Answer:
The osmotic pressure of ocean water is about 27 atm.
Explanation:
Pure water is water that contains no impurities. Ocean water is 96.5% pure with only about 3.5% of its content, salt water.
Osmotic pressure occurs when solutions that have different concentrations are isolated by a membrane. This osmotic pressure makes water move towards the solution that has the highest concentration, which means that if the concentration or temperature of the solution is high, the osmotic pressure becomes higher.
The equation for osmotic pressure is pi = iMRT.
What states can electrons exist in? A. Electron clouds or energy levels B. Positive and negative C. Up and down spin D. In phase and out of phase
Answer:
A. Electron clouds or energy levels
Explanation:
Electrons can exist in two states:
Stablized in electronic orbitalsFreely movingElectrons can exist in an electron cloud or energy level. Electron in an atoms have ability to change energy levels either by emitting or absorbing a photon that form the energy equal to the energy difference between the two levels.
Hence, the correct answer is A.
Answer:
Up and DOWN spin
Explanation:
Which molecule or ion has a trigonal planar shape?
Answer:B
Explanation: A P E X
oxygen get stable configuration by ____________two electrons
please give the answer as fast as you can
please
Answer:
gaining two electrons
Explanation:
electron configuration
2:6
so add two to 6 to get stable 2:8
What's the electron configuration of a Ca+2 ion?
A. [Kr]
B. [Ar]
C. [Ne]
D. [He]
Answer:
B
Explanation:
The Ca+2 ion means that 2 electrons have been given away. So when you try and find the answer, you have to count backwards from Calcium. When you do, you get K+ first and then Argon which is either column 8 orc column 18 depending on your periodic table.
The element you hit is Argon.
The answer is B
Answer:
B ar
Explanation:
pen foster answer
Based on their molecular structure, identify the stronger acid from each pair of oxyacids. Match the words in the left column to the appropriate blanks in the sentences on the right.
1) HI is a stronger acid than H2Te because iodine____than tellurium.
2) H2Te is a stronger acid than H2S because the H-Te bond is_____.
3) NaH is not acidic because hydrogen____than sodium.
a. has a more negative electron afflity
b. is more electronegative
c. has a larger atomic radius
d. stronger
e. is harder to ionize
Answer:
1)is more electronegative
2)
3) is more electronegative
Explanation:
1) for the first question, iodine is more electronegative than tellurium hence we naturally expect that HI should be more acidic than H2Te since electronegativities play a role in the acidity of chemical species.
2) the correct option is not listed because the H2Te bond is weaker than the H2S bond. This makes it easier for H2Te to dissociate releasing H^+ , thereby being more acidic than H2S.
3) Hydrogen is more electronegative than sodium hence it cannot be ionized thus NaH is not acidic.
Choose the substance with the lowest boiling point.
A. NBr3.
B. CI2H2.
C. H2O2.
D. H2S.
E. O2.
Answer:
E. O2
Explanation:
All substances has a simple molecular structure, where between their molecules are held by van der Waals' forces. But C must be incorrect because between the H2O2 molecules, they are mainly held by hydrogen bonds on top of van der Waals' forces. Hydrogen bonds are stronger than van der Waals' forces, so more energy is required to separate the H2O2 molecules.
In structures A and D, the molecules are polar. Their van der Waals' forces are stronger than Cl2H2 and O2, which are non-polar.
Between the Cl2H2 and O2, O2 has a smaller molecular size. The van der Waals' forces between the O2 molecules are hence the weakest. Least amount of energy is required to break the intermolecular forces between the O2 molecules therefore it has the lowest boiling point.
Determine the number of moles of the anhydrous salt present after heating, assuming that the contents of the aluminum cup after heating are pure anhydrous KAl(SO 4 ) 2 .
Answer:
0.2 moles, assuming weight of dried salt
Explanation:
In order to determine the number of moles, we need to be aware of the mass of the substance in question.
Assuming the mass of the dehydrated [tex]KAl(SO_{4} )_{2}.H_{2} O[/tex] is 50g.
No. of moles = mass of substance/ molar mass of the substance.
= [tex]\frac{50g}{39+27+32*2+16*4*2\\)g/mol}[/tex]
= 0.2 moles moles.
