Select the statement that is FLASE about this experiment. Group of answer choices Changing temperature will affect the redox potential. Changing concentration will affect the redox potential. Changing metal solution will affect the redox potential. To ensure stable reading, insert the conductivity meter in the solution for at least 60 seconds. To ensure consistent data, collect the redox potential values at least three times.
Answer:
To ensure stable reading, insert the conductivity meter in the solution for at least 60 seconds.
Explanation:
The False statement is ; To ensure stable reading, insert the conductivity meter in the solution for at least 60 seconds.
All other statement will affect the experiment's redox potential but inserting the conductivity meter in the solution for at least 60 seconds will not affect the Redox potential .
Redox potential is used to describe/measure a system's oxidizing capacity
Imine formation from an aldehyde and an amine proceeds reversibly under slightly acidic conditions. The reaction is reversible due to acid-catalyzed hydrolysis of the imine. Imine formation generates an equivalent of water. Given these facts, explain why the imine can be isolated from the reaction mixture.
Answer:
Imine can be isolated from the reaction mixture as water is continuously removed from the reaction chamber
Explanation:
In this reaction, a non -aqueous solvent is not used (not mentioned in the question). Thus, we can say that there is continuous removal water under suitable reacting conditions and hence the imine formed is left behind.
Gelatinization happens with what type of Carbohydrate?
Draw curved arrows to show electron reorganization for the mechanism step below. (For a resonance-stabilized anion, draw a single resonance form with the negative charge on oxygen, not on carbon.) Make the ends of your arrows specify the origin and destination of reorganizing electrons.
Answer:
Explanation:
Dear student, this question appears to be incomplete but in the diagram attached below, I have carefully drawn out the required organic structure.
The aim of this question is to show the mechanism of the given structure and to point out the resonance structure.
The first diagram shows the complete part (i.e. the structure of the organic compound) and its major product(answer).
The second diagram shows the mechanism by which we arrived at the major product.
From the second diagram attached below, the -oxo substituent of the methanol attacks the hydrogen atom as shown below, leading to the formation of the enolate ion (a resonance stabilized anion).
However, this is because the alpha carbon attaches to the hydrogens readily leaves as H⁺ to yield an enolate ion as a result of -C- that is doubly bonded to oxygen (an Electron withdrawing group).
Kindly check out the images attached below to see the mechanism.
Use the equation qliquid = m × c × ΔT to calculate the heat gained by the cold liquid. Use the specific heat for the liquid you selected.
Use the equation qwater = m × c × ΔT to calculate the heat lost by the hot water. Show your work using the problem-solving method shown in previous rubrics.
Complete question :
Use the equation qliquid = m × c × ΔT to calculate the heat gained by the cold liquid. Use the specific heat for the liquid you selected.
m = 248.9 g ; T = 72.6°C ; C = 3.73 j/g°C
Use the equation qwater = m × c × ΔT to calculate the heat lost by the hot water. Show your work using the problem-solving method shown in previous rubrics.
M = 237 g ; T = 41.2°C ; C = 4.184 j/g°C
Answer:
Quantity of heat gained = 67401.6 Joules
Quantity of heat lost by hot water = 40854.25 Joules
Explanation:
The heat gained by cold liquid :
Q = m × c × ΔT
m = 248.9 g ; T = 72.6°C ; C = 3.73 j/g°C
Q = 248.9 * 3.73 * 72.6
Q = 67401.6 Joules
Quantity of heat gained = 67401.6 Joules
Heat lost by hot water :
m × c × ΔT
M = 237 g ; T = 41.2°C ; C = 4.184 j/g°C
Q = 237 * 41.2 * 4.184
Q = 40854.25 Joules
Quantity of heat lost by hot water = 40854.25 Joules
How many grams of hydroxide pellets, NaOh are required to prepare 50.0ml of a 0.150 M solution ?
A.0.300g
B.3.00g
C.2.00g
D.200.g
E. 0.025 M
What is the balanced net ionic reaction for the following -
CaCl2 + Na2S --> NaCl + CaS?
Answer:
CaCl2 + Na2S --> 2NaCl + CaS?
Explanation:
CaCl2 + Na2S --> 2NaCl + CaS?
When 148. mg of a certain molecular compound X are dissolved in 75.g of cyclohexane (CH), the freezing point of the solution is measured to be 59 'C. Calculate the molar mass of X If you need any additional information on cyclohexane, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to l significant digit. XS ?
