Answers
Answer:
See the answer below
Explanation:
The complete question can be seen in the attached image.
Phenolphthalein is an indicator that is often utilized in an acid-base reaction to indicate the endpoints of such reactions due to its ability to change color from pink/colorless to colorless/pink depending on if the final solution is acidic or basic.
Phenolphthalein is usually colorless in acidic solutions and appears pink in basic solutions. The more basic or alkaline a solution is, the stronger the pink color of phenolphthalein. Hence;
1. Ammonia with a pH of 11 is basic, phenolphthalein will turn pink.
2. Battery acid with a pH of 1 is acidic, it will remain colorless.
3. Lime juice with a pH of 2 is acidic, it will remain colorless.
4. Mashed avocado with a pH of 6.5 is acidic, it will remain colorless.
5. Seawater with a pH of 8.5 is basic, it will turn pink.
6. Tap water with a pH of 7 is neutral, it will remain colorless
Phenolphthalein is a chemical compound with the formula[tex]C_{20}H_{14}O_4[/tex]. Phenolphthalein is often used as an indicator in acid-base titrations. For this application, it turns colorless in acidic solutions and pink in basic solutions
Phenolphthalein works as in:-
Colorless in acidPink in baseAccording to the question, There are 5 solutions having different ph and the indication only turns basic solution to pink
The indicator only turn the basic solution pink and these solutions are as follows,
AmmoniaSea waterTap water.Hence, these are the answer.
For more information, refer to the link:-
Related Questions
Chlorine monoxide and dichlorine dioxide are involved in the catalytic destruction of stratospheric ozone. They are related by the equation:
2ClO(g) ⇌ Cl2O2(g) for which Kc is 4.96×10^11 at 273 K.
For an equilibrium mixture in which [Cl2O2] is 6.00 x 10^-6M, what is [ClO]?
Answers
Answer:
[ClO] = 3.48×10¯⁹ M.
Explanation:
The following data were obtained from the question:
Equilibrium constant (Kc) = 4.96×10¹¹
Concentration of Cl2O2, [Cl2O2] = 6x10¯⁶ M.
Concentration of ClO, [ClO] =.?
The equation for the reaction is given below:
2ClO(g) ⇌ Cl2O2(g)
The equilibrium constant for a reaction is simply defined as the ratio of the concentration of product raised to their coefficient to the concentration of the reactant raised to their coefficient.
The equilibrium constant, Kc for the reaction is given by:
Kc = [Cl2O2] / [ClO]²
Thus, we can calculate the concentration of ClO, [ClO] as follow:
Kc = [Cl2O2] / [ClO]²
4.96×10¹¹ = 6x10¯⁶ / [ClO]²
Cross multiply
4.96×10¹¹ × [ClO]² = 6x10¯⁶
Divide both side by 4.96×10¹¹
[ClO]² = 6x10¯⁶ / 4.96×10¹¹
[ClO]² = 1.21×10¯¹⁷
Take the square root of both side
[ClO] = √ (1.21×10¯¹⁷)
[ClO] = 3.48×10¯⁹ M
Therefore, the concentration of ClO, [ClO] is 3.48×10¯⁹ M.
When balancing redox reactions under acidic conditions, hydrogen is balanced by adding: Select the correct answer below:
a. hydrogen gas
b. water molecules
c. hydrogen atoms
d. hydrogen ions
Answers
Answer:
water molecules
Explanation:
Redox reactions are carried out under acidic or basic conditions as the case may be.
If the reaction is carried out in an acid medium, then we must balance the hydrogen ions on the lefthand side of the reaction equation with water molecules on the righthand side of the reaction equation.
