The question is incomplete; the complete question is;
Before running a titration, you calculate the expected endpoint. However, when performing the experiment, you pass the expected endpoint with no visible color change. What is the most likely problem with the titration set- up? Select one
a) There was an air bubble in the burette tip.
b) There is not enough indicator in the analyte.
c) The burette tip is leaking titrant into the analyte.
d) The analyte solution is being stirred too quickly
Answer:
a) There was an air bubble in the burette tip.
Explanation:
Titration involves the determination of the concentration of a solution by measuring the volumes of reactants used in the reaction. The concentration of one of the species must be known while the concentration of the other specie is to be determined by the volumetric analysis.
However, if there are air bubbles at the tip of the burette, this will cause less volume of titrant to be delivered from the burette than expected. Hence, the analyst may think that a certain volume of titrant has been delivered while in reality, a lesser volume was actually delivered due to the air bubbles present. Hence, the analyst may pass the expected endpoint without any colour change because of this problem.
From the available options to the question:
a) There was an air bubble in the burette tip.
b) There is not enough indicator in the analyte.
c) The burette tip is leaking titrant into the analyte.
d) The analyte solution is being stirred too quickly
The most likely problem with the titration setup that could make one to pass the expected endpoint with no visible color change would be if there is not enough indicator in the analyte. The correct option would be B.
A suitable quantity (in drops) of the indicator should be added to the analyte in the conical flask before carrying out a titration. The color of indicators changes quickly near their pKa.
If too few drops of the indicator is used, the color change will be too faint to be obvious and the endpoint will be exceeded. If too many drops of the indicator is used, the final pH of the reaction would be affected and the titer value will be inaccurate.
In this case, the expected endpoint has been exceeded without any color change. The most likely problem would, therefore, be that there is not enough indicator in the analyte.
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Benny Beaver wants to determine what dyesare present in his favorite sports drink. He analyzesa sample witha UV-visiblespectrophotometer and sees absorbance peaks at 415.2nm and 519.6nm. What colordyesare present in his drink
Answer:
At 415.2nm and 519.6nm, the dyes observed by the instrument are violet and green respectively.
Explanation:
In the electromagentic spectrum, visible wavelengths cover a range from approximately 400 to 800 nm. The colours of the spectrum range from red to violet (Red, Orange, Yellow, Green, Blue, Indigo and violet: a.k.a ROGBIV), in order of decreasing wavelength.
I hope this explanation would suffice.
Question 1
1 pts
2B+6HCI --
| --> 2BCl3 + 3H2
How many moles of boron chloride will be produced if you start with 8.752 moles of HCI
(hydrochloric acid)? (Round to 3 sig figs. Enter the number only do not include units.)
Answer:
2.92 mol
Explanation:
Step 1: Write the balanced equation
2 B(s) + 6 HCI(aq) ⇒ 2 BCl₃(aq) + 3 H₂(g)
Step 2: Establish the appropriate molar ratio
The molar ratio of hydrochloric acid to boron chloride is 6:2.
Step 3: Calculate the moles of boron chloride produced from 8.752 moles of hydrochloric acid
[tex]8.752molHCl \times \frac{2molBCl_3}{6molHCl} = 2.92molBCl_3[/tex]
At a particular temperature, an equilibrium mixture the reaction below was found to contain 0.171 atm of I2, 0.166 atm of Cl2 and 9.81 atm of ICl. Calculate the value of the equilibrium constant, Kp at this temperature.I2(g) + Cl2(g) <=> 2 ICl(g)
Answer: 3390
Explanation:
Since this problem already gives is the equilibrium values, all we have to do is to plug them into the formula for [tex]K_{p}[/tex].
[tex]K_{p} =\frac{[ICl]^2}{[I_{2}][Cl_{2}] }[/tex]
[tex]K_{p} =\frac{(9.81)^2}{(0.171)(0.166)} =3390[/tex]
A glass flask has a volume of 500 mL at a temperature of 20° C. The flask contains 492 mL of mercury at an equilibrium temperature of 20°C. The temperature is raised until the mercury reaches the 500 mL reference mark. At what temperature does this occur? The coefficients of volume expansion of mercury and glass are 18 ×10-5 K-1 (mercury) and 2.0 ×10-5 K-1 (glass).