Interpret the following equation for a chemical reaction using the coefficients given:
Cl2(g) + F2(g) 2ClF(g)
On the particulate level:
________ of Cl2(g) reacts with ______ of F2(g) to form______ of ClF(g).
On the molar level:
______ of Cl2(g) reacts with______ of F2(g) to form______ of ClF(g).
Answer and Explanation:
Given the following chemical equation:
Cl₂(g) + F₂(g) ⇒ 2ClF(g)
The coefficients are: 1 for Cl₂, 1 for F₂ and 2 for ClF. The coefficients indicate the number of units of each ompound that participates in the reaction. It gives the proportion of reactants and products in the reaction. These units can be molecules or moles. In this reaction, we can say:
On the particulate level: 1 molecule of Cl₂(g) reacts with 1 molecule of F₂(g) to form 2 molecules of ClF(g).
On the molar level: 1 mol of Cl₂(g) reacts with 1 mol of F₂(g) to form 2 mol of ClF(g).
Which correctly lists the three land uses that the Bureau of Land Management was originally created to manage? mining, recreation, wildlife refuges recreation, developing oil and gas, battlefields grazing, mining, developing oil and gas developing oil and gas, battlefields, wildlife refuges
Answer: C
Explanation:
Right on edge 2020
The Bureau of Land Management was originally created to manage land for grazing, mining, developing oil and gas.
What is land management?Land management refers to the activities which are done in order to protect and preserve the land as well the resources found on land.
The Bureau of Land Management was created to manage land in the US.
The Bureau of Land Management was originally created to manage land for grazing, mining, developing oil and gas.
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Assume that a nickel weighs exactly 5.038650 g for the sets of weights listed below obtained by a single weighing on the balance below
Answer:
afshkkyfugutuiryfyi
Calculate Delta G for each reaction using Delta Gf values: answer kJ ...thank you
a) H2(g)+I2(s)--->2HI(g)
b) MnO2(s)+2CO(g)--->Mn(s)+2CO2(g)
c) NH4Cl(s)--->NH3(g)+HCl(g)
Answer:
a) [tex]\Delta G=2.6kJ[/tex]
b) [tex]\Delta G=-979.57kJ[/tex]
c) [tex]\Delta G=264.21kJ[/tex]
Explanation:
Hello,
In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:
[tex]\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol[/tex]
So we proceed as follows:
a)
[tex]\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ[/tex]
b)
[tex]\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ[/tex]
c)
[tex]\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ[/tex]
Regards.
Sample gas has a volume of 3.40 L at 10°C what will be its volume at 100°C pressure remaining constant
Answer:
V2 = 4.48L
Explanation:
using charles law
V1/T1=V2/T2
3.4/283=V2/373
0.012=V2/373
V2= 0.012 x 373
V2 = 4.48L
Which of these are elimination reactions? Check all that apply.
CH3OH + CH3COOH → CH3CO2CH3 + H20
C3H7OH → C3H6 + H20
H9C2Br + NaOH → C2H4 + NaBr + H20
Answer:
C3H7OH → C3H6 + H20
Explanation:
If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.
Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.
Answer:
B and C. Just finished my lesson on Edge.
A saturated solution of lead(II) iodide, PbI2 has an iodide concentration of 3.0 x 10^-3 mol/L.
a) What is the molar solubility of PbI2?
b) Determine the solubility constant, Ksp, for lead(II) iodide.
c) Does the molar solubility of lead (II) iodide increase, decrease, or remain unchanged with the addition of potassium iodide to the solution? EXPLAIN.
Answer:
a) 1.5 x 10^-3 mol/L
b) 1.35×10^-8
c) decrease
Explanation:
The solubility of lead II iodide is given by the equation;
PbI2(s) -----> Pb^2+(aq) + 2I^-
By looking at the ICE table, I^-=2x= 3.0 x 10^-3 mol/L/2 = 1.5×10^-3 mol/L
Hence molar solubility of PbI2 = 1.5 x 10^-3 mol/L
Ksp= [Pb^2+] [2I^-]^2 =
Let the molar solubility of each ion be x, therefore;
Ksp= 4x^3
Ksp= 4(1.5 x 10^-3 mol/L)^3= 1.35×10^-8
Addition of kI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI2 in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le Chateliers principle.
a) The molar solubility of PbI₂ is [tex]1.5 * 10^{-3} mol/L[/tex]
b) The solubility constant is [tex]1.35*10^{-8}[/tex]
c) The molar solubility of lead (II) will decrease.
