Answer:
a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)
workdone (w) = -8442.6 J ≈ -8.443 KJ
heat transferred (q) of the ideal gas = - w
q = 8.443 KJ
b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0
the workdone(w) in the ideal gas= - 4567.5 J ≈ - 4.57 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
Explanation:
given
mole of an ideal gas(n) = 2.5 mol
Temperature (T) = 20°C
= (20°C + 273) K = 293 K
Initial pressure of the ideal gas(P₁) = 20 atm
Final pressure of the ideal gas(P₂) = 5 atm.
2) (a)for adiabatic reversible process,
note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.
Work done (w) = nRT ln[tex]\frac{P_{1} }{P_{2} }[/tex]
= 2.5 mol × 8.314 J/mol K × 293 K × ln[tex]\frac{5atm}{20atm}[/tex]
= 6090.01 J × [-1.3863]
= -8442.6 J ≈ -8.443 KJ
So, the work done (w) of ideal gas = -8.443 KJ
For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-8.443 KJ)
q = 8.443 KJ
heat transfer (q) of the ideal gas = 8.443 KJ
(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.
Work done (w) = -nRT(1 - ln[tex]\frac{P_{1} }{P_{2} }[/tex] )
= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]
= - 6090.01 J × 0.75
= - 4567.5 J ≈ - 4.57 KJ
∴work done(w) of an ideal gas = - 4.57 KJ
For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-4.5675 KJ)
q = 4.5675 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
Reduction occurs at which electrode?
Answer:
negative charge electrode
Explanation:
In cathode positive ions are picked up to perform reduction.At the same time negative ions are picked up at anode to get oxidized from electrolyte.
Answer:
The electrode that removes ions from the solution :) 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
Benzene boils at 80.10 °C and has a molal boiling constant, k b, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 °C. What is the molality of the solute?
Answer:
[tex]m=0.395mol/kg[/tex]
Explanation:
Hello,
This is a problem about boiling point elevation which is modeled via:
[tex]\Delta T=i*m*Kb[/tex]
Whereas for this solvent (nonpolar, nonionizing), the van't Hoff factor is one. In such a way, the molality of the solute is simply computed as shown below:
[tex]m=\frac{\Delta T}{Kb}=\frac{(81.10-80.10)\°C}{2.53\°C/m} \\\\m=0.395mol/kg[/tex]
In this manner, we can also compute the molar mass of the solute by noticing 20.0 g (0.020 kg) of benzene were used:
[tex]n=0.395mol/kg*0.020kg=7.9x10^{-3} mol[/tex]
And considering the 2.15 g of the solute:
[tex]Molar\ mass=\frac{2.15g}{7.9x10^{-3}mol}\\ \\Molar\ mass=271.975g/mol[/tex]
Best regards.
Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties.HCl(g)+NaOH(s)⟶NaCl(s)+H2O(l)
Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
A solution containing lead(II) nitrate is mixed with one containing sodium bromide to form a solution that is 0.0630 M in Pb(NO3)2 and 0.0103 M in NaBr. What is the value of Q for the insoluble product? Express the reaction quotient to three significant figures.
Answer:
Q = 6.68x10⁻⁶
Explanation:
The initial solutions are Pb(NO₃)₂ and NaBr. As sodium and nitrates salts are soluble, the insoluble product must be the formed from the other ions, PbBr₂.
The soluble product equilibrium is written as:
PbBr₂(s) ⇄ Pb²⁺(aq) + 2Br⁻(aq)
Where Q is defined as:
Q = [Pb²⁺] [Br⁻]²
As:
[Pb²⁺] = 0.0630M
[Br⁻] = 0.0103M
Q = [Pb²⁺] [Br⁻]²
Q = [0.0630M] [0.0103M]²
Q = 6.68x10⁻⁶g If attempting to dissolve both silver bromide and silver chloride in aqueous solution through complex ion formation, which data will be the most relevant?
Answer:
Kf
Explanation:
The stability constant Kf of a given complex specie is an equilibrium constant that represents the formation of that particular complex specie in solution. It measures the strength of the interaction between the ligands and metal that form the particular complex specie. The magnitude of Kf shows how easily a complex specie is formed in solution.
