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3 years ago

R spectroscopy can be used to identify which _____________________ are present in a compound.

Chemistry

1 answer:

Nesterboy [21]3 years ago

3 0

Answer:

IR spectroscopy can be used to identify chemical structures are present in compounds.

Explanation:

Infrared spectroscopy is a technique in organic chemistry that can be use use to identify chemical structures present in compounds because it is base on the ability of different functional groups to adsorb infrared light.

This work by shinning the infrared lights into the organic compounds to be identified, some of the frequencies of the infrared lights are adsorbed by the compounds and its identify groups of atoms and molecules in the compound.

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What is one difference between an electron and a neutron?

VARVARA [1.3K]

Answer: Neutron has no charge, electron has a charge and mass. Neutron occurs inside the nucleus where electron is seen outside the nucleus.

Explanation:

5 0

3 years ago

Decreasing the temperature of the reaction 3H2 + N2 2NH3. In this reaction, the product absorbs heat. WHICH WAY WILL THE REACTIO

Alex

Answer:

the reaction will shift towards the “heat”—shifts to the left

Explanation:

To summarize:

o If temperature increases (adding heat), the reaction will shift away from the “heat” term and go in the

endothermic direction.

o If temperature decreases (removing heat), the reaction will shift towards the “heat” term and go in the

exothermic direction.

o NOTE: The endothermic direction is always away from the “heat” term and the exothermic direction is

towards the “heat” term.

Therefore the reaction will shift towards the “heat”—shifts to the left

3 0

4 years ago

For a certain experiment, a student requires 100 milliliters of a solution that is 8% HCl (hydrochloric acid). The storeroom has

exis [7]

Answer:

70 mL of 5% HCl and 30 mL of 15% HCl

Explanation:

We will designate x to be the fraction of the final solution that is composed of 5% HCl, and y to be the fraction of the final solution that is composed of 15% HCl. Since the percentage of the final solution is 8%, we can write the following expression:

5x + 15y = 8

Since x and y are fractions of a total, they must equal one:

x + y = 1

This is a system of two equations with two unknowns. We will proceed to solve for x. First, an expression for y is found:

y = 1 - x

This expression is substituted into the first equation and we solve for x.

5x + 15(1 - x) = 8

5x+ 15 - 15x = 8

-10x = -7

x = 7/10 = 0.7

We then calculate the value of y:

y = 1 - x = 1 - 0.7 = 0.3

Thus 0.7 of the 100 mL will be the 5% HCl solution, so the volume of 5% HCl we need is:

(100 mL)(0.7) = 70 mL

Similarly, the volume of 15% HCl we need is:

(100 mL)(0.3) = 30 mL

3 0

3 years ago

A 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 26.6

joja [24]

<u>Answer:</u> The concentration of $CH_3COOH$ comes out to be 0.16 M.

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $CH_3COOH$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=25mL\\n_2=1\\M_2=0.175M\\V_2=26.6mL$

Putting values in above equation, we get:

$1\times M_1\times 25=1\times 0.175\times 26.6\\\\M_1=0.1862M$

Hence, the concentration of $CH_3COOH$ comes out to be 0.1862 M.

4 0

3 years ago

Please help on the questions that is left blank!! Due soon and will mark brainliest

34kurt

Answer: The strength of an acid or alkali depends on the degree of dissociation of the acid or alkali in water. The degree of dissociation measures the percentage of acid molecules that ionise when dissolved in water. He could use universal indicators or litmus paper for this.

i hope this helps you!

3 0

3 years ago