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3 years ago

Astronomers use radio waves to study__

Chemistry

1 answer:

Tcecarenko [31]3 years ago

5 0

Answer:

astronomers can learn about their composition, structure, and motion. the explanation idk i found it on google

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Draw all possible Newman projections results from the rotation of C3-C4 bond in 2-methylpentane and label all the conformations.

Elden [556K]

Answer:

See the image Below

Explanation:

Single C-C bonds have free rotation and thus Newman Projections are quite useful for understanding the different 3D conformations of a molecule.

For any Newman Projection you have 2 bonded carbons, one in front of the other, therefore you rotate the bond by rotating the groups that are bonded to the carbons. i. e. in the 2-Methylpentane (viewing the C3-C4 bond) the carbons chosen are the 3rd and the 4th then you look to which are the other groups bonded; 2 H and an isopropyl for the 3rd one, and 2 H and a Methyl for the 4th one.

The different rotations of those groups receive different names:

Anti Conformation: Main groups of the two Carbons are 180º apart from one another. Making this the most stable conformation of the molecule. (Taking into account steric effects)
Gauche Conformation: Principal groups are 60º apart from each other. Since are 6 spaces between groups (360º/6)=60º. It is less stable than an Anti conformation due to the same reason.
Eclipsed Conformation: The less stable conformation since the main groups are confronted.

8 0

4 years ago

Lithium nitride reacts with water to produce ammonia and lithium hydroxide according to the equation, Li3N(s) + 3H2O (L) -->

Gemiola [76]

Answer:

0.480 grams

Explanation:

Li₃N(s) + 3D₂O (L) --------------------------> ND₃(g) + 3LiOD (aq)

1 : 3 : 1 : 3

Number of moles (n) = Mass in gram/ Molar Mass

Mass of ND₃ = 160 mg

= 0.16 g

Molar mass of ND₃= [14 + (3 x 2.014 )]

= 14 + 6.042

= 20.042 g/mol

Number of moles of ND₃ = 0.16/20.042

= 0.007983 moles

From the reaction equation, the mole ratio between Heavy water (D₂O ) and ND₃ is 3: 1.

This implies that the number of moles of Heavy water (D₂O ) required

= 3 x 0.007983 moles

= 0.023949 moles

Molar mass of Heavy water (D₂O )= [(2.014 x 2) + 16]

= 20.028 g/mol

Mass in grams of Heavy water (D₂O )= Number of moles x Molar mass

= 0.023949 x 20.028

= 0.4797 grams

≈ 0.480 grams

3 0

4 years ago

Draw the ring chain of isomer of propene ?

aksik [14]

Explanation:

Molecular formula for Propene = C3H6

The isomer of propene is cyclopropane.

(Draw a triangle to show that it is cyclopropane)

4 0

3 years ago

How many grams of copper (II) hydroxide can be prepared from 2.4 grams of copper (II) nitrate (Cu(NO3)2 ) and excess sodium hydr

andreyandreev [35.5K]

Answer:

percentage yield = 67%

Explanation:

Mass of Cu(NO₃)₂ = 15.25 g

Mass of NaOH = 12.75 g

Percentage yield = ?

Solution:

Cu(NO₃)₂ + 2NaOH → Cu(OH)₂ + 2NaNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 15.25 g /187.56 g/mol

Number of moles = 0.08 mol

Moles of NaOH :

Number of moles = mass/ molar mass

Number of moles = 12.75 g / 40 g/mol

Number of moles = 0.32 mol

Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂. NaOH : Cu(OH)₂

2 : 1

0.32 : 1/2×0.32 = 0.16 mol

Cu(NO₃)₂ : Cu(OH)₂

1 : 1

0.08 : 0.08

The number of moles produced by Cu(NO₃)₂ are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

Mass = 7.808 g

Theoretical yield = 7.808 g

Percent yield:

percentage yield = Actual yield/ theoretical yield × 100

percentage yield = 5.23 g/ 7.808 g × 100

percentage yield = 0.67 × 100

percentage yield = 67%

5 0

3 years ago

There are four different starting molecules that one might use to synthesize the illustrated alkyl halide as the major product u

Mkey [24]

Answer:

Explanation:

An electrophilic addition reaction occurs when an electrophile attacks a substrate, with the end result being the inclusion of one or many comparatively straightforward molecules along with multiple bonds.

In the given question, the hydrogen bromide provides the electrophile while the bromide is the nucleophile. The mechanism proceeds with the attack of the electrophile on the carbon, followed by deprotonation. This process is continued with a formation of carbocation and the bromide(nucleophile) finally bonds to the carbocation to form a stable product.

The first diagram showcases the possible various starting molecules for the synthesis while the second diagram illustrates their mechanism.

6 0

3 years ago