For instance, the equation for reduction of MnO4^- under acidic condition is shown below;
MnO4^-(aq) + 5e + 8H^+(aq) --------> Mn^2+(aq) + 4H2O(l)
P.1 is a variant of SARS-CoV2. This is the so-called "Brazil variant". It has the amino acid the substitution N501Y in the spike protein on the virus surface. In cellular fluids at ~pH 7 the wild type spike protein has a net negative charge. This substitution will make the spike protein
more negative
less hydrophobic
able to absorb move UV light
less negative
Answers
There are various variants of Cov id virus. The Brazilian variant P also known as Gamma variant is the third variant of the original SARS-CoV2.
The correct answer is more negative
This variant has raised concerns since it has ability to spread more quickly then previous variants and this is more negative variant.
It is assumed that Cov id variant Gamma and Delta has ability to absorb move UV light but this is not proved yet and research is underway.
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Please Help! Use Hess’s Law to determine the ΔHrxn for: Ca (s) + ½ O2 (g) → CaO (s) Given: Ca (s) + 2 H+ (aq) → Ca2+ (aq) + H2 (g) ΔH = 1925.9 kJ/mol 2 H2 (g) + O2 (g) → 2 H2O (l) ΔH = −571.68 kJ/mole CaO (s) + 2 H+ (aq) → Ca2+ (aq) + H2O (l) ΔH = 2275.2 kJ/mole ΔHrxn =
Answers
Answer:
ΔHrxn = -635.14kJ/mol
Explanation:
We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:
(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol
(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole
(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole
Reaction (1) - (3) produce:
Ca(s) + H2O(l) → H2(g) + CaO(s)
ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol
Now this reaction + 1/2(2):
Ca(s) + ½ O2(g) → CaO(s)
ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)
ΔHrxn = -635.14kJ/molMatch each compound with its appropriate pKa value.
(a) 4-Nitrobenzoic acid, benzoic acid, 4-chlorobenzoic acid pKa=4.19,3.98, and 3.41pKa =4.19,3.98, and 3.41
(b) Benzoic acid, cyclohexanol, phenol pKa=18.0,9.95, and 4.19pK a =18.0,9.95, and 4.19
(c) 4-Nitrobenzoic acid, 4-nitrophenol, 4-nitrophenylacetic acid pKa=7.15,3.85, and 3.41pK a =7.15,3.85, and 3.41
Answers
Answer:
Explanation:
a) 4-nitrobenzoic acid pKa= 3.41
benzoic acid pKa= 4.19
4-chlorobenzoic acid pKa= 3.98
b) benzoic acid pKa= 4.19
cyclohexanol pKa= 18.0
phenol pKa= 9.95
c) 4-Nitrobenzoic acid pKa= 3.41
4-nitrophenol pKa= 7.15
4-nitrophenylacetic acid pKa= 3.85
What is the final volume V2 in milliliters when 0.551 L of a 50.0 % (m/v) solution is diluted to 23.5 % (m/v)?
Answers
Answer:
[tex]V_2=1.17L[/tex]
Explanation:
Hello,
In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:
[tex]V_1C_1=V_2C_2[/tex]
Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:
[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]
Best regards.
A certain reaction has an activation energy of 39.5 kJ/mol. As the temperature is increased from 25.0°C to a higher temperature, the rate constant increases by a factor of 5.90. Calculate the higher temperature (in °C).
Answers
Answer:
Explanation:
We shall apply Arrhenius equation which is given below .
[tex]ln\frac{k_2}{k_1} = \frac{E_a}{R} [\frac{1}{T_1} -\frac{1}{T_2} ][/tex]
K₂ and K₁ are rate constant at temperature T₂ and T₁ , Ea is activation energy .
Putting the given values
[tex]ln\frac{5.9}{1} = \frac{39500}{8.3} [\frac{1}{298} -\frac{1}{T_2} ][/tex]
[tex].000373= [\frac{1}{298} -\frac{1}{T_2} ][/tex]
T₂ = 335.27 K
= 62.27 °C
The higher temperature is 62.27°C.