Hence if I want to dissolve the bromides or chlorides of silver which are ordinarily insoluble in water by means of complex formation, the magnitude of the stability constant for each particular complex specie is important as it gives information regarding the thermodynamic feasibility of the process.
2
Select the correct answer
in a redex reaction, what folle does the reducing agent play?
OA. it gives up electrons
OB. it keeps electrons
OC. it takes electrons
OD. it takes onygen atoms
Answer:
A. it gives up electrons
Explanation:
In a redox reaction, the reducing agent is the element or compound that undergoes oxidation and gives up electrons. The oxidizing agent is the element or compound that undergoes reduction and gains electrons.
Hope that helps.
Twenty-five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH has been added
Answer:
The correct answer is 1.60.
Explanation:
Based on the given question, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be determined by using the formula,
Moles = volume * concentration of HCl
= 25/1000*0.10 = 0.0025 moles
Similarly the moles of NaOH added will be determined by using the formula,
Moles of NaOH added = volume * concentration of NaOH
= 15/1000 * 0.10 = 0.0015 moles
The reaction taking place in the given case is,
HCl + NaOH = NaCl + H2O
Now the moles of excess H+ = moles of excess HCl
= 0.0025 - 0.0015 = 0.001 moles
Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L
[H+] = moles of H+/total volume
= 0.001 / 0.040 = 0.025 M
pH = -log[H+]
= -log[0.025]
= 1.60
The pH after 15 ml of NaOH has been volume is 1.60.
Calculation of Concentration of HCl Moles
It is based on the given question that is, 25 ml of 0.10 M HCl is titrated with 0.10 M NaOH. Now moles of HCl can be specified by using the formula,
Moles is = volume * concentration of HCl
Then is = 25/1000*0.10 = 0.0025 moles
Besides, the moles of NaOH added will be determined by using the formula,
When the Moles of NaOH added is = volume * concentration of NaOH
= 15/1000 * 0.10 = 0.0015 moles
When The reaction taking place in the given case is,
HCl + NaOH = NaCl + H2O
Now the moles of excess H+ = moles of excess HCl
= 0.0025 - 0.0015 = 0.001 moles
Then It Based on the given question, the sum of the volume of the solution is 25+15 = 40 ml or 0.040 L
[H+] = moles of H+/total volume
After that = 0.001 / 0.040 = 0.025 M
pH = -log[H+]
Then = -log[0.025]
Therefore, = 1.60
Find more information about Concentration of HCl Moles here:
https://brainly.com/question/25624144
What is the empirical formula for the compound: C8H8S2?
Answer:
Empirical formula = C4H4SExplanation:
The subscripts in a formula determine the ratio of the moles of each element in the compound. To convert this formula to the empirical formula, divide each subscript by 2. This is similar to reducing a fraction to its lowest denominator.
Hydrogen Bonding with Water - Your Drug Lotensin Directions: Show the structure of your molecule below. Illustrate all ways that your molecule could form hydrogen bonds with water, either as a hydrogen donor or as a target (receiver) of hydrogen bonds from water. Do this by drawing bent water molecules as necessary and representing hydrogen bonds between water and the drug using dashed RED lines (---). Be sure that it is exactly clear which atoms on each molecule are involved in the hydrogen bonds. If your drug molecule is not capable of forming hydrogen bonds with water, fully explain why not below.
Answer:
See figure 1
Explanation:
For this question, we have to remember that a hydrogen bond is an interaction in which we have a partial attraction between a positive dipole and a negative dipole, and in this attraction, we have in the middle a hydrogen atom.
In this interaction, we can have a donor (positive dipole) or a receptor (negative dipole). The receptor is a heteroatom (an atom different to carbon or hydrogen) with high electronegativity. The donor is usually hydrogen atom bonded to the heteroatom.
I hope it helps!
Differentiate between
expansion of solid and liquid
In a liquid, the expansion is a little more than in
solids. The bonds in a liquid are weaker than in a
solid, so as you heat up a liquid, the particles can
move around each other faster and in so doing,
move further apart. Solids and liquids occupy a
'set' volume at a certain temperature.
