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: (D) Are there solvents mixed in (or is it water based)?
Answer: C) Subatomic particles
Explanation: I just learned this in ap environmental science
![\LARGE{ \boxed{ \purple{ \rm{Answer}}}}](https://tex.z-dn.net/?f=%20%5CLARGE%7B%20%5Cboxed%7B%20%5Cpurple%7B%20%5Crm%7BAnswer%7D%7D%7D%7D)
☃️ Chemical formulae ➝ ![\sf{I_2}](https://tex.z-dn.net/?f=%5Csf%7BI_2%7D)
How to find?
For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.
![\boxed{ \sf{No. \: of \: moles = \frac{Given \: weight}{Molecular \: weight} }}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Csf%7BNo.%20%5C%3A%20of%20%5C%3A%20moles%20%3D%20%20%5Cfrac%7BGiven%20%5C%3A%20weight%7D%7BMolecular%20%5C%3A%20weight%7D%20%7D%7D)
Solution:
❍ Molecular weight of ![\sf{I_2}](https://tex.z-dn.net/?f=%5Csf%7BI_2%7D)
= 2 × 126.90
= 253.80
= 254 (approx.)
❍ Given weight: 12.7
Then, no. of moles,
⇛ No. of moles = 12.7 / 254
⇛ No. of moles = 0.05 moles
⚘ No. of moles of Iodine molecule in the given weight = <u>0.05</u><u> </u><u>moles </u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
Answer:
The temperature of the system once the equilibrium has been reached = 372.55K
Explanation:
Heat capacity of gold = 129 J/Kg*c.
Heat capacity of water
4,184 J/Kg*c.
Mass of gold = 75g = 0.075Kg
Mass of water = 200g = 0.2Kg
From conservation of energy
m1×C1×(t11 - t2) = m2×C2×(t2- t21)
Substituting we have
0.075 × 129×(1000-t2) = 0.2× 4184×( t2 -300) =solving for t2, we have
933.55×t2 = 347790
or t2 = 372.55K
The temperature of the system once the equilibrium has been reached = 372.55K