer the hydrogen ion concentration at values near the acid dissociation constant. The equilibrium M n + + Y 4 − − ⇀ ↽ − MY n − 4 Mn++Y4−↽−−⇀MYn−4 is governed by the equation K ′ f = α Y 4 − ⋅ K f = [ MY n − 4 ] [ M n + ] [ EDTA ] Kf′=αY4−⋅Kf=[MYn−4][Mn+][EDTA] where K f Kf is the association constant of the metal and Y 4 − Y4− , α Y 4 − αY4− is the fraction of EDTA in the form Y 4 − Y4− , and [ EDTA ] [EDTA] is the total concentration of free (unbound) EDTA EDTA . K ′ f Kf′ is the conditional formation constant. How many grams of Na 2 EDTA ⋅ 2 H 2 O Na2EDTA⋅2H2O (FM 372.23 g/mol) should be added to 1.97 1.97 g of Ba ( NO 3 ) 2 Ba(NO3)2 (FM 261.35 g/mol 261.35 g/mol ) in a 500. mL volumetric flask to give a buffer with p Ba 2 + = 7.00 pBa2+=7.00 at pH 10.00? log K f logKf for Ba − EDTA Ba−EDTA is 7.88 7.88 and α Y 4 − αY4− at pH 10.00 is 0.30. mass Na 2 EDTA ⋅ 2 H 2 O.
Final answer:
To prepare the buffer, you would need to add 2.808 grams of Na2EDTA · 2H2O to the 500 mL volumetric flask containing 1.97 grams of Ba(NO3)2. This will give the desired buffer with a pBa2+ of 7.00 at pH 10.00.
Explanation:
To calculate the mass of Na2EDTA · 2H2O needed to prepare the buffer, we need to use the equation:
Kf' = αY4− · Kf = [MYn−4][Mn+][EDTA]
Given that log Kf = 7.88 and αY4− = 0.30, we can rearrange the equation to solve for the concentration of MYn−4:
[MYn−4] = Kf' / (αY4− · Kf) = 10^(7.00 − 7.88) / (0.30 · 10^7.88) = 0.562 M
Now, we can calculate the moles of Ba(NO3)2:
moles of Ba(NO3)2 = mass / molar mass = 1.97 g / 261.35 g/mol = 0.00754 mol
Since the stoichiometric ratio between Na2EDTA · 2H2O and Ba(NO3)2 is 1:1, the moles of Na2EDTA · 2H2O required is also 0.00754 mol.
Finally, we can calculate the mass of Na2EDTA · 2H2O:
mass = moles × molar mass = 0.00754 mol × 372.23 g/mol = 2.808 g
Warning signs are sometimes placed on aerosol cans to prevent people from throwing them into a fire. What would be true about the contents of the gas in an aerosol can just after it is placed in a fire?
Answer:
Volume stays.
Explanation:
If aerosol can throwing into the fire than the temperature of the aerosol can & the temperature of the contents of the gas in an aerosol will also increase, and it induces the pressure to rise.
Against the sides of the can, the gas molecules will smash rapidly with each other which are present inside an aerosol can, and the proportion of gas will stay the same as earlier. According to the law of conservation of matter, as a result of the heat, the gas particles can not be eradicated or expanded.
The total size of the container would need to be able to hold moles of gas.
Answer:
11.88
Explanation:
The capacity of container to be able to hold any amount of moles is dependent on size of container.
What is a mole?Mole is defined as the unit of amount of substance . It is the quantity measure of amount of substance of how many elementary particles are present in a given substance.
It is defined as exactly 6.022×10²³ elementary entities. The elementary entity can be a molecule, atom ion depending on the type of substance. Amount of elementary entities in a mole is called as Avogadro's number.
It is widely used in chemistry as a suitable way for expressing amounts of reactants and products.For the practical purposes, mass of one mole of compound in grams is approximately equal to mass of one molecule of compound measured in Daltons. Molar mass has units of gram per mole . In case of molecules, where molar mass in grams present in one mole of atoms is its atomic mass.