Molar Solubility:The solubility of lead II iodide is given by the equation;
[tex]PbI_2(s) ----- > Pb^{2+}(aq) + 2I^-[/tex]
By looking at the ICE table,
[tex]I^-=2x= 3.0 * 10^{-3} mol/L/2 =[/tex] [tex]1.5 * 10^{-3} mol/L[/tex]
Hence, molar solubility of PbI2 = [tex]1.5 * 10^{-3} mol/L[/tex]
[tex]Ksp= [Pb^{2+}] [2I^-]^2[/tex]
Let the molar solubility of each ion be x, therefore;
[tex]Ksp= 4x^3\\\\Ksp= 4(1.5 * 10^{-3} mol/L)^3\\\\Ksp= 1.35*10^{-8}[/tex]
The addition of KI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI₂ in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le- Ch-ateliers principle.
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The free energy obtained from the oxidation (reaction with oxygen) of glucose (C6H12O6) to form carbon dioxide and water can be used to re-form ATP by driving the above reaction in reverse. Calculate the standard free energy change for the oxidation of glucose.
Answer:
The correct answer is -2878 kJ/mol.
Explanation:
The reaction that takes place at the time of the oxidation of glucose is,
C₆H₁₂O₆ (s) + 6O₂ (g) ⇒ 6CO₂ (g) + 6H₂O (l)
The standard free energy change for the oxidation of glucose can be determined by using the formula,
ΔG°rxn = ∑nΔG°f (products) - ∑nΔG°f (reactants)
The ΔG°f for glucose is -910.56 kJ/mol, for oxygen is 0 kJ/mol, for H2O -237.14 kJ/mol and for CO2 is -394.39 kJ/mol.
Therefore, ΔG°rxn = 6 (-237.14) + 6 (-394.39) - (-910.56)
ΔG°rxn = -2878 kJ/mol
Question 7 options: The cell potential of an electrochemical cell made of an Fe, Fe2 half-cell and a Pb, Pb2 half-cell is _____ V. Enter your answer to the hundredths place and do not leave off the leading zero, if needed.
Answer: Thus the cell potential of an electrochemical cell is +0.28 V
Explanation:
The calculation of cell potential is done by :
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Fe^{2+}/Fe]}= -0.41V[/tex]
[tex]E^0_{[Pb^{2+}/Pb]}=-0.13V[/tex]
As Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative). Thus iron acts as anode and lead acts as cathode.
[tex]E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Fe^{2+}/Fe]}[/tex]
[tex]E^0=-0.13- (-0.41V)=0.28V[/tex]
Thus the cell potential of an electrochemical cell is +0.28 V
The volume of a sample of oxygen is 300mL when the pressure is 1 atm and the temperature is 27 C . At what temperature is the volume 1.00 L and the pressure.500 atm?
Answer:
T2 = 500K
Explanation:
Given data:
P1 = 1atm
V1 = 300ml
T1= 27 + 273 = 300K
T2 = ?
V2 = 1.00ml
P2 = 500atm
Apply combined law:
P1xV1//T1 = P2xV2/T2 ...eq1
Substituting values into eq1:
1 x 300/300 = 500 x 1/T2
Solve for T2:
300T2 = 500 x 300
300T2 = 150000
Divide both sides by the coefficient of T2:
300T2/300 = 150000/300
T2 = 500K
An analytical laboratory balance typically measures mass to the nearest 0.1 mg. You may want to reference (Page) Section 21.6 while completing this problem. Part A What energy change would accompany the loss of 0.1 mg in mass
Answer:
The energy change is [tex]E = 9.0 *10^{9}\ J[/tex]
Explanation:
From the question we are told that
Mass loss is [tex]m_l = 0.1 \ mg = 0.1 *10^{-3} mkg = 0.1 *10^{-6} \ kg[/tex]
Generally the energy change that would accompany this loss is mathematically represented as
[tex]E = m * c^2[/tex]
Where c is the speed of light with values [tex]c = 3.0*10^{8} \ m/s[/tex]
[tex]E = 0.1 *10^{-6} * [3.0 *10^{8}]^2[/tex]
[tex]E = 9.0 *10^{9}\ J[/tex]
Identify the Lewis acid and Lewis base from among the reactants in each of the following equations. Match the words in the left column to the appropriate blanks in the sentences on the right.
