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Crank
3 years ago
13

How many molecules are in 0.400 moles of. N2O5?

Chemistry
2 answers:
mariarad [96]3 years ago
5 0

Answer:

               2.40 × 10²³ Molecules of N₂O₅

Explanation:

As we know 1 mole of any substance contains 6.022 × 10²³ particles (also called as Avogadro's Number). Therefore, the relation between moles and number of particles can be written as,

             Moles  =  Number of Particles ÷ 6.022 ×10²³ Particles.mol⁻¹

For the molecules of Dinitrogen pentoxide (N₂O₅) it can be written as,

             Moles  =  Number of Molecules ÷ 6.022 ×10²³ Molecules.mol⁻¹

Rearranging for Number of Molecules,

             Number of Molecules  =  Moles × 6.022 ×10²³ Molecules.mol⁻¹

Putting value of Moles,

             Number of Molecules  =  0.400 mol × 6.022 ×10²³ Molecules.mol⁻¹

             Number of Molecules  =  2.40 × 10²³ Molecules of N₂O₅

SpyIntel [72]3 years ago
4 0
Because you're turning moles into molecules, all you need to use is Avogadro's Number (6.022*10^23).
The equation would be: .400 mol N2O5/1 mol*6.022*10^23 molecules. 
The answer, including significant figures, is 151 molecules. 
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What is molarity of 47.0 g KCl dissolved in enough water to give 375 mL of solution?
natita [175]

This question provides us –

  • Weight of \bf  KCl is = 47 g
  • Volume, V = 375 mL

__________________________________________

  • Molar Mass of \bf   KCl –

\qquad \twoheadrightarrow\bf  39.0983 \times 35.453

\qquad \twoheadrightarrow\bf 74.5513

<u>Using formula</u> –

\qquad \purple{\twoheadrightarrow\bf Molarity _{(Solution)} =  \dfrac{ W\times 1000}{MV}}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \dfrac{ 47 \times 1000}{74.5513\times 375}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \dfrac{47000}{27956.7375}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = \cancel{\dfrac{47000}{27956.7375}}

\qquad \twoheadrightarrow\bf Molarity _{(Solution)}  = 1.68117M

\qquad \pink{\twoheadrightarrow\bf Molarity _{(Solution)}  = 1.7M}

  • Henceforth, Molarity of the solution is = 1.7M

___________________________________________

6 0
3 years ago
The amount of moisture in the air is called
Irina18 [472]
Humidity/ Relative Humidity
5 0
3 years ago
The half-life of francium-212 is 19 minutes. How many minutes will it take for 100 gram
vagabundo [1.1K]

Answer:

57 mins

Explanation:

From the question given above, the following data were obtained:

Half life (t₁/₂) = 19 mins

Original amount (N₀) = 100 g

Amount remaining (N) = 12.5 g

Time (t) =?

Next, we shall determine the number of half-lives (n) that wound have elapse for 12.5 g of the sample to remain. This can be obtained as follow:

Original amount (N₀) = 100 g

Amount remaining (N) = 12.5 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

12.5 = 1/2ⁿ × 100

Cross multiply

12.5 × 2ⁿ = 100

Divide both side by 12.5

2ⁿ = 100 /12.5

2ⁿ = 8

Express 8 index with 2 as the base

2ⁿ = 2³

n = 3

Finally, we shall determine the time taken for 100 g of the isotope to decay to 12.5 g. This can be obtained as follow:

Half life (t₁/₂) = 19 mins

Number of half-lives (n) = 3

Time (t) =?

t = n × t₁/₂

t = 3 × 19

t = 57 mins

Thus, it will take 57 mins 100 g of the isotope to decay to 12.5 g.

7 0
3 years ago
At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration
wlad13 [49]

Answer:

  • [HOCl] = 0.00909 mol/liter
  • [H₂O] = 0.03901 mol/liter
  • [Cl₂O] = 0.02351 mol/liter

Explanation:

<u />

<u>1. Chemical reaction:</u>

H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)

<u>2. Initial concentrations:</u>

i) 1.3 g H₂O

  • Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol
  • Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

  • Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol
  • Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

<u>3. ICE (Initial, Change, Equilibrium) table</u>

            H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)

I            0.0481      0.0326            0

C              -x                 -x              +x

E          0.0481-x    0.0326-x         x

<u />

<u>4. Equilibrium expression</u>

       K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}

     0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}

<u />

<u>5. Solve:</u>

            x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0

Use the quadatic formula:

x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}

The positive result is x = 0.00909

Thus the concentrations are:

  • [HOCl] = 0.00909 mol/liter
  • [H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter
  • [Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

3 0
3 years ago
When the number of protons and electrons are equal
Tomtit [17]

Answer:

when the atom is neutral

Explanation:

8 0
3 years ago
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