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2 years ago

Who summarized Greek medical knowledge in a textbook that would dominate medical teaching until the Scientific Revolution?

Chemistry

2 answers:

ss7ja [257]2 years ago

3 0

I recently wrote a paper on animal testing

Pretty sure it was Galen

Mashcka [7]2 years ago

3 0

Answer:

Explanation:

Aelius Galenus wrote the first anatomy textbook based on human dissection. His original works gained popularity for his remarkable job.

In Europe, during Middle Age, his textbooks were the main reference for medicine.

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erastova [34]

Answer:

Electrolysis of water is the process of using electricity to decompose water into oxygen and hydrogen gas. Hydrogen gas released in this way can be used as hydrogen fuel, or remixed with the oxygen to create oxyhydrogen gas, which is used in welding and other applications.Ordinarily, the freezing point of water and melting point is 0 °C or 32 °F. The temperature may be lower if supercooling occurs or if there are impurities present in the water which could cause freezing point depression to occur. Under certain conditions, water may remain a liquid as cold as -40 to -42°

Explanation:

8 0

2 years ago

How many liters of o2 do you have if you have 5.8 moles of o2

polet [3.4K]

Answer:

130 Liters

Explanation:

if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)

3 0

2 years ago

34. An electron in the 3d state of the hydrogen atom has a radial part of the wave function of the form A r2 exp(-r/3ao). What i

nikdorinn [45]

Answer: Option (e) is the correct answer.

Explanation:

Formula to calculate radius is as follows.

p(r) = $Ar(r^{2}e^{\frac{-r}{3a_{o}}}$

= $Ar^{3}e^{\frac{-r}{3a_{o}}}$

$\frac{dp(r)}{dr}$ = 0

$Ar^{3}e^{\frac{-r}{3a_{o}}}(\frac{-1}{3a_{o}}$ + $Ae^{\frac{-r}{3a_{o}}}(3r^{2})$ = 0

$\frac{Ar^{3}}{3a_{o}}e^{\frac{-r}{3a_{o}}}$

= $3Ar^{2}e^{\frac{-r}{3a_{o}}}$

r = $9a_{o}$

Thus, we can conclude that most likely radius at which the electron would be found is $9a_{o}$ .

4 0

2 years ago

What is the volume of 120mg of water

iVinArrow [24]

Answer:

560mg

Explanation:

120mg-560mg trust dude

7 0

2 years ago

A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c

gulaghasi [49]

Answer:

The enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. Working with this equation, and assuming no heat is lost to the surroundings, we write :

qcal= Ccal × ΔT= 490 J/K × 276.7 K= <u>135,583 J</u> = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

<u> Now that we have the heat of combustion, we need to calculate the molar heat. </u>

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write the <u>conversion factor as 135.58 kJ/ 4.40 g</u>.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.

3 0

3 years ago