1. Fe3+ (aq)+6CN (aq) Fe(CN) (aq)______is the Lewis acid and_____is the Lewis base. is the Lewis
2. CI- (aq) + AlCl3 (aq) AlCl4-____is the Lewis acid and______is the Lewis base.
3. AlBr3 + NH3 H3NAlBr3______is the Lewis acid and______is the Lewis base.
A. AlCl3
B. CN-
C. AlBr3
D. Cl-
E. NH3
F. Fe3+
Answer:
1. Lewis acid: F. Fe₃⁺, Lewis base: B. CN⁻
2. Lewis acid: A. AlCl₃, Lewis base: D. Cl⁻
3. Lewis acid: C. AlBr₃, Lewis base: E. NH₃
Hope this helps.
The Lewis acid is chemical substance which possesses an empty orbital and accepts an electron pair from a Lewis base ( donor ), in order to create a Lewis adduct ( molecule created from the bonding of Lewis base and acid ).
The Lewis acid from reaction 1 is Fe₃⁺ while the Lewis base is CN⁻ also the Lewis acid from reaction 2 is AICI₃ while the Lewis base is CI⁻
Hence we can conclude that the Lewis acids and Lewis bases of the reactions in the question are as listed above.
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Calculate the frequency (Hz) and wavelength (nm)
of the emitted photon when an electron drops from
the n = 4 to the n=2 level in a hydrogen atom
Answer:
wavelength, λ = 486.6 nm
frequency, f = 6.16 * 10¹⁴ Hz
Explanation:
a. Wavelength
Using the wavelength equation; 1/λ = (1/hc) * 2.18 * 10⁻¹⁸ J * (1/nf² - 1/ni²)
Where nf is the final energy level; ni is the initial energy level; h is Planck's constant = 6.63 * 10⁻³⁴ J.s; c is velocity of light = 3 * 10⁸ m/s
1/λ = 1/(6.63 * 10⁻³⁴ J.s * 3 * 10⁸ m/s) * 2.18 * 10⁻¹⁸ J * (1/2² - 1/4²)
1/λ = 2.055 * 10⁶ m
λ = 4.866 * 10⁻⁷ m
wavelength, λ = 486.6 nm
b. Frequency
Using f = c/λ
f = (3 * 10⁸ m/s) / 4.866 * 10⁻⁷ m
frequency, f = 6.16 * 10¹⁴ Hz
True or False: Adding 4.18 joules to water will increase the temperature more than adding 1 calorie to water.
Answer:
Because one calorie is equal to 4.18 J, it takes 4.18 J to raise the temperature of one gram of water by 1°C. In joules, water's specific heat is 4.18 J per gram per °C. If you look at the specific heat graph shown below, you will see that 4.18 is an unusually large value.