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Using the Brønsted-Lowry concept of acids and bases, identify the Brønsted-Lowry acid and base in each of the following reactions:
HSO3−(aq)+H2O(l)→H2SO3(aq)+OH−(aq)
(CH3)3N(g)+BCl3(g)→(CH3)3NBCl3(s)
Drag the appropriate items to their respective bins.
Answer:
1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)
The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻
2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)
There are no Brønsted-Lowry acids and bases in this reaction.
Explanation:
According to the Brønsted-Lowry concept, when an acid (HA) and a base (B) undergoes a chemical reaction, the acid (HA) loses a proton and forms its conjugate base (A⁻), whereas the base gains (B) the proton to form its conjugate acid (HB⁺).
The chemical equation for this reaction is:
HA + B ⇌ A⁻ + HB⁺
Given reactions:
1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)
The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻
Reason: In this reaction, the acid H₂O loses a proton and forms its conjugate base, OH⁻. Whereas, the base HSO³⁻ gains a proton to form its conjugate acid, H₂SO₃.
2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)
There are no Brønsted-Lowry acids and bases in this reaction.
Reason: In this reaction, there is no exchange of proton between the acid and the base.
The shells of marine organisms contain calcium carbonate CaCO3, largely in a crystalline form known as calcite. There is a second crystalline form of calcium carbonate known as aragonite. Physical and thermodynamic properties of calcite and aragonite are given below.
Properties
(T=298K, P=1atm) Calcite Aragonite
ΔH°f (kJ/mole) -1206.87 -1207.04
ΔG°f (kJ/mole) -1128.76 -1127.71
S° (J/mole K) 92.88 88.70
C°P (J/mole K) 81.88 81.25
Density (gm/mL) 2.710 2.930
A) Based on the thermodynamic data given, would you expect an isolated sample of
calcite at T=298K and P=1 atm to convert to aragonite, given sufficient time.
Explain.
B) What pressure must be achieved to induce the conversion of calcite to aragonite at
T=298K. Assume both calcite and aragonite are incompressible at T=298K.
C) Suppose the pressure is P=1.00 atm and the temperature is increased from
T=298K to T=400K. Based on this calculation, can isolated calcite be converted
to aragonite at P=1 atm if the temperature is increased? Explain.
D) Suppose the pressure is increased from P=1.00 atm to P=1000 atm and the
temperature is increased from T=298K to T=400K. Which form of CaCO3 is more
stable under these conditions?
Expert Answer
Answer:
Check the explanation
Explanation:
A. Calcite ------> Argonite
At 298 K and 1 atm,
\DeltaGrxn = \DeltaGarg - \DeltaGcalc
= -1127.71-(-1128.76) = 1.05 kJ/mol
\DeltaGrxn is positive, so the reaction is not spontaneous.
Thus, the reaction would not happen even if enough time is given.
B. Molar mass of CaCO3 = 100 g/mol
Volume = mass/density
Vcalc = 100/2.710 = 36.900 ml/mol = 36.9*10-3 L = 36.9*10-6 m3
Varg = 100/2.930 = 34.130 ml/mol = 34.13*10-3 L = 34.13*10-6 m3
\DeltaGrxn = 1.05 kJ/mol = 1.05*103 J/mol
\DeltaG = -\DeltaP\DeltaV
\DeltaP = -\DeltaG/\DeltaV = -1.05*103/(34.13*10-6 - 36.9*10-6) = 0.379*109 Pa = 3.79*108 Pa
The pressure should be more than 3.79*108 Pa.
C. For reaction, \DeltaSrxn = Sarg - Scalc
= 88.70-92.88 = -4.18 J/mol.K
\DeltaHrxn = -1207.04-(-1206.87) = -0.17kJ/mol
At 400 K, \DeltaG = \DeltaH-T\DeltaS
= -0.17-(400*-4.18)
= +ve
So, the reaction is not favorable on increasing the temperature to 400 K.