Answer:
101.63° C
Explanation:
Volume expansivity γa = γr - γ g = 18 × 10⁻⁵ - 2.0 × 10⁻⁵ = 16 × 10⁻⁵ /K
v₂ - v₁ / v₁θ = 16 × 10⁻⁵ /K
(500 - 492 ) mL / (492 × 16 × 10⁻⁵) = θ
θ = 101.63° C
17. Write the molecular balanced equation for the recovering of copper metal. 18. Write the complete ionic balanced equation for the recovering of copper metal. 19. Write the net ionic balanced equation for the recovering of copper metal. 20. What type of reaction is this
Answer:
Explanation:
17. it goes from solid copper to aqueous copper:
Cu(s) --> Cu₂(aq) + 2e⁻
18. complete ionic:
Cu(s) --> Cu₂(aq) + 2e⁻
19. net ionic, must include only reacting species, so
Cu(s) --> Cu₂(aq) + 2e⁻
20. this type of reaction is dissolution reaction(redox reaction)
copper reduced from Cu²⁺ to Cu.
all compounds are neutral true or false
Answer:
Even all compounds are neutral.
Explanation:
Some of them exhibit polarity. Because of the difference in electron affinity of the constituent atoms, the shared electrons are pulled towards the atom with high affinity to electrons.
In the diagram below, particles of the substance are moving from the liquid phase to the gas phase at the same rate as they move from the gas phase to the liquid phase. A number of balls are loosely packed in the bottom of a container, beneath a line across the middle of the container, and a few balls above the line. 2 balls below the line have arrows pointing upward through the centerline; a few of the balls above the line have arrows pointing down through the centerline. The gas and liquid are at equilibrium. a high vapor pressure. a low vapor pressure. zero vapor pressure.
Answer:
The gas and liquid is in equilibrium.
Explanation:
liquids within a container undergoes state change, changing into gas. If this container is left open, these gases will escape into the external environment. In a situation in which the container is closed, the molecules that leave the liquid surface as gas will eventually condense on contact with the cover wall and change back into the liquid state. Some of these gases will reenter the liquid surface. At first, more of the liquid is transformed into gas and escape into the space above the liquid surface. Eventually, the available space becomes saturated with vapor, and then some of the gases start entering the liquid phase at the same rate as the liquid enters the gas phase. At this stage, the gas and liquid phase now exists in equilibrium.
calculate how many moles of CaCl2•2H2O are present in 1.50 g of CaCl2•2H2O and then calculate how many moles of pure CaCl2 are present in the 1.50 g of CaCl2•2H2O.
Answer:
[tex]0.0102~mol~CaCl_2*2H_2O[/tex]
[tex]0.0102~mol~CaCl_2[/tex]
Explanation:
For this question, we have to start with the molar mass calculation of [tex]CaCl_2*2H_2O[/tex]. For this, we have to know the atomic mass of each atom:
O: 16 g/mol
Cl: 35.45 g/mol
H: 1 g/mol
Ca: 40 g/mol
If we take into account the amount of each atom in the formula we will have:
[tex](40*1)+(35.45*2)+(1*4)+(16*2)=~147.01~g/mol[/tex]
So, in 1 mol of [tex]CaCl_2*2H_2O[/tex] we will have 147.01 g. Now we can do the conversion:
[tex]1.50~g~CaCl_2*2H_2O\frac{1~mol~CaCl_2*2H_2O}{147.01~g~CaCl_2*2H_2O}=0.0102~mol~CaCl_2*2H_2O[/tex]
Additionally, in 1 mol of [tex]CaCl_2*2H_2O[/tex] we will have 1 mol of [tex]CaCl_2[/tex]. Therefore, we have a 1:1 mol ratio . With this in mind, we will have the same number of moles for [tex]CaCl_2[/tex]
[tex]0.0102~mol~CaCl_2*2H_2O=0.0102~mol~CaCl_2[/tex]
I hope it helps!
Combustion analysis of a 13.42-g sample of estriol (which contains only carbon, hydrogen, and oxygen) produced 36.86 g CO2 and 10.06 g H2O. The molar mass of estriol is 288.38 g/mol . Find the molecular formula for estriol. Express your answer as a chemical formula.
