POH = - log [ OH⁻ ]
pOH = - log [ 1 x 10⁻⁹ ]
pOH = 9
Answer C
hope this helps!
The answer to this question would be C
Answer:
<u><em>Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.</em></u>
A chemistry student weighs out 0.0941 g of hypochlorous acid (HClo) into a 250. ml. volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits mL.
Explanation:
1 mole HClO = 74.44g
0.0941g =
= 0.00126 moles
Concentration = no. of moles/volume in L
Hence, Concentration of HClO = 0.00126/ 0.250L
= 0.005M.
C1V1 =C2V2
0.005 × 250 mL = 0.2 × V2
<u><em>Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.</em></u>
How are the conditions at which phases are in equilibrium represented on a phase diagram?
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Along the line between liquid and solid, the melting temperatures for different pressures can be found. The junction of the three curves, called the triple point, represents the unique conditions under which all three phases exist in equilibrium together. Phase diagrams are specific for each substance and mixture.
We first assume that this gas is an ideal gas where it follows the ideal gas equation. The said equation is expressed as: PV = nRT. From this equation, we can predict the changes in the pressure, volume and temperature. If the volume and the temperature of this gas is doubled, then the pressure still stays the same.