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Tems11 [23]
2 years ago
13

What is the mass of 5 mole of ammonia . Calculate the number of NH₃ molecules, nitrogen atom and hydrogen atoms in it..

Chemistry
1 answer:
Aloiza [94]2 years ago
3 0

Molar mass of NH_3

\\ \sf\longmapsto 14u+3(1u)

\\ \sf\longmapsto 14u+3u

\\ \sf\longmapsto 17g/mol

We know.

No of moles=Given mass/Molar mass

\\ \sf\longmapsto Given\;Mass=17(5)

\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g

Now

Lets write the balanced equation

\\ \sf\longmapsto N_2+3H_2=2NH_3

  • There is 2moles of Ammonia
  • 3moles of H_2
  • 1mole of N_2

Now

\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}

For Hydrogen

\\ \sf\longmapsto 3\times 6.023\times 10^{23}

\\ \sf\longmapsto 18.069\times 10^{23}

\\ \sf\longmapsto 1.8\times 10^{22}molecules

For Ammonia

\\ \sf\longmapsto 2\times 6.023\times 10^{23}

\\ \sf\longmapsto 12.046\times 10^{23}

\\ \sf\longmapsto 1.2\times 10^{22}molecules

For Nitrogen

\\ \sf\longmapsto 1\times 6.023\times 10^{23}

\\ \sf\longmapsto 6.023\times 10^{23}molecules

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Steam rises to the top and stays there
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What element are usually shiny , can be bent or stretched and conduct electricity
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gold and copper

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If a balloon containing 3000 L of gas at 39 C and 99 k Pa rises to an altitude where the pressure is 45.5 kPa and the temperatur
mihalych1998 [28]

Answer:

The new volume of the balloon will be 6046.28 L

Explanation:

Initial pressure (P1) = 99 kpa

initial volume (V1) = 3000 L

Initial temperature = 39 C = 39 + 273 = 312 K

Final pressure (P2) = 45.5 kpa

Final temperature = 16 C = 16 +273 = 289K

Final volume = ????

To calculate the final volume using the general gas equation

                  P1 V1 / T1  = P2 V2 / T2

make V2 the subject of the formular

                   V2 = 99000 ×3000× 289 / 45500×312

                   V2 = 85833000 /14196

                     V2 = 6046.28 litres

5 0
3 years ago
In the reaction a b c, doubling the concentration of a doubles the reaction rate and doubling the concentration of b does not af
PolarNik [594]
The reaction must be a + b --> c

Then you can predict a reaction rate, r o the type r = k * a^n * b^m

Given that the reaction rate is not affected by the concentration of b you can state that m = 0 and r = k * a^n.

Now given, that there is a proportional relation between the reaction rate and a (double a gives double rate), then n = 1 and r = k*a. You can verify that if you dobule a r also doubles.

Answer: r = k*a
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3 years ago
A 15.0-L rigid container was charged with 0.500 atm of kryp‑ ton gas and 1.50 atm of chlorine gas at 350.8C. The krypton and chl
Alecsey [184]

Answer: 32.94 g

Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

Kr+2Cl_2\rightarrow KrCl_4

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.

for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm

V = 15.0 L

T = 350.8 + 273 = 623.8 K

For krypton, n=\frac{0.500*15.0}{0.0821*623.8}

n = 0.146 moles

for chlorine, n=\frac{1.50*15.0}{0.0821*623.8}

n = 0.439

From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.

Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.

So, the amount of product formed is calculated from moles of krypton.

Molar mass of krypton tetrachloride is 225.61 gram per mol.

There is 1:1 mol ratio between krypton and krypton tetrachloride.

0.146molKr(\frac{1molKrCl_4}{molKr})(\frac{225.61gKrCl_4}{1molKrCl_4})

= 32.94 g of KrCl_4

So, 32.94 g of the product will form.

5 0
3 years ago
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