Calculating the higher temperature:Given that the activation energy of the reaction is:
Eₐ = 39.5 kJ/mol
initial temperature T₁ = 25°C = 298K
Let the final temperature be T₂
The rate constant at temperature T₁ be K₁, and at a temperature T₂ be K₂.
According to the question: K₂/K₁ = 5.9
Now, applying the Arrhenius equation we get:
[tex]\ln\frac{K_2}{K1}=\frac{E_a}{R}[\frac{1}{T_1} -\frac{1}{T_2}]\\\\\ln(5.9)= \frac{39.5}{8.3}[\frac{1}{298} -\frac{1}{T_2}]\\\\0.000373=\frac{1}{298} -\frac{1}{T_2}[/tex]
T₂ = 335.27K
T₂ = 335.27 -273
T₂ = 62.27°C
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Check 0/1 ptRetries 5 Element R has three isotopes. The isotopes are present in 0.0825, 0.2671, and 0.6504 relative abundance. If their masses are 97.62, 109.3, and 138.3 respectively, calculate the atomic mass of element R.
Answers
Answer:
Atomic mass = 127.198 amu
Explanation:
The average atomic mass is obtained by summing the masses of the isotopes each multiplied by its abundance.
Atomic mass = (97.62 * 0.0825) + (109.3 * 0.2671) + (138.3 * 0.6504)
Atomic mass = 8.05365 + 29.19403 + 89.95032
Atomic mass = 127.198 amu
What happens when an atom of sulfur combines with two atoms of chlorine to produce SCl2?
Answers
Answer:
Each chlorine atom shares a pair of electrons with the sulfur atom. ... Each chlorine atom shares all its valence electrons with the sulfur atom
What are some geographic features that could be found in the hydrosphere?
Answers
Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.
hydro = water
Answer:
Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!
Explanation:
The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)
A qualitative researcher may select a ______ case that is unusually rich in information pertaining to the research question.
A) critical
B) typical
C) deviant
D) rare
Answers
Answer:
D. Rare
Explanation:
Qualitative research has to do with non-numerical data and is used to understand the beliefs of a group of people which can be gotten from surveys, questionnaires, interviews, etc.
A qualitative researcher may select a RARE case that is unusually rich in information pertaining to the research question.
This is because he wants to get an insight into the why, how, where and when of that particular ase as it's not a usual occurrence.
If you drip an ink drop into a cup of water and wait for a few seconds, all the water will be colored with the ink. This experiment is an example of facilitated diffusion ?
true
false
Answers
Answer:
false, it is not an example of facilitated diffusion
Answer:
TrueExplanation:
When a drop of ink added into the water gradually moves in the whole quantity of water due to this entire water turns into blue color. This is nothing but the diffusion of ink particles into the water molecules. This is because water, as well as ink molecules, are in random motion due to the motion of ink substance.
PLEASE HELP!! 40 POINTS
Answers
Answer:
1) 6.524779402×10^(-17)
2)521.1g
3)113
Explanation:
Answer: 1) 6.524779402×10^(-17)
2)521.1g
Explanation:
For the cell notation shown, which species is undergoing reduction?
Answers
Answer:
option (A) is right answer
Answer:
c
Explanation:
trust me I took the exam
A solution containing a unknown ionic compound, vigorously bubbles when hydrochloric acid (HCl) is added to the solution. This might indicate that the solution contains which anion?
Answers
Answer:
CO3^2-
Explanation:
In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.
When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.
I add a 50. g piece of Al (c = 0.88 J/g-deg) that is at 225°C to 100. mL of water at 20°C. What is the final temperature of the water in °C? The density of water is approximately 1g/mL.