Answer:
C18H24O3
Explanation:
Step 1:
Data obtained from the question. This include the following:
Mass of estriol = 13.42g
Mass of CO2 = 36.86g
Mass of H2O = 10.06g
Molar mass of estriol = 288.38g/mol
Step 2:
Determination of the mass of Carbon (C), Hydrogen (H) and Oxygen (O) present in the compound. This is illustrated below:
For Carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C in CO2 = 12/44 x 36.86 = 10.05g
For Hydrogen, H:
Molar Mass of H2O = (2x1) + 16 = 18g/mol
Mass of H in H2O = 2/18 x 10.06 = 1.12g
For Oxygen, O:
Mass of O = 13.42 – (10.05 + 1.12) = 2.25g
Step 3:
Determination of the empirical formula for estriol. This is illustrated below:
C = 10.05g
H = 1.12g
O = 2.25g
Divide by their molar mass
C = 10.05/12 = 0.8375
H = 1.12/1 = 1.12
O = 2.25/16 = 0.1406
Divide by the smallest i.e 0.1406
C = 0.8375/0.1406 = 6
H = 1.12/0.1406 = 8
O = 0.1406/0.1406 = 1
Therefore, the empirical formula for estriol is C6H8O
Step 4:
Determination of the molecular formula for estriol. This is illustrated below:
Molecular formula is simply a multiple of the empirical formula i.e
Molecular formula => [C6H8O]n
[C6H8O]n = 288.38g/mol
[(12x6) + (8x1) + 16]n = 288.38
[72 + 8 + 16]n = 288.38
96n = 288.38
Divide both side by 96
n = 288.38/96 = 3
Molecular formula => [C6H8O]n
=> [C6H8O]n
=> [C6H8O]3
=> C18H24O3
Therefore, the molecular formula for estriol is C18H24O3
The compound is C18H24O3.
From the information in the question;
Mass of C = 36.86 g/44 g/mol × 12 g/mol = 10.1 g
Number of moles of carbon = 10.1 g/12 g/mol = 0.84 moles
Mass of hydrogen = 10.06 g/18 g/mol × 2 g/mol = 1.11 g
Number of moles of hydrogen = 1.11 g/1g/mol = 1.11 moles
Mass of oxygen = 13.42 - (10.1 g + 1.11 g) = 2.21 g
Number of moles of oxygen = 2.21g/16 g/mol = 0.14 moles
Dividing through by the lowest number of moles;
C - 0.84 moles/0.14 moles H - 1.11 moles/0.14 moles O - 0.14 moles/0.14 moles
C - 6 H - 8 O -1
The empirical formula is C6H8O
The molecular formula of the compound is;
[6(12) + 8(1) + 16]n = 288.38
n = 288.38/86 =3
The compound is C18H24O3
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Which of the following would describe a spontaneous process? (Select all that apply.) ΔGreaction < 0 ΔSuniverse < 0 ΔSuniverse > 0 ΔHreaction > 0 ΔHreaction < 0 ΔGreaction > 0
Answer:
ΔGreaction < 0
ΔSuniverse > 0
ΔHreaction < 0
Explanation:
A spontaneous process is one which can proceed without additional input of energy releasing free energy in the process and then moves to a lower more stable thermodynamical state.
For an isolated system, a spontaneous process proceeds with an increase in entropy.
The conditions for a spontaneous process at constant temperature and pressure, can be determined using the change in Gibbs free energy, which is given by: ∆G = ∆H - T∆S
Where ∆G is change in free energy; ∆H is change in enthalpy or Heat content; ∆S is change in entropy, T is temperature.
For a process to be spontaneous, the following conditions are necessary:
1. ∆G < 0; must be negative
2. ∆S > 0; there must be an increase in entropy
3. ∆H < 0; enthalpy change must be negative such that heat is lost to the surroundings
The above conditions ensures that ∆G is negative and the process is spontaneous.
The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperature, the chemical reactions associated with the cooking of food take place at a greater rate. (a) Some food cooks fully in 7.00 min in a pressure cooker at 113.0°C and in 49.0 minutes in an open pot at 100.0°C. Calculate the average activation energy for the reactions associated with the cooking of this food. kJ mol-1 (b) How long will the same food take to cook in an open pot of boiling water at an altitude of 10000 feet, where the boiling point of water is 89.8 °C? min
Answer:
the activation energy Ea = 179.176 kJ/mol
it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.
