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Solnce55 [7]
3 years ago
13

Doctors have fought infections with penicillin for over 50 years. Recently, penicillin has not worked as well as before. Which o

f the following explanations can account for this observation? A. Some bacteria are resistant to the drug. The continued use of the drug has killed all of these bacteria. B. Penicillin use has made all bacteria resistant to it. C. Some bacteria are resistant to penicillin. The continued use of the drug has killed all but these bacteria. D. Bacteria eat penicillin, so it is not effective anymore.
Chemistry
1 answer:
Lilit [14]3 years ago
5 0

C. "A" mentions that the drug killed all of the bacteria that are resistant to the drug, and that doesn't make any sense. "B" claims that all bacteria are resistant to the drug. This is not true. "D" mentions that bacteria eat the drug, which doesn't happen.

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A rock slide may move 250 km/h. A mudflow may move 160 km/h. How much faster might a rock slide be than a mudflow.
Nutka1998 [239]
90 km/h faster than a mudflow
3 0
3 years ago
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Draw the mechanism for the williamson ether reaction using m-cresol and benzyl bromide (use sodium methoxide as the base).
Nina [5.8K]
Below is the reaction of ether formation.

3 0
4 years ago
In full detail, explain what happens during the electrolysis of a NaCl brine? Be sure to identify what is being oxidized and wha
yuradex [85]

Answer:

Explanation:

Electrolysis of aqeous sodium chloride(NaCl)

Electrolysis is a process that converts electrical energy into chemical energy.

Electrolytic processes involves three major steps:

1. Ionization of electrolyte and water

2. Migration of ions to electrodes

3. Discharge of ions at the electrodes.

For the Electrolysis of brine, we follow these three steps:

1. Ionization of the aqeous brine solution:

NaCl → Na⁺ + Cl⁻

H₂O ⇄H⁺ + OH⁻

2. Migration of ions to the electrodes

The positive charges Na⁺ and H⁺ would both go to the cathode which is the negatively charged electrode

The negative charges Cl⁻ and OH⁻ migrates to the anode which are the positively charged electrodes. The anode is positively charged electrode.

3. Discharge of ions at the electrodes.

The preferential discharge of ions is based on the activity series and concentration of the ions.

On the activity series H is lower and it discharges preferentially to Na in the cathode:

2H⁺ + 2e⁻ → H₂

At this electrode, the cathode, reduction occurs and H⁺ ions are reduced.

At the anode Cl⁻ and OH⁻ migrates. But Cl⁻ is discharged preferentially due to its higher concentration.

2Cl⁻ ⇄ Cl₂ + 2e⁻

This is the oxidation half and Cl is oxidized

3 0
3 years ago
Mention the main applications of analytical chemistry in geology.
Minchanka [31]

Answer:

geochemistry

Explanation:

geo and chemistry

3 0
3 years ago
Hydrofluoric acid solutions cannot be stored in glass containers because HF reacts readily with silica dioxide in glass to produ
satela [25.4K]
A) Limiting reactant

You need the molar ratios (from the balanced chemical equation) and the molar masses of each compound (from the atomic masses)

a) Molar ratios:

6 mol HF : 1 mol SiO2 : 1 mol H2SiF6

2) Molar masses:

Atomic masses:
H: 1 g/mol
F: 19 g/mol
Si: 28 g/mol
O: 16g/mol

=>
HF:1g/mol + 19 g/mol = 20 g/mol
SiO2: 28g/mol + 2*16g/mol = 60 g/mol
H2SiF6: 2*1g/mol + 28g/mol + 6*19g/mol = 144g/mol

3) convert data in grams to moles

21.0 g SiO2 / 60 g/mol = 0.35 mol SiO2

70.5 g HF /  20 g/mol = 3.525 mol HF

4) Use the theorical ratios to deduce which is in excess and which is the limiting reactant.

6 mol HF / 1mol SiO2   < 3.525 mol HF / 0.35 mol SiO2 ≈ 10

=> There is more HF than the needed to react with 0.35mol of SiO2 =>

SiO2 is the limiting reactant (HF is in excess)

b) Mass of excess reactant.

1) Calculate how many grams reacted, which requires to calculate first the number of moles that reacted

0.35 mol SiO2 * 6 mol HF / 1 mol SiO2 = 2.1 mol of HF

2.1 mol HF * 20 g/mol = 42 gram of HF

2) Subtract the quantity that reacted from the original quantity:

70.5 g - 42 g = 28.5 g of HF in excess

c) Theoretical yield of H2SiF6

1 mol of SiO2 ; 1 mol of H2SiF6 => 0.35 mol SiO2 : 0.35 mol H2SiF6

Convert those moles to grams: 0.35 mol * 144 g/mol = 50.4 grams

d) % yield

% yield = actual yield / theoretical yield * 100 = 45.8 / 50.4 * 100 = 90.87%
4 0
4 years ago
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