D. \DeltaT = 400-298 = 102 K
\DeltaP = 1000-1 = 999.0 atm = 999*106 Pa
\DeltaG = \DeltaS\DeltaT - \DeltaV\DeltaP
= (-4.18*102)-(-2.77*10-6*999*106)
= -426.36+2767.23 = 2340.87 J/mol
\DeltaG = +ve
calcite would be more stable.
A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba2+. When the concentration of F- exceeds ________ M, BaF2 will precipitate. Neglect volume changes. For BaF2, Ksp = 1.7
Answer:
When the concentration of F- exceeds 0.0109 M, BaF2 will precipitate.
Explanation:
Ba²⁺(aq) + 2 F⁻(aq) <----> BaF₂(s)
When BaF₂ precipitates, the Ksp relation is given by
Ksp = [Ba²⁺] [F⁻]²
[Ba²⁺] = 0.0144 M
[F⁻] = ?
Ksp = (1.7 × 10⁻⁶)
1.7 × 10⁻⁶ = (0.0144) [F⁻]²
[F⁻]² = (1.7 × 10⁻⁶)/0.0144 = 0.0001180555
[F⁻] = √0.0001180555 = 0.01086 M = 0.0109 M
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When the concentration of F⁻ exceeds 0.0109 M in a solution containing 0.0144 M Ba²⁺, BaF₂ will begin to precipitate.
To determine when BaF₂ will precipitate from a solution containing Ba²⁺ and F⁻ ions, we need to use the solubility product constant (Ksp). For BaF₂, Ksp is given as 1.7 × 10⁻⁶.
The dissociation reaction for BaF₂ is:
BaF₂ (s) ⇌ Ba²⁺ (aq) + 2F⁻ (aq)
The Ksp expression is:
Ksp = [Ba²⁺][F⁻]²
Given [Ba²⁺] = 0.0144 M, we need to find [F⁻] when the solution just starts to precipitate.
Substitute the concentrations into the Ksp expression: 1.7 × 10⁻⁶ = (0.0144)[F⁻]².Solving for [F⁻]: [F⁻] = √(1.7 × 10⁻⁶ / 0.0144).[F⁻] ≈ √(1.1806 × 10⁻⁴).[F⁻] ≈ 0.0109 M.Therefore, when the concentration of F⁻ exceeds approximately 0.0109 M, BaF₂ will start to precipitate.
A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource. Round your answer to 2 significant digits. 1% x 5 ?
Answer:
Percentage dissociated = 0.41%
Explanation:
The chemical equation for the reaction is:
[tex]C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}[/tex]
The ICE table is then shown as:
[tex]C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}[/tex]
Initial (M) 1.8 0 0
Change (M) - x + x + x
Equilibrium (M) (1.8 -x) x x
[tex]K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}[/tex]
where ;
[tex]K_a = 3.02*10^{-5}[/tex]
[tex]3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}[/tex]
Since the value for [tex]K_a[/tex] is infinitesimally small; then 1.8 - x ≅ 1.8
Then;
[tex]3.02*10^{-5} *(1.8) = {(x)(x)}[/tex]
[tex]5.436*10^{-5}= {(x^2)[/tex]
[tex]x = \sqrt{5.436*10^{-5}}[/tex]
[tex]x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M[/tex]
Dissociated form of 4-chlorobutanoic acid = [tex]C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M[/tex]
Percentage dissociated = [tex]\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100[/tex]
Percentage dissociated = [tex]\frac{7.3729*10^{-3}}{1.8 }*100[/tex]
Percentage dissociated = 0.4096
Percentage dissociated = 0.41% (to two significant digits)
Answer:
0.00091%
Explanation:
The fraction of 4-chlorobutanoic acid that would dissociatr in aqueos solution is a function of Ionization percentage. It is obtained by
Ka = [dissociated acid] / [original acid] x 100%
The equation of the reaction is
C₂HClCOOH + H20 (aq) = C₂HClCOO⁻ + H₃O⁺ pka =4.69
But pKa = - log Ka
4.69 = - Log Ka
10⁻⁴⁶⁹ = Ka
Taking the antilog of the equation we get the ionization constant
Ka = 0.0000204 M
=2.04 x 10⁻⁵M
At the beginning of the reaction we have the following concentrations
