Answer:
4.00 is the pH of the mixture
Explanation:
The ethyl amine reacts with HNO3 as follows:
C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻
To solve this question we need to find the moles of ethyl amine and the moles of HNO3:
Moles C2H5NH2:
0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine
Moles HNO3:
0.201L * (0.025mol/L) = 0.005025 moles HNO3
That means HNO3 is in excess. The moles in excess are:
0.005025 moles HNO3 - 0.00500 moles ethyl amine =
2.5x10⁻⁵ moles HNO₃
In 50 + 201mL = 251mL = 0.251L:
2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]
As pH = -log [H+]
pH = -log 9.96x10⁻⁵M
pH = 4.00 is the pH of the mixture
A 1.5-m 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa. Now paddle-wheel work is done on the system until the pressure in the tank rises to 150 kPa. Determine the entropy change of carbon dioxide during this process. Assume constant specific heats
Answer:
The entropy change of carbon dioxide = 0.719 kJ/k
Explanation:
Given:
1.5 m - 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa
The objective is to determine the entropy change of carbon dioxide
Formula used:
ΔS=
Solution:
On considering,
[tex]C_{P} =0.846 kJ/kg K\\C_V=0.657 kJ/kg k\\[/tex]
ΔS=[tex]mc_{v} lu\frac{p_{2} }{P_{1} }[/tex]
On substituting the values,
ΔS=[tex]2.7*0.657lu\frac{150}{100}[/tex]
ΔS=0.719 kJ/k
The entropy change is "0.719 kJ/K".
Given values are:
Mass of tank,
m = 2.7 kgPressure,
P₁ = 100 kPaRised pressure,
P₂ = 150 kPaAssumption of constant specific heat is,
[tex]C_v = 0.657 \ kJ/kgK[/tex]As we know the formula,
→ [tex]\Delta S = mC_v \ ln(\frac{P_2}{P_1} )[/tex]
[tex]= (2.7)(0.657) \ ln (\frac{150}{100} )[/tex]
[tex]= 1.7739\times 0.4055[/tex]
[tex]= 0.7193 \ kJ/K[/tex]
Thus above answer is right.
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The Hammond Postulate describes the relationship between the energy of the transition state and the productdistribution in a reaction that is capable of following more than one pathway.
What are the appropriate labels for the sentence?
The Hammond postulate states that reactions which are thermodynamically endothermic and kinetically ___________ have transitions states that occur ___________ in the reacton time frame, and resemble the ___________ in terms of energy and structure.
What type of a liquid will have a pH value equal to 12? (1 point)
Basic
Neutral
Strong acid
Weak aci
Answer: it will be basic
pH that ranges from 0-6 are acid
pH of EXACTLY 7 is neutral
pH greater than 7 are strongly basic or base
Suppose you analyze a 30.3 g sample of bleach and determine that there are 2.61 g of sodium hypochlorite present. What is the percent of sodium hypochlorite in the bleach sample
Answer:
8.61 %
Explanation:
The percent of sodium hypochlorite in the bleach sample can be calculated using the following formula:
% of sodium hypochlorite = Mass of sodium hypochlorite / mass of sample * 100%We input the data given by the problem:
% of sodium hypochlorite = 2.61 g / 30.3 g * 100 % = 8.61 %A cylinder contains 26.5 L of oxygen gas at a pressure of 1.7 atm and a temperature of 280 K. How much gas (in moles) is in the cylinder?
Answer:
The appropriate answer is "1.96 mol".
Explanation:
The given values are:
Volume,
V = 26.5 L
Pressure,
P = 1.7 atm
Temperature,
T = 280 K
Now,
The number of moles (n) will be:
= [tex]\frac{PV}{RT}[/tex]
By substituting the values, we get
= [tex]\frac{1.7\times 26.5}{0.0821\times 280}[/tex]
= [tex]\frac{45.05}{22.988}[/tex]
= [tex]1.96 \ mol[/tex]
The density of aluminum is 2.7 g/cm3.
Part A
What is its density in kilograms per cubic meter?
Express your answer in kilograms per cubic meter to two significant figures.
