CHEMISTRY COLLEGE

Which group is the most reactive?*
Alkaline Earth Metals
Alkali metals
Noble Gases
Lanthanides

Answers

Answer 1

Answer:

Answer:

alkali metals- Group 1

Explanation:

they have less valence electrons and therefore are more reactive

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The entropy of a substance above absolute zero will always be: a. Negative
b. Positive
c. Neither Negative nor positive

Answers

i will be positive. just because it’s positive

Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V

An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:

Ni(s)VO2+(aq,0.083M)+2H+(aq,1.1M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l)

Calculate the cell potential under these nonstandard concentrations.

Express the cell potential to two decimal places and include the appropriate units.

Answers

Answer:

Cell potential under non standard concentration is 4.09 v

Explanation:

To solve this problem we need to use Nernst Equation because concentrations of the components of the chemical reactions are differents to 1 M (normal conditions: 1 M , 1 atm).

Nernst equation at 25ºC is:

$E = E^(0) - [((0.0592)/(n) · log Q)]$

where

E: Cell potential (non standard conditions)

$E^(0)$ = Cell potential (standard conditions)

n: Number of electrons transfered in the redox reaction

Q: Reaction coefficient (we are going to get it from the balanced chemical reaction)

For example, consider the following general chemical reaction:

aA + bB --> cC + dD

where

a, b, c, d: coefficient of balanced chemical reaction

A,B,C,D: chemical compounds in the reaction.

Using the previous general reaction, expression of Q is:

$Q = (C^(c) * D^(d) )/(A^(a)*B^(b) )$

Previous information is basic to solve this problem. Let´s see the Nernst equation, we need to know: $E^(0)$ , n and Q

Let´s calcule potential in nomal conditions ( $E^(0)$ ):

1. We need to know half-reactions (oxidation and reduction), we take them from the chemical reaction given in this exercise:

Half-reactions: Eo (v):

$(VO_(2))^(2+)$ + e- --> $(VO_(2))^(+)$ -0.23

$Ni --> Ni^(2+)$ + 2 e- +0.99

Balancing each half-reaction, first we are going to balance mass and then we will balance charge in each half-reaction and then charge between half-reactions:

Half-reactions: Eo (v):

2 * [ $(VO_(2))^(2+)$ + e- --> $(VO_(2))^(+)$ ] -0.23

1 * [ $Ni --> Ni^(2+)$ + 2 e- ] +0.99

------------------------------------------------------------------- -------------

2 $(VO_(2))^(2+)$ + 2e- + Ni --> 2 $(VO_(2))^(+)$ + $Ni^(2+)$ + 2e- 0.76 v

Then global balanced chemical reaction is:

2 $(VO_(2))^(2+)$ + Ni --> 2 $(VO_(2))^(+)$ + $Ni^(2+)$

and the potential in nomal conditions is:

$E^(0)$ = 0.76 v

Also from the balanced reaction, we got number of electons transfered:

n = 2

2. Calculate Q:

Now using previous information, we can establish Q expression and we can calculate its value:

$Q = ([(VO_(2)+]^(2)* [Ni^(2+) )/([VO_(2+)]^(2) * Ni )]$

From the exercise we know:

$[VO_(2) ^(2+)] = 2.5 M$

$[VO_(2)+] = 0.083 M$

$[Ni^(2+)] = 2.5 M$

$[Ni] = 1 M, it is a pure solid, so its activity in Q is unit (1). It is also applied for pure liquids.$

$Q = ((2.5)^(2)* 2.5 )/((0.083)^(2) * 1) = 2,268.11$

3. Use Nernst equation:

Finally, we replace all these results in the Nernst equation:

$E = E^(0) - ((0.0592)/(n) - log Q)\n \nE = 0.76 - ((0.0592)/(2)-log (2,268.11) \nE = 0.76 - (0.0296 - 3.36)\nE = 4.09 v$

Cell potential under non standard concentration is 4.09 v

Final answer:

To calculate the cell potential under nonstandard conditions, we need to apply the Nernst Equation. This involves finding the reaction quotient (Q) from the given concentrations and then subtracting a value derived from Q and the number of electrons transferred, from the cell potential under standard conditions.

Explanation:

For calculating the cell potential under nonstandard conditions for an electrochemical cell, we need to use the Nernst equation. In this case, the Nernst Equation is Ecell = E∘cell - (0.0592/n) * logQ, where Q, the reaction quotient, is the ratio of the concentrations of the products to the reactants raised to their stoichiometric coefficients.

