The percentage composition of Nitrogen in the compound is 35.71%
<h3>How to determine the mole of nitrogen</h3>
- Volume (V) = 100 mL = 100 / 1000 = 0.1 L
- Pressure (P) = 700 mmHg
- Temperature (T) = 250 K
- Gas constant (R) = 62.364 torr.L/Kmol
- Number of mole (n) =?
Using the ideal gas equation, the mole of nitrogen can be obtained as follow
PV = nRT
n = PV / RT
n = (700 × 0.1) / ( 62.364 × 250)
n = 0.00448 mole
<h3>How to determine the mass of nitrogen</h3>
- Mole of nitrogen = 0.00448 mole
- Molar mass of nitrogen = 28 g/mol
- Mass of nitrogen = ?
Mass = mole × molar mass
Mass of nitrogen = 0.00448 × 28
Mass of nitrogen = 0.125 g
<h3>How to determine the percentage of nitrogen</h3>
- Mass of nitrogen = 0.125 g
- Mass of compound = 0.35 g
- Percentage of nitrogen =?
Percentage = (mass of element / mass of compound) × 100
Percentage of nitrogen = (0.125 / 0.35) × 100
Percentage of nitrogen = 35.71%
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Answer:
Like most other metals, Gallium is solid at room temperature (or liquid if it is too hot in your room). But, if it is held [in hands] for long enough, it melts in your hands, and doesn't poison you like Mercury would. This is because of its unusually low melting point of (~29 degree Centigrade).
- It melts once it reaches its melting point.
:)
Answer:
8
Explanation:
The duckweed is highest at pH of 8.
- Hope that helped! Please let me know if you need further explanation.
An atom of vanadium
(V) has 23 electrons.
Given :
A vanadium (V) atom with 23 protons and has a net charge of 0.
To find:
The number of electrons in a vanadium atom
Solution:
Number of protons in vanadium atom = 23
The 0 net charge on the atom indicates that the atom has an equal number of protons and electrons which makes it a neutral atom.
Number of electrons in vanadium atom = Number of protons in vanadium atom = 23
An atom of vanadium
(V) has 23 electrons.
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Answer is: XeF₄.
p₁ (F₂) = 8,0 atm.
p₁ (Xe) = 1,7 atm.
p₂ (F₂) = 4,6 atm.
p₂ (Xe) = 0 atm, all reacted.
p - partial pressure.
pressure of F₂ reacted = Δp (F₂) = 8.0 - 4.6 = 3.4 atm.
pressure of Xe reacted = Δp(Xe) = 1,7 - 0 = 1,7 atm.
Δp (F₂) / Δp(Xe) = 3,4 atm / 1,7 atm = 2:1.
Reaction: 2F₂ + Xe → XeF₄ (xenon-tetrafluoride)
