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2 years ago

Which is an example of practical pursuit of alchemy

Chemistry

2 answers:

Oduvanchick [21]2 years ago

8 0

dezoksy [38]2 years ago

3 0

A practical pursuit of alchemy was the development of metallurgy practices. As alchemists always tried to turn various metals into other things, metalworking techniques were developed even though their experiments were often unsuccessful.
Explanation:
The previous alchemists created vital contributions to science although there beginning premises was off a touch. Their conclusions once failure would steer them into higher directions for investigation. Science isn't any quite shut observation and also the initial mistakes end in rethinking and retrying experiments.
converting "base metal" into gold. 
Creating a "Philosopher's stone" that provides life eternal.

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Question 17 In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas t

mars1129 [50]

Answer:

Explanation:

N₂ + 3H₂ = 2 NH₃

1 vol 2 vol

786 liters 1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

$\frac{.35\times 1572}{495} =\frac{1\times V_2}{ 273 }$

$V_2 =303.44$ liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

= 230.28 gm

=.23 kg / sec .

Rate of production of ammonia = .23 kg /s .

5 0

2 years ago

A liquid with a high viscosity cannot be a mixture.

enot [183]

It can be made true by changing "cannot" to "can".

5 0

1 year ago

At 4.00 L, an expandable vessel contains 0.864 mol of oxygen gas. How many liters of oxygen gas must be added at constant temper

fgiga [73]

Answer:

We have to add 2.30 L of oxygen gas

Explanation:

Step 1: Data given

Initial volume = 4.00 L

Number of moles oxygen gas= 0.864 moles

Temperature = constant

Number of moles of oxygen gas increased to 1.36 moles

Step 2: Calculate new volume

V1/n1 = V2/n2

⇒V1 = the initial volume of the vessel = 4.00 L

⇒n1 = the initial number of moles oxygen gas = 0.864 moles

⇒V2 = the nex volume of the vessel

⇒n2 = the increased number of moles oxygen gas = 1.36 moles

4.00L / 0.864 moles = V2 / 1.36 moles

V2 = 6.30 L

The new volume is 6.30 L

Step 3: Calculate the amount of oxygen gas we have to add

6.30 - 4.00 = 2.30 L

We have to add 2.30 L of oxygen gas

4 0

2 years ago

In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)

Maru [420]

Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 3.6 L

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm (From correct source)

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}$

${V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L$

${V_2}=0.27\ L$

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

7 0

2 years ago

Use the following half-reactions to design a voltaic cell: Sn4+(aq) + 2 e− → Sn2+(aq) Eo = 0.15 V Ag+(aq) + e−→ Ag(s) Eo = 0.80

AVprozaik [17]

Answer:

E° = 0.65 V

Explanation:

Let's consider the following reductions and their respective standard reduction potentials.

Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V

Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:

Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V

The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V

4 0

2 years ago