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Nonamiya [84]
3 years ago
14

How many moles are in 2.98x10^23 particles?

Chemistry
1 answer:
Novosadov [1.4K]3 years ago
3 0

Answer:

\boxed {\boxed {\sf 0.495 \ mol}}

Explanation:

We are given a number of particles and asked to convert to moles.

<h3>1. Convert Particles to Moles </h3>

1 mole of any substance contains the same number of particles (atoms, molecules, formula units) : 6.022 *10²³ or Avogadro's Number. For this question, the particles are not specified.

So, we know that 1 mole of this substance contains 6.022 *10²³ particles. Let's set up a ratio.

\frac { 1 \ mol }{6.022*10^{23 } \ particles}}

We are converting 2.98*10²³ particles to moles, so we multiply the ratio by that value.

2.98*10^{23} \ particles *\frac { 1 \ mol }{6.022*10^{23 } \ particles}}

The units of particles cancel.

2.98*10^{23}  *\frac { 1 \ mol }{6.022*10^{23 } }}

\frac { 2.98*10^{23}}{6.022*10^{23 } }}  \ mol

0.4948522086 \ mol

<h3>2. Round</h3>

The original measurement of particles (2.98*10²³) has 3 significant figures, so our answer must have the same.

For the number we found, 3 sig figs is the thousandth place.

The 8 in the ten-thousandth place (0.4948522086) tells us to round the 4 up to a 5 in the thousandth place.

0.495 \ mol

2.98*10²³ particles are equal to approximately <u>0.495 moles.</u>

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Answer: An atom has either tendency of accepting or losing electron, on the basis of this virtue it is named as

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  • Electropositive: An atom that has tendency to give the shared pair of electron towards an electronegative atom is called electropositive.

For the existence of dipole within a molecule there must be a difference in electronegativity of the atoms participating in it. All the molecules here have dipole moment because the atoms participating in them have a difference in electronegative.

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A voltaic cell is based on the reduction of ag (aq) to ag(s) and the oxidation of sn(s) to sn2 (aq).
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Half-reaction for the cell's anode is given below:

Anode : \\Sn(s) \rightarrow Sn^{2+} (aq) +2e^{-}

The anode is defined as the electrode at which electrons leave the cell and oxidation occurs, and the cathode as the electrode at which electrons enter the cell and reduction occurs. The anode is usually the positive side.

Learn more about anode here:

brainly.com/question/4052514

Your given question is quite incomplete here is complete question.

A voltaic cell is based on the reduction of _ Agt(aq) to Ag(s) and the oxidation of Sn(s) to Sn2+(aql) : Part 1  Include the phases of all species in the chemical equation: (aqh Anode: Sn(s) Sn?+ (aq)

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Silver-108 has how many protons, neutrons, and electrons
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Hydrogen iodide decomposes according to the equation: 2HI(g) H 2(g) + I 2(g), K c = 0.0156 at 400ºC A 0.660 mol sample of HI was
Ira Lisetskai [31]

Answer:

[HI] = 0.264M

Explanation:

Based on the equilibrium:

2HI(g) ⇄ H₂(g) + I₂(g)

It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:

<em>Kc = 0.0156 = [H₂] [I₂] / [HI]²</em>

<em />

As initial concentration of HI is 0.660mol / 2.00L = <em>0.330M, </em>the equlibrium concentrations will be:

[HI] = 0.330M - 2X

[H₂] = X

[I₂] = X

<em>Where X is reaction coefficient.</em>

<em />

Replacing in Kc:

0.0156 = [X] [X] / [0.330M - 2X]²

0.0156 = X² / [0.1089 - 1.32X + 4X² ]

0.00169884 - 0.020592 X + 0.0624 X² = X²

0.00169884 - 0.020592 X - 0.9376 X² = 0

Solving for X:

X = - 0.055 → False solution, there is no negative concentrations

X = 0.0330 → Right solution.

Replacing in HI formula:

[HI] = 0.330M - 2×0.033M

<h3>[HI] = 0.264M</h3>

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