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8_murik_8 [283]
3 years ago
9

Please help me on 9 and 10

Chemistry
1 answer:
Fantom [35]3 years ago
4 0
C and A because the product of a matter will always be the same and just google A.
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Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

                             =  (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

                                                                                           = 75.3%

4 0
3 years ago
OMG PLEASE HELP. WILL MARK U BRAINLIEST IF YOU ANSWER ALL 3 QUESTIONS. THANKS SO MUCH.
Readme [11.4K]
1.) Moe will feel pain and he will see one side go to the right and disappear
4 0
3 years ago
How do you think you results would be affected if using K2Cr2O7/H2SO4 as the oxidizing agent instead of KMnO4
makkiz [27]

Answer:

Due to weak oxidizing agent.

Explanation:

In my opinion the results would be affected if using K2Cr2O7/H2SO4 as the oxidizing agent instead of KMnO4 because K2Cr2O7/H2SO4 is weak oxidizing agent as compared to KMnO4. An oxidizing agent is a substance that has the ability to oxidize other substances means that accept their electrons so that's why the results of strong oxidizing agent is different than weak oxidizing agent.

3 0
3 years ago
What is the electron configuration of the element in period 3, group 6A?
yulyashka [42]
The electron configuration of the element in period 3, group 6A is [Ne]3s-2 3p-4. 
3 0
3 years ago
Read 2 more answers
Calculate the change in energy of an atom that emits a photon of wavelength 2.21 meters. (Planck’s constant is 6.626 x 10-34 jou
aivan3 [116]

Answer : The correct option is, 8.988\times 10^{-26}joules

Solution :

Formula used :

E=\frac{h\times c}{\lambda}

where,

E = change in energy

h = Planck’s constant = 6.626\times 10^{-34}J/s

c = speed of light = 2.998\times 10^{8}m/s

\lambda = wavelength = 2.21 m

Now put all the given values in the above formula, we get the energy of an atom.

E=\frac{(6.626\times 10^{-34}J/s)\times (2.998\times 10^{8}m/s)}{2.21m}

E=8.988\times 10^{-26}joules

Therefore, the energy of an atom is, 8.988\times 10^{-26}joules

6 0
3 years ago
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