Answers
Answer:
THE FINAL TEMPERATURE OF WATER IS -4.117 °C
Explanation:
Mass of the aluminium = 50 g
c = 0.88 J/g C
Initial temperature of aluminium = 225 °C
Volume of water = 100 ml
Density of water = 1 g/ml
Mass of water = density * volume of water
Mass of water = 1 * 100 = 100 g of water
Initial temperature of water = 20 C
It is worthy to note that the heat of a system is constant and conserved as no heat is lost or gained by a closed system,
So therefore,
heat lost by aluminium = heat gained by water
H = mass * specific heat capacity * temeprature change
So:
m c ( T2- T1) = m c (T2-T1)
50 * 0.88 * ( T2 - 225) = 100 * 4.18 *( T2 - 20)
44 ( T2 - 225 ) = 418 ( T2 - 20)
44 T2 - 9900 = 418 T2 - 8360
-9900 + 8360 = 418 T2 - 44 T2
-1540 = 374 T2
T2 = - 4.117
So therefore the final temperature of water is -4.117 °C
The decomposition of ethylene oxide(CH₂)₂O(g) → CH₄(g) + CO(g)is a first order reaction with a half-life of 58.0 min at 652 K. The activation energy of the reaction is 218 kJ/mol. Calculate the half-life at 629 K.
Answers
Answer:
Half-life at 629K = 252.4min
Explanation:
Using Arrhenius equation:
[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]
And as Half-life in a first order reaction is:
[tex]t_{1/2}=\frac{ln2}{K}[/tex]
We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:
[tex]58.0min=\frac{ln2}{K}[/tex]
K = 0.01195min⁻¹ = K₁
[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]
[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]
[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]
K₂ = 2.75x10⁻³ min⁻¹
And, replacing again in Half-life expression:
[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]
Half-life at 629K = 252.4minThe half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.
The activation energy of a reaction is related to its rate constant as follows:
[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex] (1)
Where:
k: is the rate constant A: is the pre-exponential factor[tex]E_{a}[/tex]: is the activation energy of the reaction = 218 kJ/mol R: is the gas constant = 8.314 J/(K*mol)T: is the temperature
We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:
[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex] (2)
Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min
Hence, the rate constant at 652 K is:
[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]
Now, from equation (1) we can find the pre-exponential factor (A):
[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]
With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):
[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]
Finally, the half-life at 629 K is (eq 2):
[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]
Therefore, the half-life at 629 K is 251.1 min.
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Click on ALL of the following that use sound waves to communicate with their surroundings and find their way! bats cars dolphins whales birds submarines buses school
Answers
How should the magnetic field es be drawn for the magnets shown below!
Answers
Answer:
The magnetic field shows (D)
Explanation:
Because like poles repel eachother
The magnetic field is represented by two poles of a magnetic with magnetic filed lines from the poles. The image C represents the magnetic filed .
What is magnetic field?A magnetic filed is a generated by strong filed lines from a magnet with two poles. A magnet have two poles namely magnetic south and north poles.
The magnetic south poles of of a magnet is always attracted towards the magnetic north poles. Magnetic north poles of a magnet is generally shown in the left side and the field lines will be outward towards the geographic north pole.
The magnetic field lines from the south pole is inward and attracted towards the north. Therefore, right figure representing the magnetic filed is option C.
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Which of the following provides a characteristic of
MgO(s) with a correct explanation?
Choose 1 answer:
А
It is hard because its ions are held together by strong
electrostatic attractions.
B
It is malleable because its atoms can easily move past
one another without disrupting the bonding.
It is a poor conductor of electricity because its
electrons are tightly held within covalent bonds and
lone pairs.
It has a high melting point because its molecules
interact through strong intermolecular forces.
Answers
Answer:
А It is hard because its ions are held together by strong electrostatic attractions.
B It is malleable because its atoms can easily move past one another without disrupting the bonding.
Explanation:
These are correct explanations of the properties of magnesium.
C is wrong. Mg is a good conductor of electricity and it has metallic bonds.
D is wrong. Mg has no molecules. It has no intermolecular forces.
Briefly workout the relationship between these constants:
[tex]{ \bf{K _{sp} \: and \: K _{c} }}[/tex]
In consideration of the decopmposition of hydrogen iodide.