Answer:
2700 kg/m³
Explanation:
First let's convert 2.7 g/cm³ to kg/cm³, keeping in mind that 1 kilogram equals 1000 grams:
2.7 g/cm³ * [tex]\frac{1kg}{1000g}[/tex] = 0.0027 kg/cm³Finally we need to convert 0.0027 kg/cm³ to kg/m³, keeping in mind that 1 meter equals 100 centimeters, as follows:
0.0027 kg/cm³ * [tex](\frac{100cm}{1m} )^3[/tex] = 2700 kg/m³The answer is 2700 kg/m³.
Please help me ASAP I’ll mark Brainly
Answer:
cell
chloroplast and cell wall
nucleus
life processes
cell membrane
shape and size
vacuole
Hope it helps
Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in and the atom emits a photon of light with a wavelength of 486 nm. Group of answer choices
Complete Question
Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n=4 and the atom emits a photon of light with a wavelength of 486 nm. Group of answer choices
Answer:
[tex]n=2[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=486nm=>486*10^{-9}[/tex]
Generally the equation for Atom Transition is mathematically given by
[tex]\frac{1}{\lambda}=R_{\infty }(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]
Where
Rydberg constant [tex]R_{\infty}=1.097*10^7[/tex]
Therefore
[tex]\frac{1}{486*10^{-9}}=1.097*10^7*(\frac{1}{n_1^2}-\frac{1}{4^2})[/tex]
[tex](\frac{1}{n_1^2}-\frac{1}{4^2})=\frac{1}{486*10^{-9}*1.097*10^7}[/tex]
[tex]n_1^2=3.98[/tex]
[tex]n=1.99[/tex]
[tex]n=2[/tex]
Using the Rydberg formula, the final state of the electron is n=2.
Using the Rydberg formula;
1/λ = R(1/nf^2 - 1/ni^2)
Where;
λ = wavelength
nf = final state
ni = initial state
R = Rydberg constant
When λ = 486 × 10^-9 m and ni = 4, R = 1.097 × 10^7 m-1
1/486 × 10^-9 = 1.097 × 10^7(1/nf^2 - 1/4^2)
0.188 = 1/nf^2 - 0.0625
1/nf^2 = 0.188 + 0.0625
nf = 2
Missing parts;
Determine the end (final) value of n in the hydrogen atom transition, if electron starts in n-4 and the atom emits a photon of light with a wavelength of 486.
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20.1 cm3 of metal hydroxide (MOH) containing 4.8 g/dm3 react Completely with 23.0 cm3 of the HCL and Containing 3.8g/dm3. Identify the metal of M of MOH
Answer:
A metal M readily forms water-soluble sulphate MSO4, water-insoluble hydroxide M(OH)2 and oxide MO. The oxide and hydroxide are soluble in NaOH. The M is:
Determine the theoretical yield and the percent yield if 21.8 g of K2CO3 is produced from reacting 27.9 g KO2 with 57.0 g CO2. The molar mass of KO2
Answer:
26.9 g
81%
Explanation:
The equation of the reaction is;
4 KO2(s) + 2 CO2(g) → 3 O2(g) + 2 K2CO3(s)
Number of moles of KO2= 27.9g/71.1 g/mol = 0.39 moles
4 moles of KO2 yields 2 moles of K2CO3
0.39 moles of KO2 yields 0.39 × 2/4 = 0.195 moles of K2CO3
Number of moles of CO2 = 57g/ 44.01 g/mol = 1.295 moles
2 moles of CO2 yields 2 moles of K2CO3
1.295 moles of CO2 yields 1.295 × 2/2 = 1.295 moles of K2CO3
Hence the limiting reactant is KO2
Theoretical yield = 0.195 moles of K2CO3 × 138.205 g/mol = 26.9 g
Percent yield = actual yield/theoretical yield × 100
Percent yield = 21.8/26.9 × 100
Percent yield = 81%
The rate law for the following reaction: H2SiF6(aq)+2NaF(aq)+3H2O(aq)⟶Na2SiO3(s)+8HF(aq) is second order in H2SiF6, zero order in NaF and first order in H2O. By what factor will the reaction rate change if the concentrations of all reactants are tripled?
Answer:
The rate law for the following reaction:
H2SiF6(aq)+2NaF(aq)+3H2O(aq)⟶Na2SiO3(s)+8HF(aq)
is second order in H2SiF6, zero order in NaF and first order in H2O.
