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Nitrogen need 3 electron to complete it valence

Question

What is the oxidation state of an individual phosphorus atom in P O 3 3−?

Solution 1

-3 - 3(-2) = +3

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Question

How many moles of oxygen atoms do 1.5 moles of co2 contain?

Solution 1

1 molecule CO2 has 2 atoms O.

1 mole CO2 has 2 moles O,

**1.5 mole** CO2 has 2*1.5 mole O=**3.0 mole O**

1 mole CO2 has 2 moles O,

Question

How many electrons in an atom can have a quantum number of n = 2?

Solution 1

It would be 18 electrons i think

Question

Suppose you wish to make 0.879 l of 0.250 m silver nitrate by diluting a stock solution of 0.675 m silver nitrate. how many milliliters of the stock solution would you need to use?

Solution 1

Answer is: 325.5 mL of the stock solution.

c₁(AgNO₃) = 0.675 M.

V₂(AgNO₃) = 0.879 L.

c₂(AgNO₃) = 0.250 M.

V₁(AgNO₃) = ?

c₁ - original concentration of the solution, before it gets diluted.

c₂ - final concentration of the solution, after dilution.

V₁ - volume to be diluted.

V₂ - final volume after dilution.

c₁ · V₁ = c₂ · V₂.

V₁(AgNO₃) = c₂ · V₂ ÷ c₁.

V₁(AgNO₃) = 0.250 M · 0.879 L ÷ 0.675 M.

V₁(AgNO₃) = 0.325 L · 1000 mL = 325 mL.

Question

How to make sulfur hexafluoride?

Solution 1

Sulfer and fluorine but you need special equipment to get it

Question

How many moles of water h2o are present in 75.0 g h2o?

Solution 1

4.17 moles. Good luck! :)

Question

When copper reacts with silver nitrate according to the equation, the number of grams of copper required to produce 432 grams of silver is -?

Solution 1

Answer is: mass of copper is 127 grams.

Balanced chemical reaction: Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s).

m(Ag) = 432 g.

n(Ag) = m(Ag) ÷ M(Ag).

n(Ag) = 432 g ÷ 108 g/mol.

n(Ag) = 4 mol.

From chemical reaction: n(Ag) : n(Cu) = 2 : 1.

n(Cu) = 4 mol ÷ 2 = 2 mol.

m(Cu) = n(Cu) · M(Cu).

m(Cu) = 2 mol · 63.5 g/mol.

m(Cu) = 127 g.

Balanced chemical reaction: Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s).

m(Ag) = 432 g.

n(Ag) = m(Ag) ÷ M(Ag).

n(Ag) = 432 g ÷ 108 g/mol.

n(Ag) = 4 mol.

From chemical reaction: n(Ag) : n(Cu) = 2 : 1.

n(Cu) = 4 mol ÷ 2 = 2 mol.

m(Cu) = n(Cu) · M(Cu).

m(Cu) = 2 mol · 63.5 g/mol.

m(Cu) = 127 g.

Question

How does the total volume of gas particles compare to the volume of the space between gas particles?

Solution 1

There is

This scenario can be understand by taking a very simple example. As we know that

Question

Gas stoichiometry: what volume of oxygen at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53 grams of propane?

Solution 1

Answer is: volume of oxygen is 14.7 liters.

Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

m(C₃H₈-propane) = 5.53 g.

n(C₃H₈) = m(C₃H₈) ÷ M(C₃H₈).

n(C₃H₈) = 5.53 g ÷ 44.1 g/mol.

n(C₃H₈) = 0.125 mol.

From chemical reaction: n(C₃H₈) : n(O₂) = 1 : 5.

n(O₂) = 0.625 mol.

T = 25° = 298.15K.

p = 1.04 atm.

R = 0.08206 L·atm/mol·K.

Ideal gas law: p·V = n·R·T.

V(O₂) = n·R·T / p.

V(O₂) = 0.625 mol · 0.08206 L·atm/mol·K · 298.15 K / 1.04 atm.

V(O₂) = 14.7 L.

Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

m(C₃H₈-propane) = 5.53 g.

n(C₃H₈) = m(C₃H₈) ÷ M(C₃H₈).

n(C₃H₈) = 5.53 g ÷ 44.1 g/mol.

n(C₃H₈) = 0.125 mol.

From chemical reaction: n(C₃H₈) : n(O₂) = 1 : 5.

n(O₂) = 0.625 mol.

T = 25° = 298.15K.

p = 1.04 atm.

R = 0.08206 L·atm/mol·K.

Ideal gas law: p·V = n·R·T.

V(O₂) = n·R·T / p.

V(O₂) = 0.625 mol · 0.08206 L·atm/mol·K · 298.15 K / 1.04 atm.

V(O₂) = 14.7 L.

Question

A gas occupies 2240.0 l at 373 k. what are the volumes at standard temperature answers

Solution 1

Answer is: the volumes at standard temperature is 1639.46 L.

V₁(gas) = 2240.0 L.

T₁(gas) = 373 K.

T₂(gas) = 273 K, standard temperature.

V₂(gas) = ?

Charles' Law: The Temperature-Volume Law - the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature:

V₁/T₁ = V₂/T₂.

2240 L/373 K = V₂/273 K.

V₂ = 1639.46 L.

V₁(gas) = 2240.0 L.

T₁(gas) = 373 K.

T₂(gas) = 273 K, standard temperature.

V₂(gas) = ?

Charles' Law: The Temperature-Volume Law - the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature:

V₁/T₁ = V₂/T₂.

2240 L/373 K = V₂/273 K.

V₂ = 1639.46 L.

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