[tex]{ \sf{2HI _{(g)} →H _{2(g)} +I _{2(g)} }}[/tex]
[tex]{ \tt{any \: help \: is \: appreciated}}[/tex]
Answers
Kc require (aqueous/gaseous) products to be on the numerator and (aqueous/gaseous) reactants to be in the denominator, whereas Ksp will require (aqueous) products to be on the numerator and (aqueous) reactants to be in the denominator. Both require products on top and reactants in the bottom.
K = [products] / [reactants]
Kc is used when a reaction reaches dynamic equilibrium, whereas Ksp is used when an insoluble ionic solid dissolved by a tiny amount in a solution, as well as in determining whether or not a precipitate will form.
Kc can be used to measure equilibrium concentration for all reactions, whereas Ksp is limited to only ionic compounds' solubility.
The decomposition of HI (g) will required the use of Kc since the species are all gaseous, and gases cannot be ionic.
A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat. If the radiation wavelength is 12.5 cm, how many photons of this radiation would be required to heat a container with 0.250 L of water from a temperature of 20.0oC to a temperature of 99oC
Answers
Answer:
The total photons required for this radiation = 5.1938 × 10²⁸ photons
Explanation:
Given that:
A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat.
If the radiation wavelength is 12.5 cm,
density of water = 1g/cm³
volume of the container = 0.250 L = 250 cm³
density = mass/volume
mass of the water = density × volume
mass of the water = 1g/cm³ × 250 cm³
mass of the water = 250 g
specific heat capacity of water = 4.182 J/g°C
The change in temperature was from 20.0° C to 99° C
ΔT =( 99 -20.0)° C
ΔT = 79.0° C
The heat absorbed in the process is calculated by using the formula,
q = mcΔT
q = 250 g × 4.182 J/g°C × 79.0° C
q = 82594.5 Joules
Recall that the radiation wavelength λ = 12.5 cm = 0.125 m
The amount of energy of one photon of the radiation wavelength is determined by using the formula:
E = hv
since v = c/λ
E = hc/λ
where;
h = Planck's constant = 6.626 × 10⁻³⁴ J.s
c = velocity of light = 3.0 × 10⁸ m/s
∴
E = (6.626 × 10⁻³⁴ J.s × 3.0 × 10⁸ m/s)/ 0.125 m
E = 1.59024⁻²⁴ Joules
The total photons required for this radiation = total heat energy/energy of radiation
The total photons required for this radiation = 82594.5 Joules/1.59024⁻²⁴ Joules
The total photons required for this radiation = 5.1938 × 10²⁸ photons
When equal moles of an acid and a base are mixed, after reaction the two are compounds are said to be at the _______________. Select one: Indicator point Stoichiometric point Equilibrium point End point
Answers
Answer:
when equal moles of an acid and base are mixed,after reaction the two are compounds are said to be at the Equivalent point.
In the reaction 2 AgI + HgI 2 → Ag 2HgI 4, 2.00 g of AgI and 3.50 g of HgI 2 were used. What is the limiting reactant?
Answers
Answer:
AgI I the limiting reactant.
Explanation:
The balanced equation for the reaction is given below:
2AgI + HgI2 → Ag2HgI4
Next, we shall determine masses AgI and HgI2 that reacted from the balanced equation.
This is illustrated below:
Molar mass of Agl = 108 + 127 = 235 g/mol
Mass of AgI from the balanced equation = 2 x 235 = 470 g
Molar mass of HgI2 = 201 + (2x127) = 455 g/mol
Mass of HgI2 from the balanced equation = 1 x 455 = 455 g
From the balanced equation above,
470 g of AgI reacted with 455 g of HgI2.
Finally, we shall determine the limiting reactant as follow:
From the balanced equation above,
470 g of AgI reacted with 455 g of HgI2.