1.8M C₂HClCOOH : 0M (zero molar) C₂HClCOO⁻ : 0M (zero molar) H₃O⁺
At equilibrium, we have,
(1.8M -x) xM C₂HClCOO⁻ and xM H₃0⁺
Therefore,
Ka = [C2HClCOO-] [H30+] / [C2HClCOOH],
Inputing the value of Ka
Ka .[C2HClCOOH] = [CHCLCOO-] [H3O+]
0.0000204 (1.8 - X) = (x).(x)
x² = (0.00003672 - 0.0000204 X)
= (3.672 x10-5 - 2.04 x10⁻⁵X)
x² +2.04x10⁻⁵ x = 3.672 x 10⁻⁵
x² + 2.04 x 10⁻⁵x - 3.672 x 10⁻⁵ = 0
x = 0.00001632
= 1.632 x 10⁻⁵
Inputing back into equation 1
1.8 - x = [H3O+]
1.8 - 0.00001632 = 1.7999837
It therefore implies that only 0.00001632M of 4-chloroutanoic acid dissciated at equilibrium, we can now calculate the percentage dissociation by
Percentage dissociation = 0.00001632 / 1.8M x 100%
= (1.632 x 10⁻⁵/1.8 ) x 100%
= 0.00090667%
= 0.00091%
An insoluble solid is placed in water and the system allowed to reach equilibrium. The ratio of the rate at which ions join the solution and the rate at which ions join the lattice will be:
a) greater than one
b) one
c) less than one
d) depends on the substance
Answer:
One
Explanation:
Because it reaches an equibrum state so it's equals to one
different student is given a 10.0g sample labeled CaBr2 that may contain an inert (nonreacting) impurity. Identify a quantity from the results of laboratory analysis that the student could use to determine whether the sample was pure.
The student can check the purity of the CaBr2 sample by performing a quantitative chemical analysis to compare the theoretical and actual mass of the compound, which will reveal the presence of any impurities.
Explanation:A student can determine whether the 10.0g sample labeled CaBr2 is pure by calculating the molar mass of CaBr2 and comparing the theoretical mass to the actual mass of the sample. If there is a discrepancy between the theoretical and actual mass, it could indicate the presence of an inert impurity.
To carry out the analysis, the student must perform a quantitative chemical analysis, such as a titration, to determine the amount of active ingredient in the sample. The student could also measure the mass of CaBr2 after a precipitation reaction, as described in the problem statement. If the resulting mass corresponds to the known molar mass of a pure CaBr2 sample, then the sample can be considered pure. On the other hand, if the mass deviates from the expected value, then an impurity might be present.
the number of moles of O2 in 2.24 L of O2 gas.
calculate at STP
Answer:
0.1 mol O2
Explanation:
1 mol of any gas at STP = 22.4 L
2.24L/22.4L = 0.1
1mol * 0.1 = 0.1 mol
you have 0.1 mol of O2
The number of moles of [tex]O_2[/tex] is required.
The number of moles of [tex]O_2[/tex] is 0.2 moles.
V = Volume = 2.24 L
[tex]\rho[/tex] = Density of [tex]O_2[/tex] at STP = 1.429 g/L
M = Molar mass of [tex]O_2[/tex] = 16 g/mol
Density is given by
[tex]\rho=\dfrac{m}{V}\\\Rightarrow m=\rho V\\\Rightarrow m=1.429\times 2.24\\\Rightarrow m=3.20096\ \text{g}[/tex]
Molar mass is given by
[tex]M=\dfrac{m}{n}\\\Rightarrow n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{3.20096}{16}\\\Rightarrow n=0.2\ \text{moles}[/tex]
The number of moles of [tex]O_2[/tex] is 0.2 moles.
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A 0.500 g sample of He at STP has a volume that is one-half that of an unknown pure gas also at STP. The unknown pure gas sample has a mass of 36.5 g. What is the molar mass of the unknown gas?
