Answer:
0.00316
Explanation:
You have to use the following equation:
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
You are given the pH and need to find the concentration of H+. Plug in the given components and solve.
![2.5=-log[H^+]\\H^+ = 10^{-2.5}\\H^+=0.00316](https://tex.z-dn.net/?f=2.5%3D-log%5BH%5E%2B%5D%5C%5CH%5E%2B%20%3D%2010%5E%7B-2.5%7D%5C%5CH%5E%2B%3D0.00316)
The concentration of H is 0.00316.
Answer:
101,37°C
Explanation:
Boiling point elevation is one of the colligative properties of matter. The formula is:
ΔT = kb×m <em>(1)</em>
Where:
ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-
kb is ebulloscopic constant (0,52°C/m)
And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):
203g × (1mol /62,07g) = <em>3,27moles of ethlyene glycol</em>
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Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m
Replacing these values in (1):
X - 100°C = 0,52°C/m×2,64m
X - 100°C = 1,37°C
<em>X = 101,37°C</em>
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I hope it helps!
Answer:
0.29mol/L or 0.29moldm⁻³
Explanation:
Given parameters:
Mass of MgSO₄ = 122g
Volume of solution = 3.5L
Molarity is simply the concentration of substances in a solution.
Molarity = number of moles/ Volume
>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.
Number of moles = mass/ molar mass
Molar mass of MgSO₄:
Atomic masses: Mg = 24g
S = 32g
O = 16g
Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol
= (24 + 32 + 64)g/mol
= 120g/mol
Number of moles = 122/120 = 1.02mol
>>>> From the given number of moles we can evaluate the Molarity using this equation:
Molarity = number of moles/ Volume
Molarity of MgSO₄ = 1.02mol/3.5L
= 0.29mol/L
IL = 1dm³
The Molarity of MgSO₄ = 0.29moldm⁻³
