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pickupchik [31]
3 years ago
14

Briefly discuss interpretations of your observations and results. Include in your discussion, any conclusions drawn from the res

ults and any sources of error in the experiment. Be sure to discuss the reasons for your measured value of the specific heat of the metal being too high or too low.
Chemistry
1 answer:
VMariaS [17]3 years ago
4 0

Answer:

In comparison to Part 1 of this experiment, we observed similar reactions when determining the make up of our unknown. When testing for Mn2+ we observed a color change that resulted in a darker brown/red color, when testing for Co2+ we observed the formation of foamy bubbles but we could not conclude that a gas had formed, when testing for Fe3+ the result was a liquid red in color, when testing for Cr3+ we observed no change, when testing for Zn2+ we observed the formation of a pink/red liquid, when testing for K+ we observed the formation of a precipitate, when testing for Ca2+ we observe the formation of a precipitate. Sources of error may have occurred when observing whether or not an actual reaction had taken place or not, using glassware that wasn't fully cleaned, or the accidental mix of various other liquids in the lab

Explanation:

You might be interested in
Propose a mechanism to account for the formation of a cyclic acetal from 4-hydroxypentanal and one equivalent of methanol. If th
timofeeve [1]

Answer:

If carbonyl oxygen of 4-hydroxypentanal is enriched with O^{18}, then the oxygen label appears in the water .

Explanation:

  • In the first step, -OH group at C-4 gives intramolecular nucleophilic addition reaction at carbonyl center to produce a cyclic hemiacetal.
  • Then, one equivalent of methanol gives nucleophilic substitiution reaction by substituting -OH group in cyclic hemiacetal to produce cyclic acetal.
  • If carbonyl oxygen of 4-hydroxypentanal is enriched with O^{18}, then the oxygen label appears in the water produced at the end of reaction.
  • Full reaction mechanism has been shown below.

8 0
3 years ago
Calculate the mass in grams of carbon dioxide produced from 11.2 g of octane (C8H18) in the reaction above.
Ray Of Light [21]

Answer:

34.6g

Explanation:

Given parameters:

Mass of Octane  = 11.2g

  Reaction expression;

      2C₈H₁₈  + 25O₂  →  16CO₂  + 18H₂O

Mass of octane = 11.2g

Unknown:

Mass of carbon dioxide produced  = ?

Solution:

From the balanced reaction equation;

         2 mole of octane produced 16 moles of carbon dioxide

From the given specie, let us find the number of moles;

    Number of moles  = \frac{mass}{molar mass}  

 Molar mass of C₈H₁₈   = 8(12) + 18(1) = 114g/mole

Number of moles of octane  = \frac{11.2}{114}   = 0.098mole

   

    2 mole of octane produced 16 moles of carbon dioxide

    0.098 mole of octane will produce \frac{0.098 x 16}{2}   = 0.79mole of CO₂

Mass of CO₂ = number of moles x molar mass

           Molar mass of CO₂ = 12 + 2(16)  = 44g/mol

Mass of CO₂  = 0.79 x 44  = 34.6g

8 0
3 years ago
A lab group was supposed to make mL of a acid solution by mixing a solution, a solution, and a solution. However, the solution w
MissTica

Question: The question is not complete. Find below the complete question and the answer.

Alab group was supposed to make 14 mL of a 36% acid solution by mixing a 20% solution, a 26% solution, and a 42% solution. However, the 20% solution was mislabeled, and was actually a 10% solution, so the lab group ended up with 14 mL of a 34% acid solution, instead. If the augmented matrix that represents the system of equations is given below, what are the volumes of the solutions that should have been mixed? mL

Volume of 20% solution= ?

Volume of 26% solution = ?

Volume of 42% solution= ?   Round to the nearest whole number ml

Answer:

Volume of 20% solution= 3 mL

Volume of 26% solution = 1 mL

Volume of 42% solution= 10 mL

Explanation:

Find attached of the calculations.

3 0
3 years ago
C2h6o how many moles of ethanol are present in a 10.0 g sample of ethanol
Dmitriy789 [7]
46 gram of ethanol ≡ 1 mole of ethanol
1 gram of ethanol ≡ 1/46 mole of ethanol
10 gram of ethanol ≡ 1*10/46 mole of ethanol
                               =0.217 mole of ethanol
5 0
3 years ago
Read 2 more answers
A gas at STP occupies 22.4 L if the temperature is changed to 260 K and the pressures changed it to 0.50 ATM what will the new v
asambeis [7]

Answer:

The new volume will be 42, 7 L.

Explanation:

We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T. The conditions STP are: 1 atm of pressure and 273 K of temperature.

P1xV1/T1 =P2xV2/T2

1 atmx 22,4 L/273K = 0,5atmx V2/260K

V2=((1 atmx 22,4 L/273K )x 260K)/0,5 atm= 42, 67L

3 0
3 years ago
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