How to do this
Q1 and Q2
Only want to know how to find molecular formula
Answer:
Question 1
A. Empirical formula is C8H8O3
B. Molecular formula is C8H8O3
Question 2.
A. Empirical formula is CH2
B. Molecular formula is C4H8
Explanation:
Question 1:
A. Determination of the empirical formula:
Carbon (C) = 63.2%
Hydrogen (H) = 5.26%
Oxygen (O) = 31.6%
Divide by their molar mass
C = 63.2/12 = 5.27
H = 5.26/1 = 5.26
O = 31.6/16 = 1.975
Divide by the smallest
C = 5.27/1.975 = 2.7
H = 5.26/1.975 = 2.7
O = 1.975/1.975 = 1
Multiply through by 3 to express in whole number
C = 2.7 x 3 = 8
H = 2.7 x 3 = 8
O = 1 x 3 = 3
Therefore, the empirical formula for the compound is C8H8O3
B. Determination of the molecular formula of the compound.
From Avogadro's hypothesis, 1 mole of any substance contains 6.02×10²³ molecules.
Now from the question given, we were told that 1 molecule of the compound has a mass of 2.53×10¯²² g.
Therefore, 6.02×10²³ molecules will have a mass of = 6.02×10²³ x 2.53×10¯²² = 152.306 g
Therefore, 1 mole of the compound = 152.306 g
The molecular formula of the compound can be obtained as follow:
[C8H8O3]n = 152.306
[(12x8) + (1x8) + (16x3)]n = 152.306
[(96 + 8 + 48 ]n = 152.306
152n = 152.306
Divide both side by 152
n = 152.306/152
n = 1
The molecular formula => [C8H8O3]n
=> [C8H8O3]1
=> C8H8O3
Question 2:
A. Determination of the empirical formula of the compound.
Mass sample of compound = 0.648 g
Carbon (C) = 0.556 g
Mass of Hydrogen (H) = mass sample of compound – mass of carbon
Mass of Hydrogen (H) = 0.648 – 0.556
Mass of Hydrogen (H) = 0.092 g
Thus, the empirical formula can be obtained as follow:
C = 0.556 g
H = 0.092 g
Divide by their molar mass
C = 0.556/12 = 0.046
H = 0.092/1 = 0.092
Divide by the smallest
C = 0.046/0.046 = 1
H = 0.092/0.046 = 2
Therefore, the empirical formula of the compound is CH2.
B. Determination of the molecular formula of the compound.
Mole of compound = 0.5 mole
Mass of compound = 28.5 g
Molar mass of compound =.?
Mole = mass /Molar mass
0.5 = 28.5/ Molar mass
Cross multiply
0.5 x molar mass = 28.5
Divide both side by 0.5
Molar mass = 28.5/0.5 = 57 g/mol
Thus, the molecular formula of compound can be obtained as follow:
[CH2]n = 57
[12 + (1x2)]n = 57
14n = 57
Divide both side by 14
n = 57/14
n = 4
Molecular formula => [CH2]n
=> [CH2]4
=> C4H8.
Calculate how much acetylene (C2H2) will be produced from 358 g of H2O and an excess of CaC2 if the percent yield for this reaction is 94.5%. CaC2 2 H2O --> C2H2 Ca(OH)2
Answer:
244.7 g of acetylene
Explanation:
The balanced reaction equation is shown below;
2H20 (l) + CaC2 (s) → Ca(OH)2 (s) + C2H2 (g)
Number of moles of were reacted = reacting mass/molar mass = 358g/18gmol-1 = 19.89 moles of water
From the balanced reaction equation;
2 moles water yields 1 mole of acetylene
19.89 moles of water will yield 19.89 × 1/2 = 9.945 moles of acetylene
Theoretical yield of acetylene = 9.945 moles of acetylene × molar mass of acetylene
Molar mass of acetylene = 26.04 g/mol
Theoretical yield of acetylene = 9.945 moles of acetylene × 26.04 g/mol
Theoretical yield of acetylene = 258.9678 g of acetylene
% yield = actual yield/ theoretical yield × 100
94.5 = actual yield/258.9678 g × 100
Actual yield= 94.5 × 258.9678 g/100
Actual yield = 244.7 g of acetylene
Compound X absorbs photons with a wavelength of 179.3 nm. What is the increase in energy of a 0.115 M solution of compound X in which a mole of photons are absorbed
Answer:
Explanation:
one mole of photon will contain
6.02 x 10²³ no of photons
energy of one photon = h x f
= h c / λ
h is plank's constant , c is velocity of light and λ is wavelength
= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 179.3 x 10⁻⁹
= .11 x 10⁻¹⁷ J
energy of one mole of photon
= 6.02 x 10²³ x .11 x 10⁻¹⁷
= .6622 x 10⁶ J
Draw the curved arrow mechanism for the reaction between (2R,3R)-3,5-dimethylhexan-2-ol and PCl3.
