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Sladkaya [172]
3 years ago
14

Consider the reaction. 2 Al ( s ) + Fe 2 O 3 ( s ) heat −−→ Al 2 O 3 ( s ) + 2 Fe ( l ) If 17.3 kg Al reacts with an excess of F

e 2 O 3 , how many kilograms of Al 2 O 3 will be produced?
Chemistry
1 answer:
IrinaVladis [17]3 years ago
6 0

Answer:

32.7 kilograms of aluminium oxide  will be produced.

Explanation:

2 Al ( s ) + Fe_2O_3 ( s ) +heat\rightarrow Al_2O_3 ( s ) + 2 Fe ( l )

Mass of aluminum = 17.3 kg = 17300 g (1 kg = 1000 g )

Moles of aluminium = \frac{17300 g}{27 g/mol}=640.74 mol

According to reaction, 2 moles of aluminum gives 1 mole of aluminum oxide,then 640.74 moles of aluminum will give:

\frac{1}{2}\times 640.74 mol= 320.37 mol of aluminum oxide

Mass of 320.37 moles of aluminum oxides:

320.37 mol × 102 g/mol = 32,677.74 g = 32.67774 kg ≈ 32.7 kg

32.7 kilograms of aluminium oxide  will be produced.

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No elements visible!

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2 years ago
Why is chromatography particularly well suited to the needs of a drug analyst
Zarrin [17]

Chromatography is used in purification. Drugs analysts may use the technique to separate the active molecule in a drug molecule, for efficacy or toxicity analysis, from the other drug components.

Explanation:

Chromatography is used to separate a mixture of different components based on the size of their molecules. In liquid chromatography, the mixture is dissolved in a solvent that acts as the mobile phase and then passed along a stationary phase with different kinds of pores, As the mixture passes through the pores, their different components are separated because they take different times to pass through the stationary phase because of their different rates in passing through the pores.

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6 0
3 years ago
An order is given to administer methylprednisolone, an anti-inflammatory drug, by IV at a rate of 39 mg every 30 min. The IV bag
Korvikt [17]

Answer:

The Flow rate = 0.0208 mL/min

Explanation:

Data provided:

Rate of dose = 39 mg every 30 min = (39/30) mg/min  = 1.3 mg/min

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125mg of methylprednisolone is present in every 2 mL

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concentration = (125/2) mg/mL = 62.5 mg/mL

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The flow rate is given as:

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on substituting the respective values, we get

Flow rate = (1.3 mg/min) / (62.5mg/mL)

or

The Flow rate = 0.0208 mL/min

3 0
3 years ago
Which solution contains the smallest number of moles of sucrose (c12h22o11, molar mass = 342.30 g/mol)? 2,000 ml of a 5.0 × 10–5
Iteru [2.4K]

> 2,000 mL of a 5.0 × 10–5% (w/v) sucrose solution 

5.0 × 10–3 g/mL * 2000 mL * (1 mol / 342.30 g) = 0.0292 mol

<span>
> 2,000 mL of a 5.0 ppm sucrose solution</span>

5 grams / 1000000 mL * 2000 mL* (1 mol / 342.30 g) = 0.0000292 mol

 <span>
> 20 mL of a 5.0 M sucrose solution </span>

5.0 M * 0.020 L = 0.1 mol

 

 

Answer:

<span>2,000 mL of a 5.0 ppm sucrose solution</span>

5 0
3 years ago
Which phase change results in atoms with the highest kinetic energy? A. Boiling B. Melting C. Freezing D. Condensing
suter [353]

Answer: Boiling because it makes the molecules in water bounce around.

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4 0
3 years ago
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