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rusak2 [61]
2 years ago
13

What is the atomic number for an element with 6 neutrons and a mass number of 10

Chemistry
2 answers:
Rasek [7]2 years ago
6 0

Answer:

Thus,thenumberofprotonsis4

Whitepunk [10]2 years ago
5 0

\bold{\huge{\blue{\underline{Answer}}}}

\bold{\underline{ Given :- }}

  • Atomic number of an element with 6 neutrons and a mass number of 10

\bold{\underline{ Let's \:Begin}}

\sf{We\: know\: that, }

  • Atomic number is the number of protons in the nucleus of an atom

\sf{Whereas ,}

  • Mass number is the sum of protons and neutrons in the nucleus of an atom.

\sf{We\: have \:number \:of \:neutrons = 6 }

\sf{ Mass\: number = 10 }

\bold{\underline{Therefore}}

\sf{ Mass\: number = Protons + neutrons}

\sf{ 10 = Protons + 6 }

\sf{ Protons = 10 - 6}

\sf{ Protons = 4 }

\sf{ Thus, \: the \: number \: of \: protons \: is \:4}

\bold{\underline{ Therefore,\:we\:know\:that, }}

\sf{ No\: of\: protons = Atomic \:number }

\sf{ 4 = Atomic \:number }

\sf { Hence,  \:atomic\: number\: of\: the\: element\: is\: 4}

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poizon [28]

Answer:

9.57 mol.

Explanation:

<em>Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.</em>

<em />

<em>M = (no. of moles of solute)/(V of the solution (L)).</em>

<em></em>

∴ M = (no. of moles of sucrose)/(V of the solution (L)).

1.1 M = (no. of moles of sucrose)/(8.7 L).

<em>∴ no. of moles of sucrose = (1.1 M)(8.7 L) = 9.57 mol.</em>

5 0
3 years ago
The stability of an isotope is based on its
sattari [20]
Option 4. ratio of electrons to protons

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KIM [24]

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5 0
2 years ago
A voltaic cell utilizes the following reaction: 4Fe2+(aq)+O2(g)+4H+(aq)→4Fe3+(aq)+2H2O(l). What is the emf of this cell under st
Leviafan [203]

Answer:

0.46 V

Explanation:

The emf for the cell is given by:

Eº cell = Eº oxidation + Eº reduction

From the  given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced  from 0 to negative 2, according to the half cell reactions:

4Fe²⁺  ⇒ Fe³⁺ + 4e⁻           oxidation

O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O   reduction

From reference tables for the standard reduction potential, we get

Eº red Fe³⁺ / Fe²⁺   Eºred = 0.77 V  

Eº red O₂ / H₂O      Eºred = 1.23 V

Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction  O₂:

Eº cell =   Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V

5 0
3 years ago
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