Answer:
Sn2 mechanism
Explanation:
In this case, our nucleophile is the "OH" on (2R,3R)-3,5-dimethylhexan-2-ol. The alcohol group will attack the [tex]PCl_3[/tex] to produce a new bond between O and P with a positive charge in the oxygen. Additionally, when the OH attacks a Br atom leaves the molecule producing a bromide ion.
In the next step, the bromide ion produced will attack the carbon bonded to the OH that now is bonded to [tex]PCl_2[/tex]. An Sn2 reaction takes place and the substitution would be made in only one step. Due to this, we will have an inversion in the stereochemistry and the absolute configuration on carbon 2 will change from "R" to "S" to produce (2S,3R)-2-bromo-3,5-dimethylhexane.
I hope it helps!
Determine if each of the following statements about chiral molecules is True or False and select the best answer from the dropdown menus.
True or False: All R stereocenters are dextrorotatory.
True or False: Chiral centers in organic molecules will have 4 different groups attached to a carbon.
True or False: A racemic mixture has an optical activity of 0.
True or False: Normal linear amines can be chiral centers.
True or False: Compound C has an optical activity of 0.
True or False: If the R isomer of a molecule has an optical rotation of + 25.7 degree then the S isomer of the molecule will have an optical rotation of -25.7 degree.
True or False: A CN is higher priority than a CH2OH.
True or False: All molecules with chiral centers are optically active.
True or False: To have an enantiomer a molecule must have at least two chiral centers.
True or False: Chiral molecules are always optically active.
True or False: A CH2CH2Br is higher priority than a CH2F.
True or False: Meso molecules with two stereocenters have a R,S configuration.
True or False: Diastereomers have the same physical properties except in a chiral environment.
True or False: Compound H has an optical activity of 0.
True or False: A C=C double bond is higher priority than a -CH(CH3)2.
Answer:
See explanation
Explanation:
-) True or False: All R stereocenters are dextrorotatory.
The absolute configuration is based is in specific rules that are not related to the ability to deflect polarized light. FALSE
-) True or False: Chiral centers in organic molecules will have 4 different groups attached to a carbon.
A chiral carbon by definition is a carbon with 4 groups. TRUE
-) True or False: A racemic mixture has an optical activity of 0.
In a racemic mixture, we have equal amounts of enantiomers, and this cancels out the optical activity. TRUE
-) True or False: Normal linear amines can be chiral centers.
In primary amines, we have 2 hydrogens. Therefore all the groups can not be different. So, is TRUE
-)True or False: Compound C has an optical activity of 0.
We need to know the structure of the compound
-)True or False: If the R isomer of a molecule has an optical rotation of + 25.7 degree then the S isomer of the molecule will have an optical rotation of -25.7 degree.
If we have the exact opposite (as we have in this case) the magnitud of the optical activity value will remain and the sign will change. TRUE
-)True or False: A CN is a higher priority than a CH2OH.
In this case, a carbon is directly bonded to the chiral carbon. The carbon on CN is bonded to a nitrogen atom and the carbon on CH2OH is bonded to an oxygen. So, the CH2OH will have more priority because O has a higher atomic number. FALSE
-)True or False: All molecules with chiral centers are optically active.
