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1 year ago

Brainiest and 10 Points

Chemistry

2 answers:

Soloha48 [4]1 year ago

8 0

I believe the answer for this is:

C. Visible Light Rays

I hope this helps! :D

Nezavi [6.7K]1 year ago

4 0

Answer:

C, visible light rays :)

Explanation:

Hope i could help

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when 0.72 g of a liquid is vaporized at 110° C and 0.967 atm, the gas occupies a volume of 0.559L. The empirical formula of the

Leno4ka [110]

Hydrogen gas mixed with sulfur

8 0

2 years ago

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r

spin [16.1K]

<u>Answer:</u> The theoretical yield of acetanilide is 6.5 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For aniline:</u>

Given mass of aniline = $4.50\times 10^0=4.50g$ (We know that: $10^0=1$ )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

$\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol$

<u>For acetic anhydride:</u>

To calculate the mass of acetic anhydride, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of acetic anhydride = $(1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL$

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

$1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g$

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol$

The chemical equation for the reaction of aniline and acetic anhydride follows:

$C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH$

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = $\frac{1}{1}\times 0.048=0.048mol$ of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = $\frac{1}{1}\times 0.048=0.048mol$ of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

$0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g$

Hence, the theoretical yield of acetanilide is 6.5 grams.

3 0

2 years ago

Use the information below to explain why the atomic radius decreases down a group.

notsponge [240]

Answer:

Detail is given below

Explanation:

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.

In A we can see that there is one positive charge and force of attraction is 2.30×10⁻⁸ N and distance is 0.10 nm

In B we can see that negative charge is further away from nucleus because of greater distance thus force of attraction will be less. 0.58×10⁻⁸ N

In C this distance further increases and force also goes in decreasing 0.26×10⁻⁸ N.

3 0

2 years ago

Help name this, it’s chemistry

vova2212 [387]

Answer:

I believe it's

4-floro-5,6,8-Tribromo-3,3,7-trimethylnonane

5 0

2 years ago

Draw the alkene formed when 1-heptyne is treated with hbr in the presence of peroxide.

Nimfa-mama [501]

<h2>Heptene formed is -</h2><h2> $CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr$ </h2>

Explanation:

The two possibilities when the peroxide is not present

$CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-C$ ≡ $CH +HBr$ → $CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_{2}$

$CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_2$ + HBr → $CH_3-CH_2-CH_2-CH_2-CH_2-CBr_2-CH_3$

In presence peroxide,

$CH_3-CH_2-CH_2-CH_2-CH_2-C$ ≡ $CH$ + HBr → $CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr$

When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.

This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
One mole of HBr adds to one mole of 1-heptane.
The structure of heptene formed is -

$CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr$

5 0

2 years ago