H₂ + ½O₂ → H₂O
2g → 18g
So, 18g water needs 2g H₂
So, 1g water needs 2/18g H₂
So, 180g water needs 2/18 x 180g H₂
→ 20 litres of H₂
Answer: 0.225 atm
Explanation:
For this problem, we have to use Boyle's Law.
Boyle's Law: P₁V₁=P₂V₂
Since we are asked to find P₂, let's manipulate the equation.
P₂=(P₁V₁)/V₂
With this equation, the liters cancel out and we will be left with atm.
P₂=0.225 atm
Answer:
Highest pH(most basic)
Sr(OH)2(aq)
KOH (aq)
NH3(aq)
HF (aq)
HClO4(aq)
Lowest pH(most acidic)
Explanation:
The concentration of H+ ion will determine the pH of a solution. The pH actually reflects the ratio of H+ ion and OH- since both of them can combine into water. Solution with more H+ ion will have a lower pH and called acidic, while more OH- will have high pH and be called basic. Strong acid/base will be ionized more than weak acid/base.
Sr(OH)2(aq) = strong base, release 2 OH- ion per mole
KOH (aq) = Strong base, release 1 OH- per mole
NH3(aq) = weak base, release less than 1 OH- per mole
HF (aq) =strong acid, release 1 H+ per mole
HClO4(aq) = stronger acid, release 1 H+ per mole
Answer:
B.Bronze is harder than copper, so it retains a cutting edge sharper than that of copper.
Explanation:
The importance of bronze in the history of the human kind can not be only measured in the durability but the help that it provided to our ancestors, the things they were allowed to do once they dicovered how to mold it, and the tools that they produced and made many every day tasks easier for them and the ability of bronze to mantain the cutting edge longer than copper was one of the main reasons why it took its places in the fabrication of many tools.
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87
