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2 years ago

What are the various ways in which an atom of an element can achieve the noble gas configuration ?

Chemistry

1 answer:

azamat2 years ago

7 0

Answer:

<h3><em>Atoms attain noble gas configuration </em><em> by obtaining or donating and sharing of electrons present in their outermost shell.</em><em> </em></h3>

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What did timbuktu contributed to mali's important as a kingdom

Alex73 [517]

Timbuktu was considered a MASSIVE trade center in Mali. That is how all of the foreign goods came into the kingdom. It also had education centers which was very important in Mali.

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3 years ago

Read 2 more answers

How many moles of S would I have if I had 11 grams? (Stoichiometry) HELP

zepelin [54]

<h3>Answer:</h3>

0.34 mol S

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

11 g S

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of S - 32.07 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 11 \ g \ S(\frac{1 \ mol \ S}{32.07 \ g \ S})$
Multiply/Divide: $\displaystyle 0.343 \ mol \ S$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

0.343 mol S ≈ 0.34 mol S

4 0

2 years ago

g modenr vacuum pumps make it easy to attain pressures of the order of in the laboratory. at a preasusure of 6.75 atm and an ord

Ratling [72]

Answer:

Number of molecules = 1.8267×10^20

Explanation:

From the question, we can deuced that the gases behave ideally, the we can make use of the ideal gas equation, which is expressed below;

PV = nRT

where

P =pressure

V =volume

n = the number of moles

R is the gas constant equal to 0.0821 L·atm/mol·K

T is the absolute temperature

Given:

P = 6.75 atm;

T = 290.0 k,

; V = 1.07 cm³ = 0.001 L

( 6.75 atm)(0.00107 L) = n(0.0821 L·atm/mol·K)(290K)

n = 3.0335167*10^-4 moles

But there are 6.022×10²³ molecules in 1 mole,

Number of molecules = 1.8267×10^20

7 0

3 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

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3 years ago

What are the moles of silver metal produced from 0.0999 mol of copper?

julia-pushkina [17]

Answer:

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2 years ago

What are the various ways in which an atom of an element can achieve the noble gas configuration ? ​

What are the various ways in which an atom of an element can achieve the noble gas configuration ?