We can have for example mesocompounds in which the optical activity is canceled out due to symmetry planes. FALSE
-)True or False: To have an enantiomer a molecule must have at least two chiral centers.
A pair of enantiomers is made between at least 1 chiral carbon. Enantiomer R and enantiomer S. FALSE
-)True or False: Chiral molecules are always optically active.
We can have racemic mixtures or mesocompounds with chiral carbons but without optical activity. FALSE
-)True or False: A CH2CH2Br is higher priority than a CH2F.
The "Br" atom is bonded in the third carbon (respect to the chiral carbon) and the "F" atom is bonded to the second carbon. Therefore CH2F has more priority than CH2CH2Br. FALSE
-)True or False: Meso molecules with two stereocenters have a R,S configuration.
On the compounds with R and S configuration at the same time can have symmetry planes so, we will not have optical activity. TRUE
True or False: Diastereomers have the same physical properties except in a chiral environment.
All diastereomers have the same physical properties. TRUE
True or False: Compound H has an optical activity of 0.
We have to have the structure of the compound.
True or False: A C=C double bond is higher priority than a -CH(CH3)2.
In the case of C=C, we can say that is equivalent to two carbon bonds without hydrogens. Therefore C=C has higher priority. TRUE
Calculate the quantity of energy produced per gram of U-235 (atomic mass = 235.043922 amu) for the neutron-induced fission of U-235 to form Xe-144 (atomic mass = 143.9385 amu) and Sr-90 (atomic mass = 89.907738 amu). Express your answer in joules per gram to three significant figures.
Answer:
To calculate the amount of energy produced per gram from Uranium- 235 we call on the formula:
delta(m)= mass of the products - mass of reactants
Given the atomic mass of Xe-144 = 143.9385 amu; atomic mass of Sr-90= 89.907738 amu; atomic mass of U-235= 235.043922 amu
Therefore:
delta(m)= (143.9385 + 89.907738) - (235.043922) = -1.197682 amu
Recall that to calculate energy in joules, we use the formula:
Energy = mc^2
Therefore: Energy = (-1.197582/6.022 x 10^26)kg x (3 x 10^8 m/s)^2
= -1.7898104 x 10^-10 J
= (-1.7898104 x 10^-10 x 6.022 x 10^23/235.043922)
= -4.586 x 10^11 J per gram of energy released
Explanation:
To calculate the amount of energy produced per gram from Uranium- 235 we call on the formula:
delta(m)= mass of the products - mass of reactants
Given the atomic mass of Xe-144 = 143.9385 amu; atomic mass of Sr-90= 89.907738 amu; atomic mass of U-235= 235.043922 amu
Therefore:
delta(m)= (143.9385 + 89.907738) - (235.043922) = -1.197682 amu
Recall that to calculate energy in joules, we use the formula:
Energy = mc^2
Therefore: Energy = (-1.197582/6.022 x 10^26)kg x (3 x 10^8 m/s)^2
= -1.7898104 x 10^-10 J
= (-1.7898104 x 10^-10 x 6.022 x 10^23/235.043922)
= -4.586 x 10^11 J per gram of energy released
Need help finding major products
Answer:
Explanation:
RX + AgNO₃ = R⁺ ( carbocation ) + AgX + NO₃⁻
C₂H₅OH ( a nucleophile ) + R⁺ = ROC₂H₅
C₅H₁₁X + AgNO₃ = C₅H₁₁⁺ + AgX + NO₃⁻
In the first case carbocation produced is CH₃CH₂CH₂CH₂CH₂⁺
CH₃CH₂CH₂CH₂CH₂⁺ ⇒ CH₃CH₂CH₂C⁺HCH₃ ( secondary carbocation more stable )
CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃
Hence option D is correct .
b )
In the second case carbocation produced is
CH₃CH₂CH₂CH⁺CH₃
CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃
The product formed is same as in case of first
Option B is correct
How does a balanced chemical equation show the conservation of mass?
A. It shows that the number of each type of atom stays the same.
B. It shows that the mass of the products is greater than the mass of
the reactants when a reaction increases the moles of substances.
