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Aleksandr-060686 [28]
2 years ago
11

How can you recognize an alcohol?

Chemistry
1 answer:
gavmur [86]2 years ago
5 0

Answer:

O D. It has an -OH group attached to the end of the molecule.

Explanation:

Some alcohols have hydroxyl group (OH) attached to the end of a molecule and some have it attached as a branch on the molecule

{ \rm{R - OH \:  : \: primary \: alcohol   \: (1 \degree)}}  \\  \\ { \rm{R -(OH) - R {}^{i}  \:  : secondary \: alcohol \: (2 \degree)}} \\  \\ { \rm{R - R {}^{i} (OH) - R {}^{ii}  \:  : tertiary \: alcohol \: (3 \degree)}}

  • R is aryl or alkyl group
  • OH is hydroxyl group
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Is pyrophoric a chemical or physical property
Neporo4naja [7]

Answer:

Pyrophoricity is a property of metals and oxides of lower oxidation states, including radioactive ones, in which they spontaneously ignite during or after stabilization.

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3 years ago
What is the electron configuration of Mn3+
Mumz [18]

Answer:

The electron configuration for a

Mn3+ ion is [Ar]3d4

Explanation:

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3 years ago
_______is the division of the cytoplasm at the end of mitosis.
tankabanditka [31]

Answer: D. CYTOKINESIS

Explanation:

Mitosis<em> </em>ends<em> with telophase, or the stage at which the chromosomes reach the poles. ... Telophase is followed by </em>CYTOKINESIS<em>, or the division of the cytoplasm into two daughter </em><em>cells.</em>

6 0
2 years ago
Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
lakkis [162]

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

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3 years ago
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