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Nonamiya [84]
2 years ago
10

Help!! Hurry!! I will mark brainliest if you get it correct. NO SPAM!!!!!!

Chemistry
2 answers:
Lelechka [254]2 years ago
6 0

Answer:

216/84 Po + 4/2 He a

Explanation:

Ahat [919]2 years ago
3 0

Answer:

Problem: Polonium-210, 210Po, decays to lead-206, 206Pb, by alpha emission according to the equation 21084Po → 20682Pb + 42He

Explanation:

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In a solution, when the suspension of one kind of molecules is blended or dissolved with another, the ________ properties are no
Sidana [21]

Answer:

chemical

Explanation:

The chemical properties determine the identity of a substance and are not changed when there is no reaction. In the solution, what takes place is a physical change, which is a change of state (for example, going from solid to aqueous form).

8 0
3 years ago
Temperate deciduous trees lose their leaves in Fall. Explain why trees in temperate rainforest and tropical rainforest don’t los
andrey2020 [161]

Answer:

So trees in temperate don't lose their leaves because the weather events aren't harsh enough.

Trees in tropical rainforest don't lose their leaves because they are a different type of tree known as evergreens that are green all year round.

Explanation:

Ok so first we'll define some things

Deciduous Trees=  Trees that lose all of their leaves for part of the year.

Trees shed their leaves trees to try and survive harsh weather events.

Temperate deciduous trees lose their leaves in fall to better survive the winter conditions of extreme cold and reduced daylight.

Temperate rainforests = An area that doesn't experience extremely cold or extremely hot temperatures or what we would call harsh weather events.

Broad-leaved trees in tropical rainforests are known evergreen, they are known as this as they are green all year round.

6 0
3 years ago
Intermolecular forces dipole differences london dispersion
loris [4]
33233728793278237876548742787874578378572098-2932-=93788784787489
8 0
3 years ago
Chemistry! Help! Please!! Thanks
Vladimir79 [104]

Answers:

1. 3-ethyl-3-methylheptane; 2. 2,2,3,3-tetramethylpentane; 3. hexa-2,4-diene.

Explanation:

<em>Structure 1 </em>

  1. Identify and name the longest continuous chain of carbon atoms (the main chain has 7 C; ∴ base name = heptane).
  2. Identify and name all the substituents [a 1C substituent (methyl) and a 2C substituent (methyl).
  3. Number the main chain from the end closest to a substituent.
  4. Identify the substituents by the number of the C atom on the main chain. Use hyphens between letters and numbers (3-methyl, 3-ethyl).
  5. Put the names of the substituents in alphabetical order in front of the base name with no spaces (3-ethyl-3-methylheptane)

<em>Structure 2</em>

  1. 5C. Base name = pentane
  2. Four methyl groups.
  3. Number from the left-hand end.
  4. If there is more than one substituent of the same type, identify each substituent by its locating number and use a multiplying prefix to show the number of each substituent. Use commas between numbers (2,2,3,3-tetramethyl).
  5. The name is 2,2,3,3-tetramethylpentane.

<em>Structure 3 </em>

  1. Identify and name the longest continuous chain of carbon atoms that passes through as many double bonds as possible. Drop the <em>-ne</em> ending of the alkane to get the root name <em>hexa-</em>.
  2. (No substituents).
  3. Number the main chain from the end closest to a double bond.
  4. If there is more than one double bond use a multiplying prefix to indicate the number of double bonds (two double bonds = diene) and use the smaller of the two numbers of the C=C atoms as the double bond locators (2,4-diene)
  5. Put the functional group name at the end of the root name (hexa-2,4-diene).

<em>Note</em>: The name 2,4-hexadiene is <em>acceptable</em>, but the <em>Preferred IUPAC Name</em> puts the locating numbers as close as possible in front of the groups they locate.

7 0
3 years ago
1. An object was carefully weighed on three different balances. Each balance was zeroed
Hoochie [10]

Answer:

10.335

Explanation:

An object was carefully weighed on three different balances

Each of these balances were zeroed before weighing

The masses that were weighed are as follows

10.35 g , 10.355 g, 10.30 g

Therefore the average value of these measurements can be calculated as follows

The total number of mass is 3

= 10.30 + 10.355 + 10.30/3

= 31,005/3

= 10.335

Hence the average value of these measurements is 10.335

7 0
3 years ago
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