C. It shows that the total number of moles of substances stays the
same.
D. It shows that the mass of the reactants is greater than the mass
Answer:
A. It shows that the number of each type of atom stays the same.
Explanation:
Though you may see a change in the way they are arranged, the same number of atoms are present before and after. Balanced chemical equations show equal numbers of atoms of each element on each side of the equation.
What mass of Si, in grams, can be produced from the reaction below starting with 225 g SiCl4 and 101 g Mg? SiCl4 + Mg Si + MgCl2 Given: 1 mol SiCl4 = 169.8963 g SiCl4 1 mol Mg = 24.3050 g Mg 1 mol Si = 28.0855 g Si
Answer:
[tex]m_{Si}=37.2gSi[/tex]
Explanation:
Hello,
In this case, for the undergoing balanced chemical reaction:
[tex]SiCl_4 + 2Mg \rightarrow Si + 2MgCl_2[/tex]
We must first identify the limiting reactant given the 225 g of SiCl4 and 101 g of Mg. Thus, we compute the available moles of SiCl4:
[tex]n_{SiCl_4}=225gSiCl_4*\frac{1molSiCl_4}{169.8963gSiCl_4}=1.324molSiCl_4[/tex]
Next, by using the 1:2 mole ratio between SiCl4 and Mg, we compute the moles of SiCl4 consumed by 101 g of Mg:
[tex]n_{SiCl_4}^{consumed}=101gMg*\frac{1molMg}{24.3050gMg} *\frac{1molSiCl_4}{2molMg} =2.08molSiCl_4[/tex]
Thus, since less moles of SiCl4 are available, we can infer it is the limiting reactant whereas the Mg is in excess. In such a way, the produced grams of Si are computed considering the 1:1 molar ratio between SiCl4 and Si:
[tex]m_{Si}=1.324molSiCl_4*\frac{1molSi}{1molSiCl_4} *\frac{28.0855gSi}{1molSi} \\\\m_{Si}=37.2gSi[/tex]
Best regards.
Danial has a simple of pure copper.its mass 89.6 grams (g),and its volume is 10 cubic centermeters (cm3) whats the destiny of the sample?
Answer:
8.96g\ cm3
Explanation:
D = ( 89.6g \ 10cm3)
( 89.6\ 10) ( g\ cm3) = 8.96g\cm3
Identify some other substances (besides KCl) that might give a positive test for chloride upon addition of AgNO3. Based on the reactant used in your experiment, do you think it is reasonable to exclude these types of substances as contaminants that would give a false positive when you tested your reaction residue to verify that it is KCl?
Answer:
Other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide. However iodides and bromides have different colours hence they will not give a false positive test for KCl. Other chlorides present may lead to a false positive test for KCl.
Explanation:
In the qualitative determination of halogen ions, silver nitrate solution is used. Various halide ions give various colours of precipitate with silver nitrate. Chlorides yield a white precipitate, bromides yield a cream precipitate while iodides yield a yellow precipitate. All these ions or some of them may be present in the system.
However, if other chlorides are present, they will also yield a white precipitate just as KCl leading to a false positive test for KCl. Since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. We can exclude other halides from the tendency to lead us to a false positive test for KCl but not other chlorides.
In the qualitative analysis test of chloride upon addition of AgNO₃, presence of diffrerent chloride salts other than KCl will gives false test.
How do we get on addition of KCl in AgNO₃?On adding halogens on the silver nitrate solution we will get the precipitate of diffrent colors of diffrent halides.
Chlorides will gives white color precipiate in the silver nitrate solution, so KCl and other chlorides will also give white color precipitate.Bromides will gives cream color precipitate.Iodides will gives yellow color precipitate.So, presence of diffrent chloride salt in the silver nitrate solution in addition with KCl will gives a false positive result for the test.
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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 pH of a buffer solution obtained by dissolving 18.0 g of KH2PO4(s) and 35.0 g of Na2HPO4(s) in water and then diluting to 1.00 L.
