Many, many substances ... when they are surrounded by air and get very
hot ... they combine with the oxygen in the air, forming new substances.
The process is called "burning", and the new substances are often called
"ashes".
If the tungsten filament [were] surrounded by air, then when you turned on
the light and the filament got hot, it would burn, turn to ash, and fall to the
bottom of the bulb in a little pile of dust. This would all happen so fast that
you would see a short, bright '<em>flash</em>', then the light would go out forever, and
you would say "OH ! The bulb burned out. We need to go to the store and
buy another one."
1) Reaction: 3Mg + N₂ → Mg₃N₂.
m(Mg) = 0,225 g
n(Mg) = 0,225 g ÷ 24,3 g/mol = 0,009 mol.
n(Mg) : n(N₂) = 3 : 1
n₁(N₂) = 0,003 mol.
n₂(N₂) = 0,5331 ÷ 28 = 0,019 mol.
n₃(N₂) = 0,019 mol - 0,003 mol = 0,016, m(N₂) = 0,016mol·28g/mol=0,4467g.
or simpler: m(N₂) = 0,225 g + 0,5331 - 0,3114 g = 0,4467 g.
2) Answer is: 6 <span>of fluorine atoms are combined with one uranium atom.
</span>m(U) = 209 g.
m(F) = 100 g.
n(U) = m(U) ÷ M(U)
n(U) = 209 g ÷ 238 g/mol.
n(U) = 0,878 mol.
n(F) = m(F) ÷ M(F)
n(F) = 5,263 mol
n(U) : n(F) = 0,878 mol : 5,263 mol /:0,878.
n(U) : n(F) = 1 : 6.
n - amount of substance
Answer:
It's balanced but not in the lowest form
Explanation:
The actual equation is given by
Through manipulation of equations, we are able to obtain the equation:
![-pOH= log [ OH^{-}]](https://tex.z-dn.net/?f=-pOH%3D%20log%20%5B%20OH%5E%7B-%7D%5D%20)
Then we can transform the equation into:
![[ OH^{-}]= 10^{-pOH}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D%2010%5E%7B-pOH%7D%20%20)
Then we are able to plug in the pOH and directly get [OH-]:
![[ OH^{-}] = 10^{-6.48}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%20%3D%2010%5E%7B-6.48%7D%20)
![[ OH^{-}]=3.31* 10^{-7} M](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D3.31%2A%2010%5E%7B-7%7D%20M%20%20)
Answer:
0.074
Explanation:
15% means that in 100 g of solution 15 g sodium hydroxide is present.
Mass of water = 100 - 15 = 85 g
Number of moles of sodium hydroxide:
Number of moles = 15 g/40 g/mol
Number of moles = 0.375 mol
Number of moles of water:
Number of moles = 85 g/18 g/mol
Number of moles = 4.7 mol
Moles fraction of NaOH:
moles of NaOH/ moles of solvent + moles of solute
0.375 mol/ 0.375 mol+4.7mol
0.375 mol / 5.075 mol
0.074
