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4 years ago

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More than 70% of all legislation passed in Washington, DC addresses issues related to science; most of these issues pertain chem

istry. True or False

1 answer:

Zigmanuir [339]4 years ago

8 0

The answer is False.

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A beam of light passes though a liquid in a test tube without scattering. Which type of mixture is most likely in the test tube?

butalik [34]

The answer would be letter C - solution.

A mixture should be homogeneous for a light not to be scattered. This is because particles are distributed evenly throughout the mixture which allows light to pass directly. In your choices, the solution allows a beam of light to pass through a liquid in a test tube without scattering.

6 0

4 years ago

Read 2 more answers

Define the following and give an example of each:_______.

Ber [7]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2><h2> $\huge\textbf{Van de Waal}$ </h2>

London Dispersion force or Van de Waals force is a temporary attractive force which are the weakest and occur between nonpolar noble gases and same charges. This force is weaker because they have more electrons that are farther from the nucleus and are able to move around easier.

Example: $N_2$ , $F_2, \ I_2$

<h2>_____________________________________</h2>

$\huge\textbf{Dipole - Dipole Forces}$

Dipole force is present between the polar molecules. Polar molecules are those molecules which have slightly negative and slightly positive charge. Dipole-dipole forces are attractive forces between the positive end of one polar molecule and the negative end of another polar molecule.

Example: HCl, HF, $CHCl_3$

<h2>_____________________________________</h2>

$\huge\textbf{Hydrogen Bonding}$

It is a special type of dipole force present between polar molecules, it is formed between Hydrogen atom which forms positive ion, and the other negative ion. It results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom such as a N, O, or F atom. The hydrogen bond is one of the strongest intermolecular attractions, but weaker than a covalent or an ionic bond.

Example: Every polar molecule which has hydrogen has hydrogen bonding i.e. $H_2O, NH_3$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h2 />

6 0

3 years ago

How much mass would a mole of hydrogen molecules contain?

dsp73

Answer:

2.016 g/mol

Explanation:

Avogadro number

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For H₂:

2.016 g of H₂ = one mole of hydrogen gas = 6.022 × 10²³ molecules of hydrogen

6 0

3 years ago

do you think it is likely that a major volcanic eruption will occur on the mainland of the United States

ANEK [815]

Because the whole west coast is on the "ring of fire" witch has high volcanic activety and junk like that

8 0

3 years ago

Read 2 more answers

Identify the most likely reaction product(s) in the monobromination of 2-pentanone by bromine in the presence of acid in the dar

irinina [24]

The most likely reaction product in the monobromination of 2-pentanone by bromine in the presence of acid in the dark is known to be 1-Bromo-2-pentanone

<h3>What is a Monobromination?</h3>

The term is known to be the bringing in of one bromine atom into what we call an organic compound.

Note that the brominating of benzene is known to be a good example of what we know as electrophilic aromatic substitution reaction. In this kind of reaction, the electrophile (which is bromine) is said to forms a kind of sigma bond to that of the benzene ring, bringing in an intermediate. Then, a proton is said to be deleted from the intermediate to create a substituted of benzene ring.

Hence, The most likely reaction product in the monobromination of 2-pentanone by bromine in the presence of acid in the dark is known to be 1-Bromo-2-pentanone

Learn more about monobromination from

brainly.com/question/14356147

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See full question below

Identify the most likely reaction product(s) in the monobromination of 2-pentanone by bromine in the presence of acid in the dark.

a.) 1-Bromo-2-pentanone and 3-bromo-2-pentanone

b.) 1-Bromo-2-pentanone, 3-bromo-2-pentanone, and 4-bromo-2-pentanone

c.) 1-Bromo-2-pentanone, 3-bromo-2-pentanone, 4-bromo-2-pentanone, and 5-bromo-2-pentanone

d.) 3-Bromo-2-pentanone

e.)1-Bromo-2-pentanone

7 0

2 years ago