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3 years ago

The part of an atom that is responsible for the bonding of atoms is

Chemistry

1 answer:

Cerrena [4.2K]3 years ago

3 0

Answer:

the answer is electrons

Explanation:

electrons are responsible for the chemical bonding.

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Calculate the mass in grams of 1.13 x 102 molecules of propane. The chemical formula for propane is C3H8.

g100num [7]

Moles = (6.74*10^23)/(6.02*10^23) =1.119 moles

1.119*44.09=49.36g

3 0

3 years ago

10 moles of carbon dioxide has a mass of 440 g. What is the relative formula mass of carbon dioxide?

Ksenya-84 [330]

Answer: 44g

Explanation: The formular for finding Moles is ;

Moles = Mass / Molar Mass or Formular Mass.

Base on this question; Moles = 10, Mass = 440g, and Formular Mass = ?

Making 'Formular Mass', subject of the formular; we thus have;

Formular mass = Mass / Moles = 440/ 10 = 44g

7 0

3 years ago

5 of 10

Roman55 [17]

Answer:

Peer review is a quality control measure for medical research. It

is a process in which professionals review each other's work to

make sure that it is accurate, relevant, and significant. Scientific

researchers aim to improve medical knowledge and find better

ways to treat disease.

if I am right mark the answer as brainliest

6 0

2 years ago

Read 2 more answers

True or False a chemical formula is a representation of the smallest unit of a compound?

trapecia [35]

Answer:

true. I think

Explanation:

A chemical formula shows the atoms a molecule is made of.

3 0

3 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