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IrinaVladis [17]
3 years ago
14

Grignard reactions are usually performed in ether solvent. Which statement below does not identify a safety concern associated w

ith the use of diethyl ether in the lab?
A. Diethyl ether has a very low miscibility in waterB. Diethyl ether can form explosive peroxides when exposed to air and sunlightC. Diethyl ether has a powerful anesthetic effect if inhaledD. Diethyl ether is highly flammable, with a flash point of -40 degrees Celsius
Chemistry
1 answer:
nikdorinn [45]3 years ago
6 0

Answer: A.  Diethyl ether has a very low miscibility in wate

The fact that the diethyl ether is miscible or not in water <u>does not imply a ris</u>k for the person who is working with this reagent in the laboratory.

However, the fact that diethyl ether forms explosive peroxides and that it is highly flammable implies that there is a risk of explosion when exposed to air and sunlight. On the other hand, as option C mentions, if a person inhales a large quantity of this reagent, they may lose consciousness and suffer some injury when fainting, due to the powerful anesthetic effect of this reagent.<u> In conclusion, options B, C and D are statements that imply safety problems associated with the use of diethyl ether in the laboratory, while option A does not imply it.</u>

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Answer:Chemistry problems can be solved using a variety of techniques.

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2 years ago
How many liters of hydrogen are needed to produce 34 grams NH3
polet [3.4K]
You would need 1000 liters
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2. Which of the following is true about the total number of reactants and the total number of
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9 moles of reactants chemically change into 11 moles of product.

Explanation:

In the reaction above, 9 moles of reactants chemically change into 11 mole of products. The coefficients in a reaction is the number of moles of the reacting atoms .

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8 0
2 years ago
Read 2 more answers
Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a
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Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

E^0_{[H^{+}/H_2]}=+0.00V

E^0_{[Ag^{+}/Ag]}=+0.799V

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : H_2\rightarrow 2H^{+}+2e^-    

Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

Now we have to calculate the standard electrode potential of the cell.

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V

Therefore, the standard cell potential will be +0.799 V

4 0
3 years ago
Convert the following Grams: 0.200 moles of H2S
ANEK [815]

Answer:

6.82 g H₂S

General Formulas and Concepts:

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

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Explanation:

<u>Step 1: Define</u>

0.200 mol H₂S

<u>Step 2: Identify Conversions</u>

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

<u>Step 3: Convert</u>

  1. Set up:                    \displaystyle 0.200 \ mol \ H_2S(\frac{34.09 \ g \ H_2S}{1 \ mol \ H_2S})
  2. Multiply:                  \displaystyle 6.818 \ g \ H_2S

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

6.818 g H₂S ≈ 6.82 g H₂S

7 0
3 years ago
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