Explanation:
From the given information
[tex]T_1 = 100^0 C = 100+273 = 373 \ K \\ \\ T_2 = 113^0 C = 113 + 273 = 386 \ K[/tex]
[tex]R_1 = \dfrac{1}{7}[/tex]
[tex]R_2 = \dfrac{1}{49}[/tex]
Thus; [tex]\dfrac{R_2}{R_1} = 7[/tex]
Because at 113.0°C; the rate is 7 time higher than at 100°C
Hence:
[tex]In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})[/tex]
1.9459 = [tex]\dfrac{Ea}{8.314}* 9.0292 *10^{-5}[/tex]
[tex]1.9459*8.314 = Ea * 9.0292*10^{-5}[/tex]
[tex]16.1782126= Ea * 9.0292*10^{-5}[/tex]
[tex]Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}[/tex]
Ea = 179.176 kJ/mol
Thus; the activation energy Ea = 179.176 kJ/mol
b)
here;
[tex]T_2 = 386 \ K \\ \\T_1 = (89.8 + 273)K = 362.8 \ K[/tex]
[tex]In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})[/tex]
[tex]In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})[/tex]
[tex]In (\dfrac{R_2}{R_1}) = 0.00357[/tex]
[tex]\dfrac{R_2}{R_1}= e^{0.00357}[/tex]
[tex]\dfrac{R_2}{R_1}= 1.0035[/tex]
where ;
[tex]R_2 = \dfrac{1}7{}[/tex]
[tex]R_1 = \dfrac{1}{t}[/tex]
Now;
[tex]\dfrac{t}{7}= 1.0035[/tex]
t = 7.0245 mins
Therefore; it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.
a). The activation energy given by the reactions related to the cooking of food in the pressure cooker would be:
[tex]Ea = 179.176 kJ/mol[/tex]
b). The time duration that is taken by the same food to cook in an open vessel would be:
[tex]7.0245 mins[/tex]
Activation Energya). Given that,
Temperature [tex]1[/tex] [tex]= 100[/tex]° C
Temperature [tex]2[/tex] [tex]= 113[/tex]° C
In Kelvin,
Temperature [tex]1[/tex] [tex]= 100 + 273[/tex]
[tex]= 373 K[/tex]
Temperature [tex]2[/tex] [tex]= 113 + 273[/tex]
[tex]= 386 K[/tex]
[tex]R_{1} = 1/7\\R_{2} = 1/49[/tex]
∵ [tex]R_{2}/R_{1} = 49/7 = 7[/tex]
It is given that at [tex]113[/tex] rate [tex]=[/tex] [tex]7[/tex] × [tex]100[/tex]°C
Therefore,
[tex]Ea/8.314 (1/373 - 1/386) =[/tex] [tex]In(7)[/tex]
so,
[tex]Ea[/tex] [tex]= 16.1782126/(9.0292 * 10^{-5})[/tex]
∵ Activation energy [tex]= 179.176 kJ/mol[/tex]
b). As we know,
[tex]T_{2}[/tex] [tex]= 386 K[/tex]
[tex]T_{1}[/tex] [tex]= (89. 8 + 273)[/tex]
[tex]= 362.8 K[/tex]
by employing the formulae,
[tex]In(\frac{R_{2} }{R_{1} }) = \frac{Ea}{R} (1/T_{1} - 1/T_{2})[/tex]
[tex]In(\frac{R_{2} }{R_{1} }) = 179.176/8.314 (1/362.8 - 1/386)[/tex]
By solving this, we get
[tex]R_{2}/R_{1} = 1.0035[/tex]
Thus,
[tex]R_{2} = 1/7[/tex]
[tex]R_{1} = 1/t[/tex]
∵ t [tex]= 7.0245 min[/tex]
Thus, the time duration would be [tex]7.0245 minutes[/tex].
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Given a gas whose temperature is 418 K at a pressure of 56.0 kPa. What is the pressure of the gas if its Temperature changes to 64°C?
Answer: P₂=0.44 atm
Explanation:
For this problem, we are dealing with temperature and pressure. We will need to use Gay-Lussac's Law.
Gay-Lussac's Law: [tex]\frac{P_{1} }{T_{1} } =\frac{P_{2} }{T_{2} }[/tex]
First, let's do some conversions. Anytime we deal with the Ideal Gas Law and the different laws, we need to make sure our temperature is in Kelvins. Since T₂ is 64°C, we must change it to K.
64+273K=337K
Now, it may be uncomfortable to use kPa instead of atm, so let's convert kPa to atm.
