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Elis [28]
2 years ago
11

A 2.0 mol sample of ammonia is introduced into a 1.00 L container. At a certain temperature, the ammonia partially dissociates a

ccording to the equation: NH_3(g) N_2(g) + H_2(g) At equilibrium 1.00 mol of ammonia remains. Calculate the value for K.
Chemistry
1 answer:
aleksley [76]2 years ago
8 0

Answer:

K = 1.69M

Explanation:

For the reaction:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

K is defined as:

k = [N₂] [H₂]³ / [NH₃]² <em>(1)</em>

Molarity in equilibrium for each specie is:

NH₃(g): 2.0mol - 2x = 1.0mol/1L = 1M → X = 0.5mol

N₂(g): X = 0.5mol/1L = 0.5M

H₂(g): 3X = 1.5mol/1L = 1.5M

Replacing:

k = [0.5] [1.5]³ / [1]²

<em>k = 1.69M</em>

<em></em>

I hope it helps!

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(5 points)
ExtremeBDS [4]

<u>Answer 2 :</u> The given electronic configuration for a neutral atom of phosphorous in its ground state is incorrect.

Explanation :

A neutral atom of phosphorous has 15 electrons.

The given electronic configuration is incorrect.

The reason is, According to Aufbau principle, the electrons will be first filled in the sub-shell having lower orbital energy. As from the given configuration, 3p sub-shell has lower orbital energy than 4s sub-shell. So, the electrons will be filled in 3p sub-shell first. Hence, the ground state electronic configuration of neutral atom of phosphorous is,

1s^22s^22p^63s^23p^3

<u>Answer 3 :</u>

Element                     Rubidium              Magnesium                Aluminium

Symbol                             Rb                         Mg                              Al

Group number                  1                             2                               13

Number of valence          1                             2                                3

electrons

The order of general reactivity on the basis of number of valence electrons.

Rb > Mg > Al

Reason : The reactivity is determined by the number of electrons present in the outermost shell that means the element which have 1 valence electron will be more reactive because they can easily lose electrons.

5 0
2 years ago
If a gas in a box with fixed volume has a pressure of 913 torr at 15 C, what temperature should the gas be heated to double the
algol13
<span>700 mL hop it helps u ok</span>
4 0
3 years ago
Please help
Mashutka [201]

Answer:

Potential

Explanation:

The answer is potential!

6 0
1 year ago
Read 2 more answers
For each of the following unbalanced chemical equations suppose that exactly 50.0 g of each reactant is taken. Determine which r
Helen [10]

Answer:

1) Br2 is the limiting reactant.

Mass NaBr produced = 64.4 grams

2) CuSO4 is the limiting reactant

Mass Cu = 19.89 grams

Mass ZnSO4 = 50.54 grams

3) NH4Cl is the limiting reactant

Mass NaCl = 54.6 grams

Mass NH3 =15.9 grams

Mass H2O =16.8 grams

4) Fe2O3 is the limiting reactant

Mass Fe = 35.0 grams

Mass CO2 = 41.3 grams

Explanation:

1) Na+br2 ------------->Nabr

Step 1: Data given

Mass Na = 50.0 grams

Mass Br2 = 50.0 grams

Molar mass Na = 22.99 g/mol

Molar mass Br2 = 159.81 g/mol

Step 2: The balanced equation

2Na + Br2 → 2NaBr

Step 3: Calculate moles

Moles = mass / molar mass

Moles Na = 50.0 grams / 22.99 g/mol = 2.17 moles

Moles Br2 = 50.0 grams / 159.81 g/mol = 0.313 moles

Step 4: Calculate limiting reactant

Br2 is the limiting reactant. It will completely be consumed (0.313 moles).

