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2 years ago

How does the size of the atoms relate to how reactive it is?

Chemistry

2 answers:

Nata [24]2 years ago

6 0

Answer:

The larger an atom is, the more reactive it will be.

It helps to think about the size of an atom with the location of the electrons and nucleus in mind. You know that protons are positively charged and electrons are negatively charged, therefore there is a level of attraction between these two oppositely charged subatomic particles. This attraction helps stabilize the atom.

Now, in a larger atom, with more electrons are those electrons (specifically the valence electrons) going to be closer or further from the nucleus than in a smaller atom?

They will of course be further from the nucleus and therefore further from the positively charged nucleus. This makes it easier for the valence electrons to leave the atom and form an ionic bond or to interact with another atom to form a covalent bond.

If electrons can leave more easily, then the atom is more likely to react.

pishuonlain [190]2 years ago

5 0

Answer: nucleus

Explanation: the bigger the nucleus there more reactive the atom is

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Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

2 years ago

The density of a pure liquid at 25 degrees Celsius was calculated by determining the mass and volume of a sample of the liquid.

adelina 88 [10]

Answer

The density of the liquid is 0.82 g/mL.

Explanation:

From the question given above, the following data were obtained:

Mass of empty flask = 18.990 g

Mass of flask + liquid = 39.439 g

Volume of liquid = 25 mL

Density of liquid =..?

Next, we shall determine the mass of the liquid. This can be obtained as follow:

Mass of empty flask = 18.990 g

Mass of flask + liquid = 39.439 g

Mass of liquid =.?

Mass of liquid = (Mass of flask + liquid) – (Mass of empty flask)

Mass of liquid = 39.439 – 18.990

Mass of liquid = 20.503 g

Finally, we shall determine the density of the liquid as follow:

Mass of liquid = 20.503 g

Volume of liquid = 25 mL

Density of liquid =..?

Density = mass / volume

Density of liquid = 20.503 / 25

Density of liquid = 0.82 g/mL

8 0

2 years ago

No spam or links! Thanks!

Monica [59]

Answer:

364 K or 91°C

Explanation:

Applying,

V₁/T₁ = V₂/T₂................ Equation 1

Where V₁ = Initial Volume, V₂ = Final volume, T₁ = initial Temperature, T₂ = final Temperature.

make T₂ the subject of the equation,

T₂ = V₂T₁/V₁................. Equation 2

From the question,

Given: V₁ = 375 mL, V₂ = 500 mL, T₁ = 0.0°C = (273+0) K = 273 K

Substitute these values into equation 2

T₂ = (500×273)/375

T₂ = 364 K

T₂ = (364-273) °C = 91 °C

5 0

2 years ago

Consider the equation: 2 nan3 (s) 2 na (s) + 3 n2. what is the mass of nan3 needed to produce 17.2 l nitrogen at stp?

Lerok [7]

The mass of NaN3 needed to produce 17.2 L nitrogen at STP is calculated as follows

find the moles of N2 produced at STP
At STP 1mole of gas = 22.4 L , what about 17.2 L of nitrogen

by cross multiplication

= (1 mole x17.2 L)/ 22.4 L= 0.768 moles

2NaN3 =2Na +3 N2
by use of mole ratio between NaN3 to N2 (2:3) the moles of NaN3 = 0.768 x2/3 = 0.512 moles of NaN3

mass of NaN3 is therefore =moles of NaN3 xmolar mass of NaN3

=0.512moles x 65 g/mol =33.28 grams of NaN3

3 0

2 years ago

Was the first person t propose the idea of moving continents as a scientific hypothesis

BlackZzzverrR [31]

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6 0

2 years ago

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