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Aleksandr [31]
3 years ago
13

Advantages for building on the green hill

Chemistry
2 answers:
Ksenya-84 [330]3 years ago
8 0
Flooding - Some places in Green Hill have steep areas and don't have any steep areas.
Minchanka [31]3 years ago
6 0
I think that building on a green hill
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What is the name of the molecule below?
SIZIF [17.4K]

Answer:

C.) 2-butyne

Explanation:

Since the molecule has 4 central carbons, it has the prefix but-.

Since the molecule has a triple bond between central carbons, it has an ending of -yne.

Since the triple bond starts on the second carbon, it has a 2 - prefix.

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3 years ago
1.As per the Aufbau principle, which of the following orbitals have the lowest energy?
madam [21]

The correct answer attached in file, Thank you for joining brainly community.

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2 years ago
How much mass defect is required to release 5.5 x 1020 J of energy?
Montano1993 [528]

Answer : The mass defect required to release energy is 6111.111 kg

Explanation :

To calculate the mass defect for given energy released, we use Einstein's equation:

E=\Delta mc^2

E = Energy released = 5.5\times 10^{20}J

\Delta m = mass change = ?

c = speed of light = 3\times 10^8m/s

Now put all the given values in above equation, we get:

5.5\times 10^{20}Kgm^2/s^2=\Delta m\times (3\times 10^8m/s)^2

\Delta m=6111.111kg

Therefore, the mass defect required to release energy is 6111.111 kg

5 0
3 years ago
In a spacecraft tge carbon dioxide exhaled by the astronaut can be removed by the reaction with lithium hydroxide, LiOH accordin
quester [9]

Answer:

30 moles

Explanation:

From the equation it is a one - to - one reaction

3 0
2 years ago
When Iodine-131 emits a β particle (beta particle), what nuclide is produced? *
k0ka [10]

When Iodine-131 emits a β particle will produce Xe-131

<h3>Further explanation </h3>

Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,  

  • alpha α particles ₂He⁴
  • beta β ₋₁e⁰ particles
  • gamma particles ₀γ⁰
  • positron particles ₁e⁰
  • neutron ₀n¹

So for reaction Iodine-131 :

\tt _{53}^{131}I\rightarrow _{-1}^0\beta +_{54}^{131}Xe

5 0
3 years ago
Read 2 more answers
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