Answer: The molar mass of X is 61.3g/mol
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=K_f\times m[/tex]
[tex]\Delta T_f=T_f^0-T_f=(6.55^0C-5.9^0C)=0.65^0C[/tex] = Depression in freezing point
[tex]K_f[/tex] = freezing point constant = [tex]20.2^0C/mol[/tex]
m= molality = [tex]\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]
where,
Now put all the given values in this formula, we get
[tex]0.65^0C=20.2°C/m\times frac{0.148g}}{M\times 0.075kg}[/tex]
[tex]M=61.3g/mol[/tex]
Thus molar mass of X is 61.3g/mol
- Calculate the Standard Enthalpy of the reaction below:
NH3(g) + HCI (g) → NH4Cl(s)
Using the following Enthalpy of Reactions:
2HCI(g) → H2(g) + Cl2(g)
AH = +184.6 KJ
2H2(g) + 1/2 N2(g) + 1/2 Cl2(g) → NH4Cl(s) deltaH = -314.4 kJ
N2(g) + 3 H2(g) → 2 NH3(g)
deltaH = +184.6 kJ
Answer:
Explanation:
We have the three equations:
[tex]NH_{3(g)} + HCl_{(g)} => NH_4Cl_{(s)} ..... \Delta H = ? (1)\\2HCl_{(g)} => H_{2(g)} + Cl_{2(g)} .... \Delta H = +184.6 kJ (2)\\2H_{2(g)} + 1/2N_{2(g)} + 1/2Cl_{2(g)} => NH_4Cl_{(s)} ..... \Delta H = -314.2 kJ (3)\\ N_{2(g)} + 3H_{2(g)} => 2NH_{3(g)} .... \Delta H = +184.6kJ (4)[/tex]
(can you double check that it is 184.6kJ for both equations 2 and 4 because it seems unlikely). We need to solve for equation 1 by addition and changing equations 2, 3 and 4. After possibly some trial and error, we can find that if we flip equations 4, multiply equation 3 by 2, add the equations together, and then finally divide by 2, we can get equation 1. We will get the answer of -314.2 kJ. However, I am again skeptical about the delta H values for equation 2 and 4 so double check that. This method might be super confusing and it is really hard to explain. So what I would suggest you to watch videos on Hess' law.
What is the greatest concentration of pollution
Answer:
B
Explanation:The poisonous substances, present in the environment can easily get into the trophic level as living organism depends on each other and environment for food and nutrition. These poisonous substances may not be broken down in the body or excreted easily, efficiently and quickly. Instead, they accumulate in the tissues, and as the living organism eats more, the concentration of these substances increases and they pass from one trophic level to the next. The tertiary consumer being at the top of trophic levels receives the maximum pollutant. This phenomenon is known as biological magnification.
So, the correct answer is option B.
How many moles of BaS would be used to make 1200 mL of a 10.OM solution?
the number of particles in a mole if substance
A.avografo number
b.molar volume
c.molar mass
d.molar ratio
calculate the percentage of CL in AL(CLO3)3
Answer:
38.4%
Explanation:
We'll begin by calculating the molar mass of Al(ClO₃)₃. This can be obtained as follow:
Molar mass of Al(ClO₃)₃
= 27 + 3[35.5 + (16×3)]
= 27 + 3[35.5 + 48]
= 27 + 3[83.5]
= 27 + 250.5
= 277.5 g/mol
Next, we shall determine the mass of Cl in Al(ClO₃)₃. This can be obtained as follow:
Mass of Cl in Al(ClO₃)₃ = 3 × Cl
= 3 × 35.5
= 106.5 g
Finally, we shall determine the percentage of Cl in Al(ClO₃)₃. This can be obtained as follow:
Mass of Cl in Al(ClO₃)₃ = 106.5 g
Mass of Al(ClO₃)₃ = 277.5 g
Percentage of Cl =?
Percentage of Cl = mass of Cl / mass of Al(ClO₃)₃ × 100
Percentage of Cl = 106.5 / 277.5 × 100
Percentage of Cl = 38.4%
3. Convert the word equations below to symbolic:
Sodium + chlorine sodium chloride
• Calcium + bromine ---- calcium bromide
• Potassium + water → potassium hydroxide + hydrogen
ot
Answer:
1)2Na + Cl2 ----> 2NaCl
2)Ca + Br2 ---->CaBr2
3)K + H2O -----> KOH + H2
How many grams of H2SO4 will react 650.0 grams of NaOH?
Answer:
H2SO4 + 2NAOH➡NA2SO4+2H2O
H2SO4 = 1*2 +32+16*3= 98g
NAOH = (11+16+1)=28
Explanation:
To calculate the number of Mole = Mole *Molar Mass
To calculate for gram = Mole/ Molar Mass
20.5g NaoH/1 * 1 mole NaOH/28g NaoH = 0.73M NaoH
0.73M NaoH /1 * 1 M H2SO4 /2M NaoH = 0.37M H2SO4
0.37 M H2SO4 /1 * 98g H2SO4 /1M H2SO4 = 0.00367g H2SO4
:- 0.00367g of Sulphuric Acid will be needed to complete the reaction.