A scientist is conducting a Sanger's sequencing experiment to determine the number of polypeptides present in an oligomeric protein. The molecular weight of the protein is 18000 g/mol . After the reaction of 520 mg of the protein with 1‑fluoro‑2,4‑dinitrobenzene, the peptide bonds were hydrolyzed with an acid. As a result, the scientist obtained 39 mg of 2,4‑dinitrophenyl serine. What is the number of the polypeptide chains present in the oligomer?
Answer:
Depends on molecule.
Explanation:
The number of the polypeptide chains present in the oligomer depends on the molecule. Some molecules have more polypeptide chains whereas some of them have less polypeptide chains. For example, Hemoglobin is a oligomer that consists of four Polypeptide Chains, two of these Polypeptide Chains are α-globin molecules, each comprise of 141 amino acids, and the other two are (β, γ, δ, or ε) globins, each consist of 146 amino acids.
A Bronsted-Lowry acid is defined as a substance that ________. A Bronsted-Lowry acid is defined as a substance that ________. increases Ka when placed in H2O increases [OH-] when placed in H2O acts as a proton donor acts as a proton acceptor decreases [H ] when placed in H2O
Answer: A Bronsted-Lowry acid is defined as a substance that acts as a proton donor.
Explanation:
A substance that is able to donate a proton or hydrogen ion to another substance is a Bronsted-Lowry acid.
For example, HCl is a Bronsted-Lowry acid as it dissociates to give a hydrogen ion.
[tex]HCl \rightleftharpoons H^{+} + Cl^{-}[/tex]
Thus, we can conclude that a Bronsted-Lowry acid is defined as a substance that acts as a proton donor.
What is Heisnberg's uncertainity principle? Why it make sense only for microscopic particles.
Answer:
The uncertainty principle is one of the most famous (and probably misunderstood) ideas in physics. It tells us that there is a fuzziness in nature, a fundamental limit to what we can know about the behavior of quantum particles and, therefore, the smallest scales of nature.
Suppose of potassium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the potassium acetate is dissolved in it. Round your answer to significant digits.
The question is incomplete, the complete question is;
Suppose 0.377g of potassium acetate is dissolved in 250.mL of a 57.0mM aqueous solution of ammonium sulfate.
Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the potassium acetate is dissolved in it.
Round your answer to 3 significant digits.
Answer:
0.0152 M
Explanation:
The equation of the reaction is;
2CH3COOK(aq) + (NH4)2SO4(aq)------> K2SO4(aq) + 2CH3COONH4(aq)
Number of moles of potassium acetate = 0.377g/98.15 g/mol = 0.0038 moles
Number of moles of ammonium sulphate = 250/1000L × 57 × 10^-3 = 0.014 moles
2 moles of potassium acetate yields 2 moles of ammonium acetate
Hence;
0.0038 moles of potassium acetate yields 0.0038 moles of ammonium acetate
Also
1 mole of potassium sulphate yields 2 moles of ammonium acetate
0.014 of potassium sulphate yields 0.014 × 2/1 = 0.028 moles of ammonium acetate
So potassium acetate is the limiting reactant.
Since 0.0038 moles of ammonium acetate is produced, the final concentration of potassium acetate is = 0.0038 moles of ammonium acetate/0.25L = 0.0152 M
Hence final concentration of acetate ions =0.0152 M
Og is the noble gas after Rn. To go from [Rn] to [Og], you must fill four subshells (s, p, d, and f) with a total of 32 electrons. Thus, the atomic numbers of 6th and 7th period elements of the same group differ by 32.
a. To go from [Og] to the next noble gas, however, you would theoretically fill five subshells (s, p, d, f, and g). How many electrons are needed to fill all five subshells?
b. Element 106 in the periodic table is Sg Determine the atomic number of the element just below Sg in the periodic table.
Answer:
See explanation
Explanation:
Since we have to fill five subshells in moving from Og to the next noble gas in the eight period, we have to know the maximum electrons contained in each of those subshells;
s= 2, p=6, d= 10, f= 14, g = 18
This means that we need a total of 50 electrons to fill all the five subshells.
Hence, the element just below Sg in the eight period will have an atomic number of 156.