[tex]56.0kPa*\frac{1000Pa}{1kPa} *\frac{atm}{101325Pa} =0.55atm[/tex]
Since our units are in atm and K, we can use Gay-Lussac's Law to find P₂.
[tex]P_{2} =\frac{T_{2} P_{1} }{T_{1} }[/tex]
[tex]P_{2}=\frac{(337K)(0.55atm)}{418K}[/tex]
P₂=0.44 atm
ii)
An organic liquid having carbon, hydrogen, nitrogen and oxygen was found to
contain C = 41.37%; H = 5.75% ; N= 16.09 % and the rest oxygen. Calculate
the Empirical formula
(6 marks)
Answer:
C3H5NO2
Explanation:
C = 41.37%;
H = 5.75% ;
N= 16.09 %;
O= (100 - 41.37 - 5.75 - 16.09)% = 36.79 %.
In 100 g of substance we have
C = 41.37 g;
H = 5.75 g ;
N= 16.09 g;
O = 36.79 g.
Molar mass (C) = 12 g/mol;
Molar mass(H) =1 g/mol;
Molar mass(N)= 14 g/mol;
Molar mass(O) = 16 g/mol.
C = 41.37 g* 1 mol/12g = 3.4475 mol;
H = 5.75 g *1 mol/1g = 5.75 mol;
N= 16.09 g*1mol/14g = 1.1493 mol;
O = 36.79 g* 1mol/16g = 2.2994 mol.
The Empirical formula shows ratio of moles of elements in the substance , so
C : H : N : O = 3.4475 mol : 5.75 mol : 1.1493 mol : 2.2994 mol =
= (3.4475 mol /1.1493 mol) : (5.75 mol/1.1493 mol) : (1.1493 mol /1.1493 mol) : :( 2.2994 mol/1.1493 mol) = 3 : 5 : 1 : 2
C : H : N : O = 3 : 5 : 1 : 2
C3H5NO2
The proposed mechanism for a reaction is: Step 1: A + B X (fast) Step 2: X + C Y (slow) Step 3: Y D (fast) What is the overall reaction? A. A + B + C D B. A + X Y + D C. A + B Y D. A + Y D
Answer:
A. A + B + C --> D
Explanation:
Step 1: A + B --> X (fast)
Step 2: X + C --> Y (slow)
Step 3: Y --> D (fast)
To obtain the overall reaction, we have to sum up the reactants and products of all step and eliminate the intermediates.
Reactants:
A + B + X + C + Y
Products:
X + Y + D
So we have;
A + B + X + C + Y --> X + Y + D
Upon elimination of intermediates, we have;
A + B + C --> D
The correct option is A.
A certain mass of carbon reacts with 9.53 g of oxygen to form carbon monoxide. ________ grams of oxygen would react with that same mass of carbon to form carbon dioxide, according to the law of multiple proportions.
Answer: 9.53 *2= 19.06
Explanation:
The law of multiple proportions states that if two elements combines to form more than one compound the ratio of masses of the second element which combines to the fixed mass of the first element will always be the ratios of the small whole numbers.
in case of carbon monoxide, mass of carbon will be the same of mass of oxygen.
But in case of carbon dioxide, if carbon is 9.53 units then oxygen will be twice as that of carbon.
CO2, so 9.53*2= 19.06 grams of oxygen will combine with 9.53 grams of carbon to form carbon dioxide.
Tubes through which water flows as it is brought from 0.8 MPa, 150C to 240C at essentially constant pressure in the boiler of a power plant. The total mass flow rate of the water is 100 kg/s. Combustion gases passing over the tubes cool from 1067 to 547C at essentially constant pressure. The combustion gases can be modeled as air as an ideal gas. There is no significant heat transfer from the boiler to its surroundings. Surrounding (dead state) temperature and pressure are given as 25C and 1 atm, respectively. Determine i) the exergetic efficiency of the boiler ii) rate of exergy destruction as kW iii) mass flow rate of the combustion gases as kg/s
Answer:
The correct answer is i) 50.2 % ii) 13440.906 kW and iii) 71.986 kg/s.