The half-life of argon-39 is 269 years. It decays into krypton-39. After 1,076 years, what fraction of the original amount of argon-39 in a sample will still be argon
1/16
1/4
1/2
1/8
Answer:
1/16
Explanation:
From the question given above, the following data were obtained:
Half-life (t½) = 269 years
Time (t) = 1076 years
Fraction remaining =?
Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 269 years
Time (t) = 1076 years
Number of half-lives (n) =?
n = t / t½
n = 1076 / 269
n = 4
Thus, 4 half-lives has elapsed.
Finally, we shall determine the fraction of the original amount remaining. This can be obtained as follow:
Let N₀ be the original amount.
Let N be the amount remaining.
Number of half-lives (n) = 4
Fraction remaining (N/N₀ ) =?
N = 1/2ⁿ × N₀
N = 1/2⁴ × N₀
N = 1/16 × N₀
Divide both side by N₀
N/N₀ = 1/16
Thus, the fraction of the original amount remaining is 1/16
PLZ HELP ME ASAP!!!!(I WILL ALSO MARK BRAINLIEST!!!)
(True or False) A cold front often signals sunny warm weather.
Calculate the Standard Enthalpy of the reaction below:
NH3(g) + HCl (g) → NH4Cl(s)
Using the following Enthalpy of Reactions:
2HCl(g) → H2(g) + Cl2(g)
AH = +184.6 kJ
2H2(g) + 12 N2(g) + 2 Cl2(g) NH4Cl(s) AH = -314.4 kJ
N2(g) + 3 H2(g) → 2 NH3(g)
AH = +184.6 kJ
Answer:
(we use hess's law) it is so simple but the second reaction is not correct please right it
A 1.4639-g sample of limestone is analyzed for Fe, Ca, and Mg. The iron is determined as Fe2O3 yielding 0.0357 g. Calcium is isolated as CaSO4, yielding a precipitate of 1.4058 g, and Mg is isolated as 0.0672 g of Mg2P2O7. Report the amount of Fe, Ca, and Mg in the limestone sample as %w/w Fe2O3, %w/w CaO, and %w/w MgO
Answer:
A) w/w % of Fe - 2.44%
B) w/w % of Mg - 4.590%
C) w/w % of Ca - 96%
Explanation:
A) w/w % of Fe in limestone as Fe2O3 = (Mass of Fe2O3 /Mass of limestone) x 100
0.0357/1.4639 X 100
= 2.438 =2.44
B) w/w % of Mg in limestone as Mg2P2O7 = (Mass of Mg2P2O7 /Mass of limestone) x 100
= 0.0672/1.4639 X 100
= 4.590
C) w/w % of Ca in limestone as CaSO4 = (Mass of CaSO4/Mass of limestone) x 100
= 1.4058/1.4639 X 100
= 96
Answer:Fe =1.71% Ma=0.99% Ca=28.24%
Explanation:
Which phase change is exothermic?
Answer:
B. Solidification
Hope this helped! :)
lave
Use the drop-down menus to identify the parts of a
wave.
A
B:
C
D
Answer:
Use the drop-down menus to make your selections.
What is the difference between wave A and wave B?
✔ Wave A has higher amplitude than wave B.
What is the difference between wave C and wave D?
✔ Wave C has a lower frequency than wave D.
Explanation:
Acid-catalyzed addition of water to an alkene yields an alcohol with Markovnikov regiochemistry.
a. True
b. False
Answer:
True
Explanation:
Markovnikov rule states that the negative part of the addendum is added to the carbon atom having the least number of hydrogen atoms.
This means that during acid-catalyzed addition of water to an alkene, the -OH is added to the more substituted carbon atom while the -H is added to the least substituted carbon atom in obedience to Markovnikov regiochemistry.
Answer:
a. True
Explanation:
Acid-catalyzed addition of water to an alkene yields an alcohol with Markovnikov regiochemistry. The electrophilic H+ adds to the sp2 carbon with the most hydrogens to yield the most stable carbocation intermediate, which then adds water to give the product alcohol.
A gas in a sealed container has a pressure of 125 atm at a temperature of 303 K. If the pressure in the container is increased to 200 atm, what is the new temperature if the volume remains constant?
(Show work pls :)!)
Answer:
[tex]T_2=484.8K[/tex]
Explanation:
Hello there!
