The correct answer is C. Colligative properties only depend upon the number of solute particles in a solution but not on the identity or nature of the solute and solvent particles. I hope this anwers your question.
<h2>Question:- </h2>
A solution has a pH of 5.4, the determination of [H+].
<h2>Given :- </h2>
- pH:- 5.4
- pH = - log[H+]
<h2>To find :- concentration of H+</h2>
<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>
<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>
Take negative to other side
-pH = log H+
multiple Antilog on both side
(Antilog and log cancel each other )
Antilog (-pH) = [ H+ ]
New Formula :- Antilog (-pH) = [+H]
Now put the values of pH in new formula
Antilog (-5.4) = [+H]
we can write -5.4 as (-6+0.6) just to solve Antilog
Antilog ( -6+0.6 ) = [+H]
Antilog (-6) × Antilog (0.6) = [+H]

put the value in equation
![{10}^{ - 6} \times 4 = [H+] \\ 4 \times {10}^{ - 6} = [H+]](https://tex.z-dn.net/?f=%20%7B10%7D%5E%7B%20-%206%7D%20%20%20%5Ctimes%204%20%3D%20%5BH%2B%5D%20%5C%5C%204%20%5Ctimes%20%20%20%7B10%7D%5E%7B%20-%206%7D%20%20%3D%20%5BH%2B%5D)
Answer:
Basically, all phosphates except Sodium phosphates, Potassium phosphates and Ammonium phosphates are insoluble in water. That, of course, includes Magnesium phosphate.
Explanation:
Hope this helped!
Answer:
Rain shadow
Explanation:
Climate is the average weather condition of a place over a long period of time.
The image shows a rain shadow effect around a mountainous region. The type of rainfall controlled this way is known as an orographic rainfall.
- On the leeward side which is the opposite side of where the rain cloud forms, there is dryness and lack of rainfall.
- As moist air rises along the edge of the windward side, cloud forms.
- This windward side receives a significant amount of rainfall
- The windward side is hit with cold and dry wind and it is barren.
- 407.4 kJ of heat is released.
<u>Explanation:</u>
We have to write the balanced equation as,
2 C₂H₆(g) + 7O₂ → 4CO₂ + 6H₂O
Here 2 moles of ethane reacts in this reaction.
Now we have to find out the amount of ethane reacted using its given mass and molar mass as,
2 mol C₂H₆ × 30.07 g of C₂H₆ / 1 mol C₂H₆ = 60.14 g of C₂H₆
Heat released = ΔH × given mass / 60.14
= - 1560. 7 kj ×15.7 g / 60. 14 g = -407. 4 kJ