Explanation:
In order to find the mass flow rate of the combustion of gases, there is a need to use the energy balance equation:
Mass of water × specific heat of water (T2 -T1)w = mass of gas × specific heat of gas (T2-T1)g
100 × 4.18 × [(240 + 273) - (150 + 273)] = mass of gas × 1.005 × [(1067+273) - (547+273)]
Mass of gas = 71.986 kg/s
The entropy generation of water can be determined by using the formula,
(ΔS)w = mass of water × specific heat of water ln(T2/T1)w
= 100 × 4.18 ln(513/423)
= 80.6337 kW/K
Similarly the entropy generation of water will be,
(ΔS)g = mass of gas × specific heat of gas ln(T2/T1)g
= 71.986 × 1.005 ln (820/1340)
= -35.53 kW/K
The rate of energy destruction will be,
Rate of energy destruction = To (ΔS)gen
= T₀ [(ΔS)w + (ΔS)g]
= (25+273) [80.6337-53.53)
Rate of energy destruction = 13440.906 kW
The availability of water will be calculated as,
= mass of water (specific heat of water) [(T₁-T₂) -T₀ ln T₁/T₂]
= 100 × 4.8 [(513-423) - 298 ln 513/423]
= 13591.1477 kW
The availability of gas will be calculated as,
= mass of gas (specific heat of gas) [(T₁-T₂) - T₀ ln T₁/T₂]
= 71.986 × 1.005 × [(1340-820) - 298 ln 1340/820]
= 27031.7728 kW
The exergetic efficiency can be calculated as,
= Gain of availability / loss of availability
= 13591.1477/27031.7728
= 0.502
The exergetic efficiency is 50.2%.
Telluric acid (H2TeH4O6) is a diprotic acid with Ka1 = 2.0x10-8 and Ka2 = 1.0x10-11. A 0.25 M H2TeH4O6 contains enough HCl so that the pH is 3.00. What is the concentration of HTeH4O6
Answer:
5x10⁻⁶ = [HTeH₄O₆⁺]
Explanation:
The first dissociation equilibrium of the telluric acid in water is:
H₂TeH₄O₆ + H₂O ⇄ HTeH₄O₆⁺ + H₃O⁺
Using H-H equation for telluric acid:
pH = pKa + log₁₀ [HTeH₄O₆⁺] / [H₂TeH₄O₆]
pKa of telluric acid is -logKa1
pKa = -log 2.0x10⁻⁸
pKa = 7.699
As concentration of [H₂TeH₄O₆] is 0.25M, replacing in H-H equation:
3.00 = 7.699+ log₁₀ [HTeH₄O₆⁺] / [0.25M]
-4.699 = log₁₀ [HTeH₄O₆⁺] / [0.25M]
2x10⁻⁵ = [HTeH₄O₆⁺] / [0.25M]
5x10⁻⁶ = [HTeH₄O₆⁺]408 J of energy is transferred to a system consisting of 2.0 moles of an ideal gas. If the volume of this gas stays at a constant 4.4 L, calculate the change in internal energy of the gas.
Answer: The change in internal energy of the gas is +408 J
Explanation:
According to first law of thermodynamics:
[tex]\Delta E=q+w[/tex]
[tex]\Delta E[/tex] =Change in internal energy
q = heat absorbed or released
w = work done or by the system
w = work done on the system= [tex]-P\Delta V[/tex]
As volume is constant , [tex]\Delta V[/tex] = 0 and w = 0.
q = +408J {Heat absorbed by the system is positive}
[tex]\Delta E=+408J+(0J)=+408J[/tex]
Thus the change in internal energy of the gas is +408 J
Choose the correct option for the redox reaction below : 8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) 4H2O(l) H+ (oxidized or reduced) Answer 1 Choose... MnO4- (oxidizing or reducing agent) Answer 2 Choose... Fe2+ (oxidizing or reducing agent) Answer 3 Choose... Fe2+ (oxidized or reduced) Answer 4 Choose... O in MnO4- (oxidized or reduced) Answer 5 Choose... Mn in MnO4- (oxidized or reduced) Answer 6 Choose...
Explanation:
8H+(aq) + MnO4-(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
To determine if a substance was reduced or oxidized, the oxidation number at the reactant and product side is compared.
When oxidation occurs, an element loses electrons and its oxidation number increases (becomes more positive). When reduction occurs, an element gains electrons and its oxidation number decreases or is reduced (becomes more negative).
H+ (oxidized or reduced) Answer 1 Choose...
There is an increase in oxidation number from 0 to 1. This means H+ was oxidized.
MnO4- (oxidizing or reducing agent) Answer 2 Choose...