In this case, according to the the variable temperature and pressure and constant volume, it turns out possible for us to calculate the new temperature via the Gay-Lussac's law as shown below:
[tex]\frac{T_2}{P_2} =\frac{T_1}{P_1}[/tex]
Thus, by solving for the final temperature, T2, we obtain:
[tex]T_2 =\frac{T_1P_2}{P_1}[/tex]
So we plug in the given data to obtain:
[tex]T_2 =\frac{303K*200atm}{125atm}\\\\T_2=484.8K[/tex]
Best regards!
water and Air are needed for iron rust answer true or false
Answer:True
Explanation:
PLZ GIVE ME BRAINLIEST, I NEED IT
how we can remove temporary hardness of water?and write the chemical reaction?
Answer:
it can be removed
1. by boiling
Ca(HCO3)2 > CaCO3 + H2O +CO2
2. by treating with calcium hydroxide
Please help:
An object rolled 15m/s for 3 seconds. How far did it roll?
You and your science group have been handed five mineral samples to identity. First, you on the minerals by color None of the
minerals are white. You and your group know that you probably do not have
A) Calcite
B) gypsum
C) Calcite and Talc
D) feldspar and calcite
Answer: A
Explanation:
A 31.25 mL aliquot of weak base that has a concentration of 0.683 M will be titrated with 0.434 M HCl. Calculate the pH of of the solution upon neutralization of half of the weak base. The Kb of the base is 1.5×10-6.
Answer:
pH = 8.18
Explanation:
The weak base, X, reacts with HCl as follows:
X + HCl → HX⁺ + Cl⁻
Where 1 mole of X with 1 mole of HCl produce 1 mole of HX⁺ (The conjugate acid of the weak base).
Now, using H-H equation for bases:
pOH = pKb + log [XH⁺] / [X]
Where pOH is the pOH of the buffer (pH = 14 -pOH)
pKb is -log Kb = 5.824
And [X] [HX⁺] are the molar concentrations of each specie
Now, at the neutralization of the half of HX⁺, the other half is as X, that means:
[X] = [HX⁺]
And:
pOH = pKb + log [HX⁺] / [X]
pOH = 5.824 + log 1
pOH = 5.824
pH = 14-pOH
pH = 8.18A 15.0 mL solution of 0.10 M Fe(NO3)3 reacts with K2C204 according to the reaction 2 Fe(NO3)3 + 3 K2C2O4 → Fe2 (C2O4)3 + 6 KNO3 . What volume of 0.30 M K2C2O4 is needed to react completely?
The volume of 0.30 M potassium chromate needed to react with 15 ml of 0.10 M ferric nitrate is 5 ml.
What is molarity?Molarity of a solution is the ratio of its number of moles of solute to the volume of solution in liter. It is the common term used to express the concentration of a solution.
Let V1 and M1 be the volume and molarity of solution 1 and V2 and M2 that of solution 2 which are reacting together.
Thus, V1 M1 = V2 M2.
The volume of 0.30 M potassium chromate is calculated as follows:
volume = (15 ml × 0.10 M) / 0.30 M = 5 ml.
Therefore, the volume of the potassium chromate solution is 5 mL.
To find more on molarity, refer here:
https://brainly.com/question/8732513
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An isolated atom of a certain element emits light of wavelength 655 nm when the atom falls from its fifth excited state into its second excited state. The atom emits a photon of wavelength 435 nm when it drops from its sixth excited state into its second excited state. Find the wavelength of the light radiated when the atom makes a transition from its sixth to its fifth excited state.
Answer:
[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]
Explanation:
From the question we are told that
Light wavelength [tex]\lambda_l=655nm[/tex]
Light wavelength atom fall [tex]x_L=5th-2nd[/tex]
Photon wavelength [tex]\lambda_p=435nm[/tex]
Photon wavelength atom fall [tex]x_P=^th-2nd[/tex]
Generally the equation for the reciprocal of wavelength of emitted photon is is mathematically given by
[tex]\frac{1}{\lambda}=R(\frac{1}{ \lambda _f^2}-\frac{1}{\lambda _i^2} )[/tex]
Therefore for initial drop of 5th to 2nd
[tex]\frac{1}{\lambda_{5-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )[/tex]
Therefore for initial drop of 6th to 2nd
[tex]\frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]
Generally we subtract (5th to 2nd) from (6th to 2nd)
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )-\frac{1}R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{5^2}-\frac{1}{6^2} )[/tex]
[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=\frac{1}{\lambda_{5-6}}[/tex]
[tex]\frac{1}{\lambda_{5-6}}=\frac{1}{4350nm}-\frac{1}{655nm}[/tex]
[tex]\frac{1}{\lambda_{5-6}}=1.33*10^{-3}[/tex]
Therefore for 6th to 5th stage is mathematically given by
[tex]\frac{1}{\lambda_{6-5}}=(1.33*10^{-3})^{-1}[/tex]
[tex]\frac{1}{\lambda_{6-5}}=751.879nm[/tex]
[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]