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sdas [7]
3 years ago
11

Given the amount of camphor (200mg) we are using in this experiment, please determine how many mg of sodium borohydride to use i

n this reaction. We would like you to use 5.2 molar equivalents of this reagent. This means 5.2 times the mmol of camphor we are using. As an example: for 110.0 mg of camphor,142 mg of NaBH4 would be used (see if you can confirm this result). For complete credit, your work needs to be clearly drawn out!
Chemistry
1 answer:
swat323 years ago
7 0

Answer:

Explanation:

From the given information:

Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.

Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.

If the molecular weight of camphor = 152.24 g/mol

and it mass = 200 mg

The its no of moles = 200 mg/ 152.24 g/mol

= 1.3137 mmol

Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol

= 6.831 mmol

since the molar mass of NaBH4 = 37.83 g/mol

Then, using the same formula:

No of moles = mass/molar mass

mass = No of moles × molar mass

mass = 6.831 mmol × 37.83 g/mol

mass of NaBH4 used = 258.42 mg  

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Kamila [148]

Answer:

A channel in an ancient Martian "river bed" was not carved out by liquid water but built by molten lava

6 0
2 years ago
If MnO2 was not added to this reaction, what effect would that have on the production of H2O and O2?
maria [59]
For the answer to the question above, 
I am going to assume that MnO2 is a catalyst most likely breakdown of H2O2 

H2O2----MnO2------> H2O + 1/2 O2 


<span>in that case, B) is answered in the trump suit </span>

--- 
<span>edit: I need to have a plate account to see - which I don't </span>
is it something like 


2H2O2 ---------------> 2H2O + O2 
...............MnO2 

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8 0
3 years ago
What is the oxidation number of chromium in the ionic compound ammonium dichromate (nh4)2cr2o7?
lesya692 [45]
The oxidation number of an element is the number of electrons that are gained or lost by the element to form a chemical bond. 
the net oxidation number of the ion is its charge.
compound (NH₄)₂Cr₂O₇ is made of cation - NH₄⁺ and anion - Cr₂O₇²⁻
oxidation number of ion - -2
(oxidation number of Cr x 2 Cr atoms) +(oxidation number of O x 7 O atoms )= -2
oxidation number of Cr = y
oxidation number of O = -2
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4 0
3 years ago
Please show steps Thanks
Llana [10]

Answer:

2.5M

Explanation:

First calculate the gram-formula mass of CaCO3 (add up the mass of each element in the compound)

40.1 + 12.01 + 16(3) = 100.11

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25/100.11 = 2.497 or 2.5M

8 0
2 years ago
APPLICATION OF BOYLE'S LAW
lianna [129]

Answer:

P = 164 Atm

Explanation:

PV = nRT => P = nRT/V

n = 10.0 moles

R = 0.08206 L·Atm/mol·K

T = 27.0°C = 300 K

V = 1.50 Liters

P = (10.0 mol)(0.08206 L·Atm/mol·K )(300 K)/(1.50 Liters) = 164.12 Atm ≅ 164 Atm (3 sig. figs.)

5 0
2 years ago
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