How many electrons would a neutral atom of nitrogen need to gain in order to have a full valence electron shell? A. 3 B. 5 C. 7 D. More information is needed to figure this out.
2 months ago
Solution 1
2 months ago
Nitrogen need 3 electron to complete it valence
📚 Related Questions
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,
1.5 mole CO2 has 2*1.5 mole O=3.0 mole 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
Answer:
There is more space between gas particles than the size of the particles.
Explanation:
This scenario can be understand by taking a very simple example. As we know that 1 mole of any gas at standard temperature and pressure occupy 22.4 liters of volume. Lets take Hydrogen gas and Oxygen gas, 1 mole of each gas will occupy same volume. Why it is so? Why same volume although Oxygen is 16 times more heavier? This is because the space between gas molecules is very large. Approximately the distance between gas molecules is 300 times greater than their own diameter from its neighbor molecules.
There is more space between gas particles than the size of the particles.
Explanation:
This scenario can be understand by taking a very simple example. As we know that 1 mole of any gas at standard temperature and pressure occupy 22.4 liters of volume. Lets take Hydrogen gas and Oxygen gas, 1 mole of each gas will occupy same volume. Why it is so? Why same volume although Oxygen is 16 times more heavier? This is because the space between gas molecules is very large. Approximately the distance between gas molecules is 300 times greater than their own diameter from its neighbor molecules.
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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