Na is in excess. There will react 2*0.313 = 0.626 moles

There will remain 2.17 - 0.626 = 1.544 moles

Step 5: Calculate moles NaBr

For 1 mol Br2 we'll have 2 moles NaBr

For 0.313 moles we'll have 0.626 moles NaBr

Step 6: Calculate mass NaBr

Mass NaBr = 0.626 moles * 102.89 g/mol

Mass NaBr = 64.4 grams

2) Zn+cuso4 -------------->Znso4+Cu

Step 1: Data given

Mass Zn = 50.0 grams

Mass CuSO4 = 50.0 grams

Molar mass Zn = 65.38 g/mol

Molar mass CuSO4 = 159.61 g/mol

Step 2: The balanced equation

Zn + CuSO4 → Cu + ZnSO4

Step 3: Calculate moles

Moles = mass / molar mass

Moles Zn = 50.0 grams / 65.38 g/mol = 0.765 moles

Moles CuSO4 = 50.0 grams / 159.61 g/mol = 0.313 moles

Step 4: Calculate limiting reactant

CuSO4 is the limiting reactant. It will completely be consumed (0.313 moles).

Zn is in excess. There will react 0.313 moles

There will remain 0.765 - 0.313 = 0.452 moles

Step 5: Calculate moles products

For 1 mol Zn we need 1 mol CuSO4 to produce 1 mol Cu and 1 mol ZnSO4

For 0.313 moles CuSO4 we'll have 0.313 moles Cu and 0.313 moles ZnSO4

Step 6: Calculate mass products

Mass Cu = 0.313 moles * 63.546 g/mol = 19.89 grams

Mass ZnSO4 = 0.313 moles * 161.47 g/mol  = 50.54 grams

3) NH4cl+NaOH -------------->NH3+H2O+NaCl

Step 1: Data given

Mass NH4Cl = 50.0 grams

Mass NaOH = 50.0 grams

Molar mass NH4Cl = 53.49 g/mol

Molar mass NaOH = 40.0 g/mol

Step 2: The balanced equation

NH4Cl + NaOH → NaCl + NH3 + H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles NH4Cl = 50.0 grams / 53.49 g/mol = 0.935 moles

Moles NaOH = 50.0 grams / 40.0 g/mol = 1.25 moles

Step 4: Calculate limiting reactant

NH4Cl is the limiting reactant. It will completely be consumed (0.935 moles).

NaOH is in excess. There will react 0.935 moles

There will remain 1.25 - 0.935 = 0.315 moles

Step 5: Calculate moles products

For 1 mol NH4Cl we need 1 mol NaOH to produce 1 mol NaCl, 1 mol NH3 and 1 mol H2O

For 0.935 moles NH4Cl we'll have 0.935 moles NaCl, 0.935 moles NH3 and 0.935 moles H2O

Step 6: Calculate mass products

Mass NaCl = 0.935 moles * 58.44 g/mol = 54.6 grams

Mass NH3 = 0.935 moles * 17.03 g/mol  = 15.9 grams

Mass H2O = 0.935 moles * 18.02 g/mol = 16.8 grams

4) Fe2O3+CO ------------>Fe+CO2

Step 1: Data given

Mass Fe2O3 = 50.0 grams

Mass CO = 50.0 grams

Molar mass Fe2O3 = 159.69 g/mol

Molar mass CO = 28.01 g/mol

Step 2: The balanced equation

Fe2O3 + 3CO → 2Fe + 3CO2

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe2O3 = 50.0 grams / 159.69 g/mol = 0.313 moles

Moles CO = 50.0 grams / 28.01 g/mol = 1.785 moles

Step 4: Calculate limiting reactant

Fe2O3 is the limiting reactant. It will completely be consumed (0.313 moles).

CO is in excess. There will react 3* 0.313 = 0.939 moles

There will remain 1.785 - 0.939 = 0.846 moles

Step 5: Calculate moles products

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe, 3 moles CO2

For 0.313 moles Fe2O3 we'll have 0.626 moles Fe and 0.939 moles CO2

Step 6: Calculate mass products

Mass Fe = 0.626 moles * 55.845 g/mol = 35.0 grams

Mass CO2 = 0.939 moles * 44.01 g/mol  = 41.3 grams

8 0
3 years ago
4)a liquid is poured into a beaker in question 3 and the mass is 146g what is the mass of the liquid in kilograms
GaryK [48]
1 kg ------- 1000 g
? ----------- 146 g

146 x 1 / 1000

= 0.146 kg
5 0
2 years ago
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