Determine whether each of the properties described applies to volumetric or graduated glassware. 1. Used for applications in which great accuracy is needed_____.a. volumetricb. graduated 2. Capable of measuring a range of volumes of liquid_____.a. volumetric
b. graduated 3. Designed to measure one specific volume of liquid_____.a. volumetric
b. graduated
Answer:
1) volumetric
2) graduated
3) volumetric
Explanation:
A volumetric glassware is a glassware that is marked at a particular point. A typical example of a volumetric glassware is the volumetric flask. A volumetric glassware is capable of measuring only a specific volume of a liquid.
On the other hand, graduated glassware can measure a range of volumes of liquid. However, a volumetric glassware is still required where a high degree of accuracy is important.
According to the Arrhenius equation, changing which factors will affect the
rate constant?
A. Temperature and the ideal gas constant
B. The activation energy and the constant A
C. The constant A and the temperature
D. Temperature and activation energy
Answer:
e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature
Answer:
D. Temperature and activation energy is the correct answer
Explanation:
^_^
A vessel is filled at a rate of 3.41 cubic micrometers per minute. If the vessel has a volume of 54 liters, how many
seconds will it take to fill the vessel? provide a step buy step explanation.
Answer:
9.5x10¹⁷ s
Explanation:
First we convert 3.41 cubic micrometers (um³) to liters (L), as such:
3.41 um³ * [tex](\frac{1m}{10^6um} )^3*\frac{1000L}{1m^3}[/tex] = 3.41x10⁻¹⁵ LWith the converted rate of 3.41x10⁻¹⁵ L/min, we can calculate how many minutes it would take to fill a 54 L vessel:
54 L ÷ 3.41x10⁻¹⁵ L/min = 1.58x10¹⁶ minFinally we convert 1.58x10¹⁶ minutes to seconds:
1.58x10¹⁶ * 60 = 9.5x10¹⁷ sAccording to Newton's law of universal gravitation, how do the masses of two
objects relate to the gravitational force between them?
A. As either mass increases, the gravitational force between them
increases.
B. As either mass increases, the gravitational force between them
decreases.
C. Gravitational force increases only when both masses increase.
D. Gravitational force increases only when both masses decrease.
Answer:
As either mass increases, the gravitational force between them
increases.
Explanation:
According to Newton's law of universal gravitation;
F α m1m2/r^2
That is, the force between two masses in a gravitational field is directly proportional to the product of the two masses and inversely proportional to their distance apart.
Hence, as either of the masses increase, the force of gravitation between the two masses increases. Hence the answer.
What is the phase of water at 0.25 atm and 0°C?
Water
(liquid)
Pressure (atm)
0.5-
0.25
Ice
(solid)
Water vapor
(gas)
0
000
Temperature (°C)
O A. Gas
O B. Solid and gas
O C. Solid and liquid
D. Solid
Water is in the solid phase at 0.25 atm and 0°C.
In what phase is water at 25?A pressure of 50 kPa and a temperature of 50 °C correspond to the “water” region—here, water exists only as a liquid. At 25 kPa and 200 °C, water exists only in the gaseous state.
What phase is water in at 0 C?Under standard atmospheric conditions, water exists as a liquid. But if we lower the temperature below 0 degrees Celsius, or 32 degrees Fahrenheit, water changes its phase into a solid called ice.
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. The nucleophile in the reaction is _______ b. The Lewis acid catalyst in the reaction is ______ c. This reaction proceeds___________(faster or slower) than benzene. d. Draw the structure of product D
Answer:
a. eletrophile
b. able to impose regioselectivity and stereo selectivity.
c. faster
Explanation:
Necleophile reaction is chemical reaction in which electron rich chemical specie replaces functional group with another electron deficient molecule. Lewis acid catalyst is organic chemical reaction which lewis acid act as electron pair acceptor. Nucleophile reaction proceeds about 25 times more faster than benzene.
Draw the structure of the alkene with the molecular formula C6H10 that reacts with Br2 to give this compound.
Answer: Please, this question is not complete. I have attached the complete question.
The answer is in the attached picture below
Explanation:
The explanation is in the attached picture below
The structure of the alkene with the molecular formula [tex]C_6H_1_0[/tex] that reacts with [tex]Br_2[/tex]to give this compound is an alkene called 1-hexene
How do we explain?The alkene is called 1-hexene. It has a double bond between the first and second carbon atoms. When it reacts with Br2, the bromine atoms add to the double bond, resulting in the formation of 1,2-dibromohexane.
