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1 year ago

Please help, I’m a bit confused, lol

Chemistry

1 answer:

Mashutka [201]1 year ago

4 0

Explanation:

option A skeleton system is correct option

hope this helps you !

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What is the density of ice at -10°C?

Alisiya [41]

Answer:

3.98 C my friend you welcome

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3 years ago

Indicate the number of significant figures in the measured number 0.0020 A. 4 B. 5 C. 2

Sergio [31]

Answer:

The correct answer option is C. 2.

Explanation:

We are given the number '0.0020' and we are to indicate the number of significant figures in the given measured number.

According to the rules of significant figures, numbers that are non-zero, zeros between any two significant numbers and the ending zeros in the decimal position are categorized as significant figures.

Since there is one non-zero number and one ending zero in the decimal position, therefore 0.0020 has 2 significant figures.

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3 years ago

Which processes occur when energy is removed from a substance

ratelena [41]

Answer:

Note that melting and vaporization are endothermic processes in that they absorb or require energy, while freezing and condensation are exothermic process as they release energy.

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2 years ago

Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

Dima020 [189]

Answer: The $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

Explanation:

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

Oxidation half reaction: $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

Reduction half reaction: $Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = 0.406 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J$

Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

4 0

3 years ago

the element aluminum forms a cation with the charge fill in the blank 2 . the symbol for this ion is , and the name is fill in t

shtirl [24]

The element aluminum forms a cation with the charge of +3. The symbol for this ion is Al³⁺, and the name is aluminum ion. The number of electrons in this ion 10.

Ions are formed when an atom gains or loses electrons.

Metals form positive ions by a loss of electrons whereas non-metals form negative ions by gaining electrons.

The aluminum ion is formed when aluminum loses three electrons to form Al³⁺ ion which has 10 electrons.

in conclusion, ions are formed by a gain or loss of electrons.

Learn more about ions at: brainly.com/question/490569

#SPJ1

5 0

8 months ago