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Murljashka [212]
3 years ago
12

The Reaction between ketones and alcohol

Chemistry
1 answer:
Colt1911 [192]3 years ago
7 0
It has been demonstrated that water adds rapidly to the carbonyl function of aldehydes and ketones to form geminal-diol. In a similar reaction alcohols add reversibly to aldehydes and ketones to form hemiacetals (hemi, Greek, half). This reaction can continue by adding another alcohol to form an acetal.
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How many formula units are there in 212 grams of mgCl2
AleksandrR [38]

Formula units are there in 212 grams of MgCl₂ are 830.56

Formula is the empirical of any ionic or covalent network solid compound used as an independent entity for stoichiometric calculations and it is the lowest whole number ratio of ions represented in an ionic compound

Here given data is

MgCl₂ = 212 grams

1 mole of magnesium chloride has mass = 95.211 gram and contains 6.022×10²³formula units of magnesium chloride

Here 212 grams×6.022×10²³form unit of MgCl₂/95.211 gram = 830.56

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What is the mass of an atom that has 4 protons, 5 neutrons, and 4 electrons?
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Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
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Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

U_{2}-U_{1}=Q-W

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

U_{2}-U_{1}=500-(3*75)=275kJ

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

7 0
3 years ago
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