The molar mass of the unknown gas which has a mass of 36.5 g is 146 g/mol
We'll begin by calculating the number of mole in 0.5 g of He.Mass of He = 0.5 g
Molar mass of He = 4 g/mol
Mole of He =.?[tex]Mole = \frac{mass}{molar mass} \\\\= \frac{0.5}{4}\\\\[/tex]Mole of He = 0.125 moleNext, we shall determine the volume occupied by 0.125 mole of He at stpAt standard temperature and pressure (stp),
1 mole of He = 22.4 L
Therefore,
0.125 mole of He = 0.125 × 22.4
0.125 mole of He = 2.8 L
Next, we shall determine the volume of the unknown gas.Volume of He = 2.8 L
Volume of unknown gas =?
Volume of He = ½ × Volume of unknown gas
2.8 = ½ × Volume of unknown gas
Cross multiply
Volume of unknown gas = 2.8 × 2
Volume of unknown gas = 5.6 L
Next, we shall determine the mole of the unknown gas that occupied 5.6 L at stp.22.4 L = 1 mole of unknown gas
Therefore,
5.6 L = [tex]\frac{5.6}{22.4}\\\\[/tex]
5.6 L = 0.25 mole of unknown gas.
Finally, we shall determine the molar mass of the unknown gas.Mole of unknown gas = 0.25 mole
Mass of unknown gas = 36.5 g
Molar mass of unknown gas =?
[tex]Molar mass = \frac{mass}{mole} \\\\ = \frac{36.5}{0.25}\\\\[/tex]
= 146 g/molTherefore, the molar mass of the unknown gas is 146 g/mol
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At STP, one mole of any ideal gas occupies around 22.4 L. As the 0.500 g Helium sample occupies half the volume of the unknown gas, it means the that the molar mass of the unknown gas would be twice the mass of the helium. The molar mass of the unknown gas is calculated to be 292 g/mol.
Explanation:In order to determine the molar mass of the unknown gas, we need to observe the characteristics of gases at Standard Temperature and Pressure (STP). At STP, one mole of any ideal gas will occupy approximately 22.4 L. Given that the 0.500 g sample of Helium with a molar mass of 4 g/mol occupies half of the volume the unknown gas does, we can conclude that one mole of the unknown gas will have twice the mass of the Helium sample.
So, if 0.500 g of He occupies a volume that is equal to the volume of 0.125 moles of He (since 4 g of He = 1 mole), this means the same volume is being occupied by the unknown gas. Given the mass of the unknown gas is 36.5 g, the molar mass of the unknown gas would be 36.5 g / 0.125 moles = 292 g/mol.
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17.131) A 24.0 −mL volume of a sodium hydroxide solution requires 19.5 mL of a 0.193 M hydrochloric acid for neutralization. A 11.0 −mL volume of a phosphoric acid solution requires 34.8 mL of the sodium hydroxide solution for complete neutralization. Calculate the concentration of the phosphoric acid solution
Answer:
The concentration of Phosphoric acid required for the neutralization described = 0.165 M
Explanation:
Given,
Volume of NaOH = V = 24.0 mL
Concentration of HCl = Cₐ = 0.193 M
Volume of HCl = Vₐ = 19.5 mL
NaOH + HCl -----> NaCl + H₂O
1 mole of NaOH reacts with 1 mole of HCl
Using the equivalence point expression
(CₐVₐ)/(CV) = (nₐ/n)
where
Cₐ = concentration of acid = 0.193 M
Vₐ = volume of acid = 19.5 mL
C = concentration of base = ?
V = volume of base = 24.0 mL
nₐ = Stoichiometric coefficient of acid in the balanced equation = 1
n = Stoichiometric coefficient of base in the balanced equation = 1
(CₐVₐ)/(CV) = (nₐ/n)
(0.193 × 19.5)/(C × 24) = 1
(C × 24) = 3.7635
C = (3.7635/24) = 0.157 M
This NaOH is then reacted with phosphoric acid.
Phosphoric acid = H₃PO₄
3NaOH + H₃PO₄ --------> Na₃PO₄ + 3H₂O
3 moles of NaOH reacts with 1 mole of Phosphoric acid.
