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anastassius [24]
3 years ago
12

What is the difference between an autosome and a sex chromosome

Chemistry
1 answer:
tatyana61 [14]3 years ago
8 0
. The members of theautosome pairs are truly homologous; that is, each member of a pair contains a full complement of the same gene.The sex chromosomes, on the other hand, do not constitute a homologous pair.
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Cylinder are used to measure those things so yes correct I can’t write it all sorry
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Sonar is a technology that uses sound waves to calculate the distance to an object. Explain why you think sonar is useful for sc
Margarita [4]

Answer:

<em>Sonar can be used to measure the depth of the seabed or the distance of any object, animal or a man-made vessel any other objects </em>

Explanation:

  • A sound wave or sound pulse is projected into the water.
  • If any object tends to come in the way of the pulse, an echo is produced or the signal is reflected back.
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This is much better way of measuring distances under the sea since light alone is not able to pierce through the ocean after a certain distance.

A powerful sonar blast can help in taking multiple measurements at a time, while even a laser will find it difficult to reach below 3 km of depth.

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3 years ago
What kind of bond is formed when lithium and fluorine combine to form lithium fluoride?
cupoosta [38]

Answer:

ionic bond

Explanation:

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3 years ago
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A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
3 years ago
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