The reaction is a radical addition reaction. The first step is the formation of a radical by the homolytic cleavage of one of the bromine atoms in Br2. This radical then adds to the double bond in the alkene, forming a new radical. The second bromine atom then adds to the radical, forming 1,2-dibromohexane
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How many moles of carbon dioxide at STP will fit in a 50 liter container?
Answer:
n = 2.23 moles
Explanation:
Given the following data;
Standard temperature = 273 K
Standard pressure = 101.325 kPa
Volume = 50 liter
R = 8.314 J/mol·K
To find the number of moles, we would use the ideal gas law formula;
PV = nRT
Where;
P is the pressure.V is the volume.n is the number of moles of substance.R is the ideal gas constant.T is the temperature.Making n the subject of formula, we have;
[tex] n = \frac {PV}{RT} [/tex]
Substituting into the formula, we have;
[tex] n = \frac {101.325*50}{8.314*273} [/tex]
[tex] n = \frac {5066.25}{2269.722} [/tex]
n = 2.23 moles
Therefore, 2.23 moles of carbon dioxide at STP will fit in a 50 liter container.
The addition of chromic acid or chromate is a qualitative test for alcohols as the reaction causes a color change. However, not all alcohols react with chromic acid or chromate. Determine whether the named alcohol will react with chromic acid or chromate to cause a color change.
a. 3-hexanol
b. 1 -butanol
c. 2-pentanol
d. 3-ethyl-3-pentanol
Answer:
3-hexanol
1 -butanol
2-pentanol
Explanation:
Let us recall that chromic acid or chromate are strong oxidizing agents. When they are oxidized, their colour changes from orange to green.
This shows a reduction in chromic acid or chromate. The reaction of chromic acid or chromate with a primary alcohol yields a carboxylic acid while reaction with a secondary substrate yields an alkanal.
Note that Tertiary alkyl halides are not be oxidized hence reactions involving a point where invitation carried along occur.
3-ethyl-3-pentanol is a tertiary alkyl halide hence it can not be oxidized.
A balloon contains 0.118 mol of gas and has a volume of 2.58 L . If an additional 0.116 mol of gas is added to the balloon (at the same temperature and pressure), what will its final volume be? Can you also show the work so I can understand why is it that answer. thank you
Answer:
v2=5.11L
Explanation:
given
v1=2.58L
N1=0.118mol
N2=0.234
v2=x
according to charles law V1/N1=V2/N2
2.58/0.118=V2/0.234
21.86=V2/0.234
21.86×0.234= v2
5.116L=v2
5.116L is the
answer or u can simplify it and make 5.1 L
A chemistry student needs 90.0mL of carbon tetrachloride for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of carbon tetrachloride is 1.59g*cm3- Calculate the mass of carbon tetrachloride the student should weigh out. Be sure your answer has the correct number of significant digits.
Answer:
volume = mass/density
Here, volume = 80g/1.59gcm-3 → 50.314 cm3
Explanation:
¿soy guapo?
si dices que si te doy una galletita
Answer:
Hindi ko po ma gets sinasabi nyu pwedeng pakiayus
What is the amount of solute required if the solution is 50 ml and the solvent is 35 ml. Solve and explain
I don’t know what to do
Answer:
15 mL of the solute
Explanation:
From the question given above, the following data were obtained:
Solution = 50 mL
Solvent = 35 mL
Solute =?
Solution is simply defined as:
Solution = solute + solvent
With the above formula, we can easily obtain the solute in the solution as follow:
Solution = 50 mL
Solvent = 35 mL
Solute =?
Solution = solute + solvent.
50 = solute + 35
Collect like terms
50 – 35 = solute
15 = solute
Solute = 15 mL
Therefore, 15 mL of the solute is required.
Classify each of the following as either macroscopic, microscopic or particulate:
a. a red blood cell.
b. a sugar molecule.
c. baking powder.
Answer:
Classify each of the following as either macroscopic, microscopic or particulate:
a. a red blood cell.
b. a sugar molecule.
c. baking powder.
Explanation:
a. A red blood cell is a microscopic particle.
It can be viewed under a microscope.
b. A sugar molecule is also a microscopic substance.
It can be viewed under a microscope.
c. Baking powder is macroscopic substance.