The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. Assume that the volume of the solutions are additive . What would be the Ka for NH4
Answer:
Following are the answer to this question:
Explanation:
The value of pH solution is =5.17 So, the p^{OH}:
[tex]p^{OH}[/tex]=14-56.17
=8.823
The volume of the [tex]NH_{3}[/tex] = 40.00 ml
convert into the liter= 0.040L
The value of the concentrated [tex]NH_{3}[/tex] =0.10 M
The volume of the [tex]NH_{4}Cl[/tex]= 50.00 ml
convert into the liter= 0.050L
The value of concentrated [tex]NH_{4}Cl[/tex]= 0.10 M
The volume of the [tex]H_{2}So_{4}[/tex]= 30 ml
convert into the liter= 0.030L
The value of concentrated [tex]H_2So_4[/tex]=0.05 M
Calculating total volume=(0.40+0.050+0.030)
=0.120 L
calculating the new concentrated value of [tex]NH_3[/tex] = [tex]\frac{0.10\times 0.040}{0.120}= 0.33 \ M[/tex]
calculating the new concentrated value of [tex]NH_4Cl[/tex]= [tex]\frac{0.050\times 0.10}{0.120}= 0.04166 \ M[/tex]calculating the new concentrated value of [tex]H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M[/tex] when 1 mol [tex]H_2So_4[/tex] produced 2 mols [tex]H^{+}[/tex] so, 0.0125 in [tex]H_2So_4[/tex]produced:
[tex]=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}[/tex]
create the ICE table:
[tex]NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}[/tex]
I (m) 0.033(m) 0.025 0.04166
C -0.025 -0.025 + 0.025
E 8.3\times 10^{-3} 0 0.0667
now calculating pH:
when ph= 8.83:
[tex]P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769[/tex]
A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Answer:
V2 = 17371.43ml
Explanation:
We use Boyles laws
since temperature is constant
P1V1=P2V2
760 x 400 = 17.5 x V2
304000 = 17.5 x V2
V2 = 304000/17.5
V2 = 17371.43ml
The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at 760 torrs will be 18 ml.
What is vapor pressure?
The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.
The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -
P1 V1 / T1 = P2 V2 / T1
here, P = pressure
T = temperature
V = volume
substituting the value in the equation,
400 ×760 / 20 = 17.5× V / 20
V = 400× 760 / 20 × 17.5 / 20
V = 18 ml
Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.
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plez hurry Which is an important safety precaution that should be taken during a tornado? Stay away from doors and windows. Move to high ground to avoid flood waters. Try to avoid the storm by driving or running. Stay outside to avoid being trapped in a building.
Answer: stay away from doors and windows.
Explanation:
to aviod geting hit by glass
Answer:
Stay away from doors and windows.
Explanation:
Always stay in the center of the room during a tornado storm. Avoid windows, doors, and corners. If you’re near a window, the glass can shatter and hurt you.
Nitrogen has different oxidation states in the following compounds: nitrite ion, nitrous oxide, nitrate ion, ammonia, and nitrogen gas. Arrange these species in order of increasing nitrogen oxidation state. Select the correct answer below: A. ammonia, nitrogen gas, nitrite, nitrous oxide, nitrate B. nitrogen gas, ammonia, nitrous oxide, nitrite, nitrate C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate D. ammonia, nitrogen gas, nitrate, nitrite, nitrous oxide
Answer:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
Explanation:
To establish the oxidation number of nitrogen in each compound, we know that the sum of the oxidation numbers of the elements is equal to the charge of the species.
Nitrite ion (NO₂⁻)
1 × N + 2 × O = -1
1 × N + 2 × (-2) = -1
N = +3
Nitrous oxide (NO)
1 × N + 1 × O = 0
1 × N + 1 × (-2) = 0
N = +2
Nitrate ion (NO₃⁻)
1 × N + 3 × O = -1
1 × N + 3 × (-2) = -1
N = +5
Ammonia (NH₃)
1 × N + 3 × H = 0
1 × N + 3 × (+1) = 0
N = -3
Nitrogen gas (N₂)
2 × N = 0
N = 0
The order of increasing nitrogen oxidation state is:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the actual average molarity you measured in the standardization. What do you notice
Answer:
0.1M solution of NaOH
Explanation:
1 mole of NaOH - 40g
? moles - 1 g = 1/40 = 0.025 moles.
Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume
= 0.025/0.25
= 0.1M.
Which of the following solutions would have the highest pH? Assume that they are all 0.10 M in acid at 25°C. The acid is followed by its Ka value.
a. HCHO2, 1.8 x 10-4
b. HF, 3.5 x 10-4
c. HClO2, 1.1 x 10-2
d. HCN, 4.9 x 10-10
e. HNO2, 4.6 x 10-4
Answer:
[tex]HCN~~Ka=4.9x10^-^1^0[/tex]
Explanation:
In this case, we have to remember the relationship between the Ka value and the pH. We can use the general reaction for any acid with his Ka value expression:
[tex]HA~->~H^+~+~A^-[/tex] [tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
In the Ka expression, we have a proportional relationship between Ka and the concentration of [tex]H^+[/tex]. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher
So, if we have to find the higher pH value we need to search the smaller Ka value in this case [tex]HCN~~Ka=4.9x10^-^1^0[/tex].