Answer:
pH of the buffer is 7.48
Explanation:
The H₂PO₄⁻/HPO₄²⁻ buffer has a pKa of 7.21. You can find pH of this buffer following H-H equation:
pH = pKa + log [A⁻] / [HA]
pH = 7.21 + log [HPO₄²⁻] / [ H₂PO₄⁻]
Where [] represents molarity of each specie of the buffer and, as volume is 1.00L, also represents its moles.
Thus, to find pH of the buffer we need to calculate moles of each specie, thus
Moles of 18.0g of KH₂PO₄(Molar mass: 136.086g/mol) = moles of H₂PO₄⁻ are:
18.0g KH₂PO₄ ₓ (1mol / 136.086g) = 0.132 moles of KH₂PO₄= H₂PO₄⁻
Moles of 35.0g of Na₂HPO₄(Molar mass: 141.96g/mol) = moles of HPO₄²⁻ are:
35.0g Na₂HPO₄ ₓ (1mol / 141.96g) = 0.2465 moles of Na₂HPO₄= HPO₄²⁻
Replacing in H-H equation:
pH = 7.21 + log [HPO₄²⁻] / [ H₂PO₄⁻]
pH = 7.21 + log [0.2465] / [0.132]
pH = 7.48
pH of the buffer is 7.48
Calculate the amount of HCl in grams required to react with 3.75 g of CaCO3 according to the following reaction: CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Answer:
The correct answer is 2.75 grams of HCl.
Explanation:
The given balanced equation is:
CaCO₃ (s) + 2HCl (aq) ⇒ CaCl₂ (aq) + H₂O (l) + CO₂ (g)
Based on the given information, one mole of calcium carbonate is reacting with two moles of HCl. The molecular mass of HCl is 36.5 grams, thus, the mass of 2 moles of HCl will be, 36.5 × 2 = 73 grams
The molecular mass of CaCO₃ is 100 gram per mole, that is, the mass of 1 mole of CaCO₃ is 100 grams, therefore, the mass of HCl required for reacting with 3.75 grams of CaCO₃ will be,
= 3.75 × 2 × 36.5 / 100 = 2.74 grams of HCl.
what is the meaning of the word tetraquark?
Answer:
A tetraquark in physics is an exotic meson composed of four valence quarks.
Explanation:
It has been suspected to be allowed by quantum chromodynamics, the modern story of strong interactions.
Hope it helps.
A. Identify the structure drawn below.
Answer:
Hexane
Explanation:
You have a carbon structure with only single bonds. This means that the name will end in -ane.
There are 6 carbon atoms. This means that the name will begin with hex-.
The structure is hexane.
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
Consider the reaction between two solutions, X and Y, to produce substance Z: aX + bY → cZ When 500. mL of a 1.8 M solution of X is combined with 500. mL of a 1.8 M solution of Y, the resulting solution has a concentration of 0.60 M Y and 0.60 M Z. No more of substance X remains in the flask. 1. How many moles each of X and Y are present before the reaction occurs? 2. How many moles each of Y and Z are present after the reaction occurs? 3. How many moles each of X and Y have reacted? 4. What is the balanced equation for this reaction?
Answer:
1. 0.90 are the initial moles of X and Y
2. 0.60 moles are the moles of Y and Z after the reaction
3. 0.90 moles of X and 0.30 moles of Y
4. 3X + 1Y → 2Z
Explanation:
1. For the reaction, initial moles of X and Y are:
500mL = 0.500L × (1.8 moles / L) = 0.90 are the initial moles of X and Y
2. After the reaction. The total volume is 500mL + 500mL = 1L
Moles Y and Z = 1L × (0.60 moles / 1L) = 0.60 moles are the moles of Y and Z after the reaction
3. As there is no moles of X after the reaction, all X reacts, that is 0.90 moles of X. And moles of Y that reacts are 0.90 mol - 0.60mol = 0.30 moles of Y
4. That means 3 moles of X reacts per mole of Y 0.90/0.30 = 3. Also, 2 moles of Z are produced per mole of Y 0.60/0.30 = 2.
That means balanced equation is:
aX + bY → cZ
3X + 1Y → 2ZQ1. 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.