Using the equivalence point expression
(CₐVₐ)/(CV) = (nₐ/n)
where
Cₐ = concentration of acid = ?
Vₐ = volume of acid = 34.8 mL
C = concentration of base = 0.157 M
V = volume of base = 11.0 mL
nₐ = Stoichiometric coefficient of acid in the balanced equation = 1
n = Stoichiometric coefficient of base in the balanced equation = 3
(CₐVₐ)/(CV) = (nₐ/n)
(Cₐ × 11)/(0.157 × 34.8) = (1/3)
Cₐ × 11 × 3 = 0.157 × 34.8 × 1
33Cₐ = 5.457
Cₐ = (5.457/32)
Cₐ = 0.165 M
Hence, the concentration of Phosphoric acid required for the neutralization described = 0.165 M
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Maria needs to dilute the stock of 16.0 M HCl solution for a lab.
How many mL of water must she add to 5.00 mL of 16.0 M HCI
solution to prepare a 1.00 M HCl solution? __ mL
Answer: 80mL
Explanation:
Its on ck12
5*16=80
To prepare a 1.00 M HCl solution from a 16.0 M HCl stock solution, Maria needs to add 75 mL of water to the 5.00 mL of the stock solution.
Explanation:To prepare a 1.00 M HCl solution from a 16.0 M HCl stock solution, Maria needs to dilute the stock. She should add water to the 5.00 mL of the stock solution. Let's calculate the volume of water she needs to add.
First, we can use the equation (stock concentration) × (stock volume) = (final concentration) × (final volume) to find the volume of the diluted solution. Plugging in the values, we have (16.0 M) × (5.00 mL) = (1.00 M) × (final volume). Solving for the final volume, we get:
final volume = (16.0 M × 5.00 mL) / 1.00 M
final volume = 80 mL
To find the volume of water Maria needs to add, we subtract the volume of the stock solution from the final volume:
volume of water = final volume - volume of stock solution = 80 mL - 5.00 mL = 75 mL
Therefore, Maria needs to add 75 mL of water to the 5.00 mL of 16.0 M HCl solution to prepare a 1.00 M HCl solution.
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Which of the following statements will decrease the amount of work the system could perform? (a) Add to the solution (b) Add solid NaOH to the reaction (assume no volume change) (c) Increase the concentration of the (d) Selectively remove (e) Add to the solution
Final answer:
The addition of solid NaOH to a reaction can decrease the amount of work a chemical system can perform by shifting the equilibrium position and altering the concentration of reactants and products involved in work
Explanation:
The statement that will decrease the amount of work the system could perform is (b) Add solid NaOH to the reaction (assume no volume change). In a chemical system, work is related to changing conditions that affect the direction of equilibrium according to Le Chatelier's principle. Adding solid NaOH to a reaction mixture could shift the equilibrium position in a way that may result in the consumption of reactants or products involved in performing work (such as electrical or mechanical work in an electrochemical cell or in muscle contraction). For example, if the reaction were exergonic and NaOH was a product, adding more of it would shift the equilibrium to the left, thereby reducing the amount of work the system could do. This is because there would be a greater proportion of reactants and less of the energy-containing products required to perform work.
What is the pH of a buffer that consists of 0.254 M CH3CH2COONa and 0.329 M CH3CH2COOH? Ka of propanoic acid, CH3CH2COOH is 1.3 x 10-5 Enter your answer with two decimal places.
Answer:
4.77 is the pH of the given buffer .
Explanation:
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=-\log[K_a]+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=-\log[K_a]+\log(\frac{[CH_3CH_2COONa]}{[CH_3CH_2COOH]})[/tex]
We are given:
[tex]K_a[/tex] = Dissociation constant of propanoic acid = [tex]1.3\times 10^{-5}[/tex]
[tex][CH_3CH_2COONa]=0.254 M[/tex]
[tex][CH_3CH_2COOH]=0.329 M[/tex]
pH = ?
Putting values in above equation, we get:
[tex]pH=-\log[1.3\times 10^{-5}]+\log(\frac{[0.254 M]}{[0.329]})[/tex]
pH = 4.77
4.77 is the pH of the given buffer .
