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3 years ago

How are an atoms chemical properties determined?

1 answer:

5 0

Electrons determine the chemical properties. Protons and neutrons are in the nucleus the number of protons determines the number of electrons of an atom

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Given the equation representing a reversible reaction:

tester [92]

Answer:

- Acetic acid (CH₃COOH) and hydronium ion (H₃O⁺)

Explanation:

Hello,

In this case, based on the acid-base theory which states that acids are known as H⁺ donors, if we consider the direct reaction:

$CH_3COOH(aq) + H_2O \rightarrow CH_3COO^-(aq) + H_3O^+(aq)$

It is clear that the acetic acid is the first H⁺ donor as it losses one H⁺ to turn into the acetate ion. Moreover, if we consider the inverse reaction:

$CH_3COO^-(aq) + H_3O^+(aq)\rightarrow CH_3COOH(aq) + H_2O$

It is also clear that the hydronium ion is the second H⁺ donor as it losses one H⁺ to turn into water.

Best regards.

5 0

3 years ago

Why do atoms become anion and cations (remember the octet rule and how 8 is a very important number)?

Kryger [21]

Answer:

The atoms become cations because they have lost electrons and the atoms and has a positive charge then atoms become a ions because they have gained electrons from cations and has a negative charge

8 0

3 years ago

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8

Lubov Fominskaja [6]

Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is $pK_a =7.82$

Explanation:

From the question we are told

The volume of base is $V_B = 26.mL = 0.0260L$

The pH of solution is $pH = 7.82$

The concentration of the acid is $C_A = 0.1M$

From the pH we can see that the titration is between a strong base and a weak acid

Let assume that the the volume of acid is $V_A = 18mL= 0.018L$

Generally the concentration of base

$C_B = \frac{C_AV_A}{C_B}$

Substituting value

$C_B = \frac{0.1 * 0.01800}{0.0260}$

$C_B= 0.0692M$

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

$HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}$

Now before the reaction the number of mole of base is

$No \ of \ moles[N_B] = C_B * V_B$

Substituting value

$N_B = 0.01300 * 0.0692$

$= 0.0009 \ moles$

Now before the reaction the number of mole of acid is

$No \ of \ moles = C_B * V_B$

Substituting value

$N_A = 0.01800 *0.1$

$= 0.001800 \ moles$

Now after the reaction the number of moles of base is zero i.e has been used up

this mathematically represented as

$N_B ' = N_B - N_B = 0$

The number of moles of acid is

$N_A ' = N_A - N_B$

$= 0.0009\ moles$

The pH of this reaction can be mathematically represented as

$pH = pK_a + log \frac{[base]}{[acid]}$

Substituting values

$7.82 = pK_a +log \frac{0.0009}{0.0009}$

$pK_a =7.82$

8 0

3 years ago

Explain why scientists find the particulate matter of theory useful

BARSIC [14]

This idea has historical significance. The ancient Greek philosopher Democritus (born 460 BCE), who held that everything is composed of small particles moving in empty space, is credited with developing the first hypothesis we have about the microscopic universe. He had some concrete proof for this, such the fact that items like a new loaf of bread or a rose may give off a scent even when they are far from the source. Being a materialist, he thought that these odors originated from actual material particles released by the bread or the rose, rather than being purely a type of magic. He reasoned that these particles must float through the air, with some of them maybe landing in your nose where you can smell them immediately. This still makes sense in modern times. But many of us now have quite different perspectives on these "particles."

Thank you,

Eddie

6 0

2 years ago

Really need help!

WARRIOR [948]

The magnitude of a star as it would appear to a hypothetical observer at a distance of 10 parsecs or 32.6 light-years. This rates how visible celestial bodies are when they are all viewed from the same distance. Luminosity: The brightness of a star in comparison with that of the sun.

7 0

3 years ago

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