I hope helps!
HCN has the highest pH among all the acids listed in the question.
The Ka is called the acid dissociation constant. It shows the extent to which an acid is ionized in water. The pH shows the hydrogen ion concentration of water. The higher the Ka, the higher the hydrogen ion concentration and the lower the pH.
Hence, HCN has the lowest Ka and the lowest hydrogen ion concentration. Therefore, HCN has the highest pH among all the acids listed in the question.
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Q1. Calculate the amount of copper produced in 1.0 hour when aqueous CuBr2 solution was electrolyzed by using a current of 4.50 A. Q2. In another electroplating experiment, if electric current was passed for 3 hours and 2.00 g of silver was deposited from a AgNO3 solution, what was the current used in amperes
Answer:
[tex]\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}[/tex]
Explanation:
Q1. Mass of Cu
(a) Write the equation for the half-reaction.
Cu²⁺ + 2e⁻ ⟶ Cu
The number of electrons transferred (z) is 2 mol per mole of Cu.
(b) Calculate the number of coulombs
q = It
[tex]\text{t} = \text{1.0 h} \times \dfrac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} \times \text{ 3600 s} = \textbf{10 800 C}[/tex]
(c) Mass of Cu
We can summarize Faraday's laws of electrolysis as
[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\& = &\dfrac{10 800 \times 63.55}{2 \times 96 485}\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}[/tex]
Note: The answer can have only two significant figures because that is all you gave for the time.
Q2. Current used
(a) Write the equation for the half-reaction.
Ag⁺ + e⁻ ⟶ Ag
The number of electrons transferred (z) is 1 mol per mole of Ag.
(a) Calculate q
[tex]\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\2.00& = &\dfrac{q \times 107.87}{1 \times 96 485}\\\\q &=& \dfrac{2.00 \times 96485}{107.87}\\\\& = & \textbf{1789 C}\\\end{array}[/tex]
(b) Calculate the current
t = 3 h = 3 × 3600 s = 10 800 s
[tex]\begin{array}{rcl}q&=& It\\1789 & = & I \times 10800\\I & = & \dfrac{1789}{10800}\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}[/tex]
Note: The answer can have only one significant figure because that is all you gave for the time.
Aluminum and oxygen react according to the following equation: 4Al + 3O2 -> 2Al2O3 In a certain experiment, 4.6g Al was reacted with excess oxygen and 6.8g of product was obtained. What was the percent yield of the reaction?
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
4 moles of Al produce 2 moles of Al₂O₃
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.0852 moles of Al₂O₃,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
8.7g of Al₂O₃ can be produced (Theoretical yield)Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
78.2%Explain why o-vanillin does not fully protonate p-toluidine. Reference appropriate pKa values and include a balanced chemical reaction and an appropriate reaction arrow in your answer.
Answer:
Here's what I get
Explanation:
pKₐ of o-vanillin = 7.81; pKₐ of p-toluidine = 4.44
The higher the pKₐ, the weaker the acid.
Thus, o-vanillin is the weaker acid and has a stronger conjugate base.
The conjugate acid of p-toluidine is the stronger and has the weaker conjugate base.
The equation for the equilibrium is
H-OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₂ ⇌ ⁻OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₃⁺
weaker acid weaker base stronger base stronger acid
The reaction between the stronger acid and the stronger base pushes the position of equilibrium to the left.
Thus, o-vanillin does not fully protonate p-toluidine.
O-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base; hence, o-vanillin does not fully protonate p-toluidine.
The pKa is defined as the negative logarithm of Ka. The dissociation constant of an acid Ka shows the extent of dissociation of an acid in solution. The higher the pKa, the lower the Ka and the weaker the acid.
The pKₐ of o-vanillin is 7.81 while the pKₐ of p-toluidine is 4.44. This means that o-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base. Hence, o-vanillin does not fully protonate p-toluidine.
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