<u>Answer:</u>
<em>To raise the pH of the solution to 3.10 we have to add 2.34 L of water.</em>
<u>Explanation:</u>
<em>Given that the pH of the solution of HCl in water is 2.5.</em> Here the solution’s pH is changing from 2.5 to 3.10 which means the acidic nature of the solution is decreasing here on dilution. ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
ions contribute to a solution’s acidic nature and ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
contribute to a solution’s basic nature.
The equation connecting the concentration of and pH of a solution is pH= ![-log[H^+]](https://tex.z-dn.net/?f=-log%5BH%5E%2B%5D)
<em>![[H^+]= 10^(^-^p^H^)](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%2010%5E%28%5E-%5Ep%5EH%5E%29)
![[H^+]= 10^(^-^2^.^5^)=0.00316](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%2010%5E%28%5E-%5E2%5E.%5E5%5E%29%3D0.00316)
</em>
<em>When the pH is ![3.1 [H^+ ]= 10^(^-^3^.^1^)=0.000794](https://tex.z-dn.net/?f=3.1%20%5BH%5E%2B%20%5D%3D%2010%5E%28%5E-%5E3%5E.%5E1%5E%29%3D0.000794)
</em>
<em>On dilution the concentration of a solution decreases and volume increases.</em>
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<em>Volume of water to be added 
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Answer:
can u give us the options
I think 43.12 I’m not that good with math
Answer:
Change in internal energy (ΔU) = -9 KJ
Explanation:
Given:
q = –8 kJ [Heat removed]
w = –1 kJ [Work done]
Find:
Change in internal energy (ΔU)
Computation:
Change in internal energy (ΔU) = q + w
Change in internal energy (ΔU) = -8 KJ + (-1 KJ)
Change in internal energy (ΔU) = -8 KJ - 1 KJ
Change in internal energy (ΔU) = -9 KJ
Answer:
d) 8.01 E23 atoms
Explanation:
∴ mass C = 16 g
∴ molar mass C = 12.0107 g/mol
⇒ mol C = (16 g)*(mol/12.0107 g) = 1.332 mol
⇒ atoms C = (1.332 mol)*(6.022 E23 atoms/mol)
⇒ atoms C = 8.02 E23 atoms
