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stira [4]
3 years ago
9

Why do you think the soccer ball fell more slowly than the golf ball

Chemistry
1 answer:
alexgriva [62]3 years ago
3 0

Answer:

kinetic energy than the potential energy it carries

Explanation:

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identify the reagents necessary to achieve each of the following transformations o3 dms, pcc, ch2cl, h2so4,h2o,hgso4
hoa [83]

the reagents necessary to convert alcohol to ketoneNa_2Cr_2O_7 , H_2O which involves oxidation of alcohols.

<h3>What is oxidation of alcohols?</h3>
  • Alcohol oxidation is a significant organic chemistry process. Secondary alcohols can be oxidized to produce ketones, while primary alcohols can be oxidized to produce aldehydes and carboxylic acids.
  • In contrast, tertiary alcohols cannot be oxidized without the C-C bonds in the molecule being broken.
  • In order to cause primary alcohols to oxidize into aldehydes
  1. Cr_2O_7 ^-^2 (dichromate)
  2. CrO_3/pyridine (Collins reagent)
  3. Chromium pyridinium compound (PCC)
  4. Dichromate of pyridinium (PDC, Cornforth reagent)
  5. Periodinane by Dess-Martin
  6. Oxalyl chloride with dimethylsulfoxide (DMSO) for Swern
  • oxidation of secondary alcohols to ketones
  1. Cr_2O_7 ^-^2 (dichromate)
  2. CrO_3/pyridine (Collins reagent)
  3. Chromium pyridinium compound (PCC)
  4. Dichromate of pyridinium (PDC, Cornforth reagent)
  5. Periodinane by Dess-Martin
  6. Oxalyl chloride and dimethyl sulfoxide (DMSO) (Swern oxidation)
  7. CrO_3, H_2SO_4/acetone (Jones oxidation)
  8. Acetone with aluminum isopropoxide (Oppenauer oxidation)

To learn more about oxidation of alcohols with the given link

brainly.com/question/7207863

#SPJ4

<u>Question:</u>

Identify the reagents necessary to achieve each of the following transformations

a. O_3 , DMS

b.H_2SO_4 , H_2O , HgSO_4

c.Na_2Cr_2O_7 , H_2O

d.Fe ^ {2+}, NaOH

3 0
2 years ago
A student conducts an experiment by placing a wooden stick in a cup of sugar solution. Over the next few weeks, sugar crystals f
Ipatiy [6.2K]

Answer: In order to be alive there must be cells present and since sugar does not contain cells it is not alive.

Explanation:

8 0
3 years ago
A 50 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 300°C. How many moles of argon gas does the cylinder co
KengaRu [80]
In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.

Given 
P = 10130.0 kPa 
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol

Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.

PV = nRT                      PV
----    ------    --->    n = --------
 RT     RT                       RT

          10130.0 kPa  x 50 L
n= ---------------------------------------------
       8.314 L. kPa/K.mol  x 573.15 K
             506,500 
  =  ----------------------------
         4,765.17  mol K

=106.29 mol Ar

So the moles of argon gas is 106.29 moles 
8 0
3 years ago
Read 2 more answers
What word chemical equation describes Cavendish’s experiment with zinc?
LiRa [457]
The chemical reaction of Cavendish involving zinc would be a reaction between hydrochloric acid and zinc yielding zinc chloride and hydrogen gas. The balanced chemical equation would be:

2Zn + 2HCl = 2ZnCl + H2

This is an example of a single replacement reaction where zinc replaces hydrogen in the acid molecule.
3 0
3 years ago
Read 2 more answers
A 2.00 L sample of gas at 35C is to be heated at constant pressure until it reaches a volume of 5.25 L. To what Kelvin temperatu
Anastaziya [24]

Answer:

The sample will be heated to 808.5 Kelvin

Explanation:

Step 1: Data given

Volume before heating = 2.00L

Temperature before heating = 35.0°C = 308 K

Volume after heating = 5.25 L

Pressure is constant

Step 2: Calculate temperature

V1 / T1 = V2 /T2

⇒ V1 = the initial volume = 2.00 L

⇒ T1 = the initial temperature = 308 K

⇒ V2 = the final volume = 5.25 L

⇒ T2 = The final temperature = TO BE DETERMINED

2.00L / 308.0 = 5.25L / T2

T2 = 5.25/(2.00/308.0)

T2 = 808.5 K

The sample will be heated to 808.5 Kelvin

7 0
4 years ago
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