Final answer:
To determine the pH of a buffer with given concentrations of propanoate and propanoic acid, apply the Henderson-Hasselbalch equation using the given Ka value for propanoic acid and the concentrations of the acid and its conjugate base.
Explanation:
To calculate the pH of a buffer consisting of 0.254 M sodium propanoate (CH3CH2COO-Na+) and 0.329 M propanoic acid (CH3CH2COOH), we use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pKa is the negative logarithm of the acid dissociation constant (Ka) for propanoic acid.[A-] is the concentration of the conjugate base, sodium propanoate.[HA] is the concentration of the weak acid, propanoic acid.First, we calculate the pKa:
pKa = -log(Ka) = -log(1.3 x 10-5)
Then plug the values into the Henderson-Hasselbalch equation:
pH = pKa + log(0.254/0.329)
After calculating the above expression, we find the pH of the buffer solution.
Be sure to answer all parts.A sample of natural gas contains 6.816 moles of methane (CH4), 0.589 moles of ethane (C2H6), and 0.381 moles of propane (C3H8). If the total pressure of the gases is 3.93 atm, what are the partial pressures of the gases?
Answer:
Explanation:
Given parameters:
number of moles of CH₄ = 6.816 moles
number of moles of C₂H₆ = 0.589 mole
number of moles of C₃H₈ = 0.381 moles
Total pressure of the gases = 3.93 atm
Unknown:
Partial pressure of the gases = ?
Solution:
Dalton's law of partial pressure states that " the total pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases".
P[tex]_{T}[/tex] = P[tex]_{1}[/tex] + P[tex]_{2}[/tex] + P[tex]_{3}[/tex]...............
where P[tex]_{T}[/tex] = total pressure of the gas mixture
P₁, P₂, P₃...... = partial pressure of gas 1, 2, 3............
The partial pressure of a gas is the pressure it would exert if confined alone in the volume of the gas mixture.
Partial pressure of gas = mole fraction of gas x total pressure of mixture
Now let us solve the problem.
Total number of moles of gases = 6.816 moles + 0.589 mole + 0.381 moles
= 7.786moles
Partial pressure of CH₄ = [tex]\frac{6.816}{7.786} x 3.93 = 3.44 atm[/tex]
Partial pressure of C₂H₆ = [tex]\frac{0.589}{7.786}[/tex] x 3.93 = 0.30atm
Partial pressure of C₃H₈ = [tex]\frac{0.381}{7.786}[/tex] x 3.93 = 0.19atm
If the hydroxide ion concentration is 10-7 M, what is the pH of the solution
Answer:
pH= 7
Explanation:
pOH=-log [OH-] = -log (10^-7)= 7
pH= 14-pOH= 14-7
=7
Why did hurricane Katrina slow down at data point 7 ?
Answer:
Because the hot air from the equator is balance with the cold air from the polar region, meaning the temperature is the right degree, therefore it causes the slowing down of that hurricane.
Explanation:
From your science class you do study the convectional current right? that's what happen on the outside real life
As more than just a result of the equator's hot air being in balance with the poles' cold air, which means the temperature is at the right level, the hurricane is slowed down.
What is hurricane ?Hurricanes are low-pressure systems with organized thunderstorm activity that originate over tropical or subtropical waters. They are sometimes referred to as tropical cyclones in general. The warm ocean waves provide them energy.
When warm, humid air begins to rise over water, hurricanes emerge. Cooler air replaces the rising air. Large clouds and thunderstorms continue to grow as a result of this process. Thanks to the earth's Coriolis Effect, these thunderstorms are still expanding and starting to rotate.
The greatest wind speeds during Katrina's impact near Grand Isle, Louisiana, may have reached 140 mph. The NWS Doppler Radar at Mobile (KMOB) recorded winds of up to 132 mph between 3,000 and 4,000 feet above ground level in the morning as Katrina moved farther north and made a second impact along the Mississippi/Louisiana border.