Given the half-cell reduction potentials, we can calculate the cell potential under standard conditions (E°cell) by subtracting the potential of the anode from the potential of the cathode (E°cell = Ecathode - Eanode = 0.99V - (-0.23V), resulting in E°cell = 1.22V.

Next, Q = [Ni2+]/([VO2+]×[H+]²), substituting the given concentrations, Q = (2.5)/(0.083×1.1²).

After calculating Q, we substitute all known values into the Nernst Equation and solve for Ecell. Hence, the cell potential under these nonstandard conditions can be calculated.

Learn more about Nernst Equation here:

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Two samples of carbon tetrachloride were decomposed into their constituent elements. One sample produced 38.9 g of carbon and 463 g of chlorine, and the other sample produced 14.8 g of carbon and 144 g of chlorine. Part A Are these results consistent with the law of definite proportions?

Answers

Explanation:

Firstly, defining Law of definite proportion which states that in any chemical compound, the elements are the same and are in the same proportion by mass. It is also called Proust's law.

In sample A;

38.9 g of carbon

463 g of chlorine

In sample B;

14.8 g of carbon

144 g of chlorine.

Sample A, Ratio of mass of Carbon to mass of Chlorine

= 38.9 : 463

= 0.084

Sample B, Ratio of mass of Carbon to mass of Chlorine

= 14.8 : 144

= 0.103

These results above show that the ratio of the masses of Carbon to Chlorine in both samples A and B are not the same so therefore, the results are not consistent with the law of definite proportion.

Identify the type of bonding within each substance. Co ( s ) ionic covalent metallic CoCl 2 ( s ) covalent ionic metallic CCl 4 ( l ) metallic covalent ionic

Answers

Answer:

1. Co ( s ) - metallic bonding

2. CoCl₂ ( s ) - ionic bonding

3. CCl₄ ( l ) - covalent bonding

Explanation:

Metallic bonding -

It is the type of bonding present between the atoms of the metals , via the electrostatic interaction between the metal and the delocalized electrons , is known as metallic bonding .

For example ,

Mostly metals show metallic bonding .

Ionic bonding -

It is the type of bonding present between the ions i.e. , the cation and the anion is known as ionic bonding .

For example ,

Mostly ionic compound , like salts show ionic bonding .

Covalent bonding -

It is the type of bonding which is present between shared pair of electrons , is known as covalent bonding .

For example ,

Most of the carbon compounds are capable to show covalent bonding .

Hence , from the question ,

1. Co ( s ) - metallic bonding

2. CoCl₂ ( s ) - ionic bonding

3. CCl₄ ( l ) - covalent bonding

Final answer:

Co(s) forms metallic bonds, CoCl2(s) forms ionic bonds, and CCl4(l) forms covalent bonds.

Explanation:

The type of bonding within each substance can be identified by understanding the nature of the substances.

Co(s): Co (solid) denotes a metal, and metals typically form metallic bonds.
CoCl2(s): CoCl2 is a compound formed between a metal (Co) and a nonmetal (Cl), making it an ionic compound.
CCl4(l): CCl4 is a molecular compound composed of carbon and chlorine atoms, so it forms covalent bonds.

Learn more about Bonding here:

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Nf3 Rotate the molecule until you have a feeling for its three-dimensional shape. How many atoms are bonded to the central atom?

Answers

Answer:

Three atoms are attached to the central atom in NF3.

Explanation:

The central atom is always regarded as the atom having the least electronegativity in a molecule or ion. We can decide on what atom should be the central atom by comparing the relative electro negativities of the atoms in the molecule or ion.

If we consider NF3, we can easily see that nitrogen is less electronegative than fluorine, hence nitrogen is the central atom in the molecule. We can also observe from the molecular model that three atoms of fluorine were attached to the central atom. Hence there are three atoms attached to the central atom in the molecule NF3.

I have a cup of hot coffee at 140 oC but I want to cool it to 110 oC. My cup holds about 0.3 kg of coffee. Fortunately, I have a bunch of aluminum cubes in the freezer that I can drop into my hot coffee to cool it down. If each aluminum cube has a mass of 1 g (not 1 kg!) and my freezer keeps its contents at a temperature of –10 oC, how many cubes do I have to drop into my coffee? The specific heat of water is around 4000 joules/kg/oC and aluminum is about 900 joules/kg/oC. (Pick the answer closest to the true value and ignore any thermal losses to surroundings.)

A. 200
B. 330
C. 400
D. 110
E. 88