In chemistry, an oxidizing agent is a substance that has the ability to oxidize other substances — in other words to accept their electrons
MnO4- is a strong oxidizing agent.
Fe2+ (oxidizing or reducing agent) Answer 3 Choose...
A reducing agent means it will itself get oxidized. In this reaction, Fe2+ is the reducing agent.
Fe2+ (oxidized or reduced) Answer 4 Choose...
Oxidized
O in MnO4- (oxidized or reduced) Answer 5 Choose...
There is no change in oxidation number of , so it was neither reduced nor was it oxidized.
Mn in MnO4- (oxidized or reduced) Answer 6 Choose...
There is a decreaase in oxidation number from +7 to +2. This means reduction.
What type of chemical reaction occurs between AgNO3 (sliver nitrate) and Cu (Copper)?
Answer:
The answer is option c.
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In the first 15.0 s of the reaction, 1.7×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.440 L . What is the average rate of the reaction over this time interval?
Answer:
[tex]Rate=2.57x10^{-3}\frac{M}{s}[/tex]
Explanation:
Hello,
In this case, for the reaction:
[tex]2N_2O(g) \rightarrow 2N_2(g)+O_2(g)[/tex]
We can easily compute the average rate by firstly computing the final concentration of oxygen:
[tex][O_2]=\frac{0.017mol}{0.440L}=0.0386M[/tex]
Then, we compute it by using the given interval of time: from 0 seconds to 15.0 seconds and concentration: from 0 M to 0.0386M as oxygen is being formed:
[tex]Rate=\frac{0.0386M-0M}{15.0s-0s}\\ \\Rate=2.57x10^{-3}\frac{M}{s}[/tex]
Regards.
According to the question,
Volume = 0.440 LTime = 15.0 sMol of O₂ = 1.7×10⁻²The reaction will be:
[tex]2 N_2 O (g) \rightarrow 2 N_2 (g) +O_2 (g)[/tex]Now,
The final concentration of O₂ will be:
→ [tex][O_2] = \frac{0.017}{0.440}[/tex]
[tex]= 0.0386 \ M[/tex]
hence,
The rate of reaction will be:
= [tex]\frac{0.0386-0}{15.0-0}[/tex]
= [tex]2.57\times 10^{-3} \ M/s[/tex]
Thus the above approach is right.
Learn more about volume here:
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A weather balloon is inflated to a volume of 27.6 L at a pressure of 755 mmHg and a temperature of 29.9 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 385 mmHg and the temperature is -14.1 ∘C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.
Answer: The volume of the balloon at this altitude is 46.3 L
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 755 mm Hg
[tex]P_2[/tex] = final pressure of gas (at STP) = 385 mm Hg
[tex]V_1[/tex] = initial volume of gas = 27.6 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = [tex]29.9^0C=(29.9+273)K=302.9K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]-14.1^0C=((-14.1)+273)K=258.9K[/tex]
Putting all the values we get:
[tex]\frac{755\times 27.6}{302.9}=\frac{385\times V_2}{258.9}[/tex]
[tex]V_2=46.3L[/tex]
Thus the volume of the balloon at this altitude is 46.3 L
An empty beaker is weighed and found to weigh 23.1 g. Some potassium chloride is then added to the beaker and weighed again. The second weight is 24.862 g. What is the mass of the potassium chloride
Answer:Mass of Potassium chloride =1.762g
Explanation:
Mass of empty beaker = 23.100 g
Mass of beaker with Potassium chloride = 24.862g
Mass of Potassium chloride = Final weight - initial weight = Mass of beaker with Potassium chloride - Mass of empty beaker = 24.862-23.100 = 1.762g
need helpp asapp please
Answer:
B. None of these
Explanation:
Sulfur has less ionization energy than phosphorus because sulfur has a pair of electron in its 3p subshell that increases electron repulsion in sulfur and sulfur electrons can easily remove from its sub-level.
While, there are no electron pairs in 3p subshell of phosphorus, therefore it requires more energy to remove an electron from 3p subshell.
Hence, the reason is electron repulsion and the correct answer is B.
Methane (CH4) and the perchlorate ion (ClO4â) are both described as tetrahedral. Methane and the perchlorate ion are both described as tetrahedral. What does this indicate about their bond angles?
a. Bond angles in the methane molecule are greater than those in the perchlorate ion.
b. Bond angles in the perchlorate ion are greater than those in the methane molecule.
c. Bond angles in the methane molecule and in the perchlorate ion are approximately equal
Answer:
c
Explanation:
The correct answer would be that the bond angles in the methane molecule and in the perchlorate ion are approximately equal.