Thus, the equator's hot air being in balance with the poles' cold air, which means the temperature is at the right level, the hurricane is slowed down.
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An ideal gas is confined to a cylinder by a piston. The piston is slowly pushed in so that the gas temperature remains at 20 degrees Celsius. During the compression, 730 J of work is done on the gas. Part A) Calculate the entropy change of the gas. Part B) Describe clearly why isn’t the result a violation of the entropy statement of the second law, ΔS 0 ?
Answer:
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Explanation:
(a) It is given that the gas is ideal. Formula for change in entropy of the gas according to the first law of thermodynamics is as follows.
[tex]\Delta U = dQ - dW[/tex]
For isothermal process, [tex]\Delta U = 0[/tex] at constant temperature.
So, [tex]\Delta U = dQ - dW[/tex]
[tex]0 = dQ - dW[/tex]
or, dQ = dW = 730 J
Now, according to the second law of thermodynamics the entropy change is as follows.
[tex]\Delta S = \frac{dQ}{dT}[/tex]
= [tex]\frac{730 J}{293.15 K}[/tex]
= 2.490 J/K
Therefore, the entropy change of the gas is 2.490 J/K.
(b) In the given process, at constant temperature the gas will be compressed slowly because then kinetic energy of the gas molecules will also be constant. The volume decreases so that the movement of molecules increases as a result, entropy of molecules will also increase.
This means that [tex]\Delta S = 2.490 J/K > 0[/tex]
An ideal gas is a hypothetical gas in which there are no intermolecular attractions between the molecules of a gas. The collision between the molecules is perfectly elastic.
The answers are as follows:
(a) The ideal gas is given, in which the change of entropy of the gas can be calculated by the first law of thermodynamics.
[tex]\Delta[/tex] U = dQ - dW
In an isothermal process, the \Delta U = 0 at constant temperature, such that:
[tex]\Delta[/tex] U = dq - dW
0 = dQ - dW
dQ = dW = 730 J (given)
Now, based on the second law of thermodynamics, the entropy change will be:
[tex]\Delta S = \dfrac {\text {dQ}}{\text {dW}}\\\\\Delta S = \dfrac {730 \text J}{293.15 \text K}[/tex]
[tex]\Delta[/tex]S = 2.490 J/K
Thus, the entropy change for the gas is 2.490 J/K.
(b) At constant temperature, the gas will be compressed slowly because the kinetic energy of the molecules will be constant. The decrease in the volume will be due to the increase in the movement of the molecules, which will cause an increase in the entropy. This means [tex]\Delta[/tex] S will be greater than 2.490 J/K.
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The common laboratory solvent diethyl ether (ether) is often used to purify substances dissolved in it. The vapor pressure of diethyl ether , CH3CH2OCH2CH3, is 463.57 mm Hg at 25 °C. In a laboratory experiment, students synthesized a new compound and found that when 7.745 grams of the compound were dissolved in 159.9 grams of diethyl ether, the vapor pressure of the solution was 457.87 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight of this compound?
Answer: The molecular weight of this compound is 288.4 g/mol
Explanation:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure
i = Van'T Hoff factor = 1 (for non electrolytes)
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\text {moles of solute}}{\text {total moles}}[/tex]
Given : 7.745 g of compound is present in 159.9 g of diethyl ether
moles of solute = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.745g}{Mg/mol}[/tex]
moles of solvent (diethyl ether) = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{159.9g}{74.12g/mol}=2.157moles[/tex]
Total moles = moles of solute + moles of solvent = [tex]\frac{7.745g}{Mg/mol}+2.157[/tex]
[tex]x_2[/tex] = mole fraction of solute =[tex]\frac{\frac{7.745g}{Mg/mol}}{\frac{7.745g}{Mg/mol}+2.157}[/tex]
[tex]\frac{463.57-457.87}{463.57}=1\times \frac{\frac{7.745g}{Mg/mol}}{\frac{7.745g}{Mg/mol}+2.157}[/tex]
[tex]M=288.4g/mol[/tex]
Thus the molecular weight of this compound is 288.4 g/mol