Therefore, 2 g of AgI will react with
= (2 x 455)/470 = 1.94 g of HgI2.
From the calculations made above, we can see that only 1.94 g out of 3.50 g of HgI2 given is needed to react completely with 2 g of AgI.
Therefore, AgI I the limiting reactant.
How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)->MgCl2(s)
Answers
The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.
How to calculate moles in stoichiometry?Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.
According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:
Mg + Cl₂ → MgCl₂
Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.
This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.
Next, we convert moles of magnesium chloride to mass as follows:
molar mass of magnesium chloride = 95.211g/mol
mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.
Therefore, 218.99 grams of magnesium chloride will be formed.
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Are acids harmful to work with.
Answers
Answer:
yes it is
Explanation:
because there are some acid which really harm skin.
A gas mixture contains 3.50 moles of helium, 5.00 moles of krypton and 7.60 moles of neon. A) What is the mole fraction for each gas
Answers
Answer:
.217, .311, and .472, respectively.
Explanation:
The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).
X of helium=3.50/16.10 = .217
X of krypton=5.00/16.10 = .311
X of neon=7.60/16.10 = .472
Rubric #2
Forensic Science
1. Define Nucleus.
2. Define Cytoplasm.
3. Define Cell Membrane.
4. Define DNA.
5. Define Plant.
6. Define Chlorophyll.
7. Define Photosynthesis.
8. How do Plant cells and Animal cells differ?
9. Define Cell Wall.
10. Define Vacuole.
11. Why do cells differentiate in multicellular organisms?
12. Define Multicellular.
13. Complete the Eukaryotic cells and Cell Differentiation assessment.
https://clever.discoveryeducation.com/learn/techbook/units/95c20a43-6d3d-40d3-
848d-89929101140d/concepts/co0fef01-33e7-4116-8819.
143e289e15ba/tabs/6e1551ab-57b8-42d4-8e5b-25549791c760/pages/de4182af-aa 60-
454f-ae5e-28df6f4eb3ac
Answers
Explanation:
1. Nucleus is a memberane bound organelle that contains cell,s chromosomes.
Calculate the pH of a 0.20 M NH3/0.20 M NH4Cl buffer after the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Answers
Answer:
Explanation:
For pH of a buffer solution , the formula is
pH = pKa + log [ Base ] / [ conjugate acid ]
= pKa + log [ NH₃ ] / [ NH₄⁺ ]
Ka = Kw / Kb
Kb for NH₄OH = 1.8 x 10⁻⁵
Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵
= 5.6 x 10⁻¹⁰
pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2
= 10 - log 5.6
= 9.25
Effect of addition of HCl
H⁺ of HCl will react with NH₃ to produce NH₄⁺
25 mL of .1 HCl = 2.5 mM of HCl
25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺
65 mL of .2 M NH₃ = 13 mM of NH₃
65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺
NH₃ + H⁺ = NH₄⁺
NH₄⁺ formed = 2.5 + 13 mM
15.5 mM of NH₄⁺
NH₃ = 13 mM
Concentration of NH₃ = 13 / 90
Concentration of NH₄⁺ = 15.5 / 90
pH of final buffer mixture
= 9.6 + log 13 / 15.5
= 9.25 - .076
= 9.174
The pH value is mathematically given as
pH= -6.332.
What is the pH of a 0.20 M NH3/0.20 M NH4Cl buffer?Question Parameters:
the pH of a 0.20 M NH3/0.20 M NH4Cl buffer
the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Generally, the equation for the Chemical Reaction is mathematically given as
HCl + NH3 --> NH4^+ + Cl^-
Therefore
pH= pka + log(13/14).
pH= -6.3 + log 0.93.
pH= -6.3+ (-0.032).
pH= -6.332.
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Three structural isomers have the formula C5H12.C5H12. Draw and name the isomers using IUPAC names. Draw the isomer with five carbon atoms in main chain.
Answers
Answer:
Explanation:
Answer in attached file .