A tetrahedral shape or geometry means both molecules are sp3 hybridized with the hydrogen atoms and the oxygen ions surrounding the central carbon and chlorine atoms respectively being as far away from one another as possible. This results in an approximate angle of 109.5 degrees in both molecules.
The correct option is c.
With methyl, ethyl, or cyclopentyl halides as your organic starting materials and using any needed solvents or inorganic reagents, outline syntheses of each of the following. More than one step may be necessary and you need not repeat steps carried out in earlier parts of this problem. (a) CH3I (b) I (c) CH3OH (d) OH (e) CH3SH (f) SH (g) CH3CN (h) CN (i) CH3OCH3 (j) OMe
Answer:
In the attachment you can find all the possible chemical reactions.
Some reaction can not be obtained by using alkyl halides because halides are weak leaving group which can leave compound during reaction easily but hydroxyl groups is a strong nucleophile which can not leave compound easily. So we can obtain alcohol from ethyl bromide, but we can not obtain hydroxyl ion from ethyl bromide.
Explanation:
The methyl of ethyl halides as the organic starting materials are using the needed solvents or the inorganic reagents. These can be not repeated in steps that arrive out in earlier parts.
The reaction can not be taken by the use of alkyl halides as the halides are the weakest leaving group which leave the compound during reaction easily.the hydroxyl group is the strong nucleophile that cannot leave the compound easily. Thus we can get alcohol from the ethyl bromide, but we can not obtain the hydroxyl ion from the ethyl bromide.Learn more about the methyl or the cyclopentyl.
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This ether can, in principle, be synthesized by two different combinations of haloalkane and metal alkoxide. Draw the combination of alkyl chloride and potassium alkoxide that forms the higher yield of ether. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Include counter-ions, e.g., Na , I-, in your submission, but draw them in their own separate sketcher. Separate multiple reactants using the sign from the drop-down menu.
Answer:
The 2 different combinations are given in the attached figure. The 2nd combination has a higher yield due to less hindrance by the alkyl halide.
Explanation:
The first reaction is between an alkyl halide and metal alkoxide. In this case, the alkyl chloride would be a secondary component and thus will cause greater hindrance to the yield. The reaction is as given in the attached figure
The second reaction is between a benzene ring containing halide and metal alkoxide. Now as the reaction is via alkyl chloride being the primary agent, there is less hindrance and thus greater yield in this case.
can a kind human being help me with this table at least only with the first burning fire wood someone please
Answer:
See below
Explanation:
* Burning fire wood is given to be our first option. Now burning tends to be a property of wood, and it does effect the chemical compositions of it. Wood, in the presence of fire / oxygen, turns into ash and carbon dioxide.
* Decomposition is recognized as a chemical change, and heating copper carbonate is a perfect example of decomposition. When energy is added to this chemical process, the copper carbonate decomposes into copper oxide.
* Mixing sodium chloride solution and silver nitrate solution. When this reaction takes place, a white precipitate of AgCl is formed. And of course, this is a chemical reaction.
* When acids or bases come in contact with citric acid, the pH degree changes much. Due to this, carbon dioxide bubbles are formed.
* When eggs are fried they absorb the heat in the pan. Doing so the egg starts to curl a bit, resulting in the formation of new particles.
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When 1-iodo-1-methylcyclohexane is treated with NaOCH2CH3 as the base, the more highly substituted alkene product predominates. When KOC(CH3)3 is used as the base, the less highly substituted alkene predominates. Give the structures of the two products and offer an explanation.
Answer:
See explanation
Explanation:
In this case, we have 2 types of reactions. [tex]CH_3CH_2ONa[/tex] is a strong base but only has 2 carbons therefore we will have less steric hindrance in this base. So, the base can remove hydrogens that are bonded on carbons 1 or 6, therefore, we will have a more substituted alkene (1-methylcyclohex-1-ene).
For the [tex]KOC(CH_3)_3[/tex] we have more steric hindrance. So, we can remove only the hydrogens from carbon 7 and we will produce a less substituted alkene (methylenecyclohexane).
See figure 1
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Which of the following is evidence of a physical change?
A) burning
B) melting
C) decomposing
D) rusting