Answer:
2
b= they are grouped differently, but all the atoms are still there.
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)
Al
Explanation:
The limiting reactant will be Al:
4Al + 3O₂ → 2Al₂O₃
The limiting reactant is the reactant in short supply in a chemical reaction.
Given parameters:
Mass of Al = 30g Molar mass = 27g/mol
Number of moles =
= 
Number of moles of Al = 1.111 mole
Mass of O₂ = 30g, molar mass = 32g/mol
Number of moles =
= 0.94mol
In the reaction:
4 moles of Al reacted with 3 moles of O₂
1.11moles of Al will require
= 0.83mole to react
But we have been given 0.94mole of O₂. This is more than required.
Therefore O₂ is in excess and Al is the limiting reactant.
Learn more:
Limiting reagents brainly.com/question/6078553
#learnwithBrainly
Answer:
Q = 90,000 J
Explanation:
Given data:
Mass skillet = 2000 g
Specific heat capacity = 0.450 J/g.°C
Energy required to raise temperature = ?
Initial temperature = 25°C
Final temperature = 125°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 125°C - 25°C
ΔT = 100°C
Q = 2000 g × 0.450 J/g.°C × 100°C
Q = 90,000 J
Answer:
16.0 g; 3.1 mol
Explanation:
(a) Mass of O atoms
Mass = 6.022 × 10^23 atoms × (2.66 × 10^-23 g/1 atom) = 16.0 g
(b) Moles of O atoms
0.050 kg = 50 g
Moles = 50 g × (1 mol/16.0 g) = 3.1 mol