The quality of being easily dissolved in liquid.
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Answer:
Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.
Explanation:
Molarity of the solution is the moles of compound in 1 Liter solutions.
Mass of NaOH = x
Molar mass of NaOH = 40 g/mol
Volume of the NaOH solution = 450 mL =- 0.450 L ( 1 ml = 0.450 L)
Molarity of the solution of NaOH = 0.250 M
Solving for x:
x = 4.5 g
Weigh 4.5 grams of sodium hydroxide and add it to the dry volumetric flask of 450 mL followed by small amount of water to dissolve all the NaOH .After this add the water upto tye mark of 450 mL.
Answer:
See explanation
Explanation:
Crystals can be made from methanol by recrystallizing the plant extract from methanol.
The methanol/water system is heated rapidly using a hot plate and the plant extract dissolves in the heating solution until a clear solution is obtained.
The solution is now cooled rapidly. The interior of the flask used for the re crystallization may even by scratched to assist the quick formation of crystals. Large crystals of plant compounds may be obtained using this method. This process should be carried out in a fume hood because of the toxicity of methanol.
Conjugate base pairs are acid and bases having common features. These features are the equal gain or loss of protons of the pairs. Conjugate pairs should always be one base and one acid. One would not exist without the other. Conjugate acids are the substances that gains protons while conjugates bases are those that loses protons. <span>The substances in the equilibrium reaction that is given is identified as follows:
HCO3^- + H2O <-----> CO3^2- + H3O^+
acid base conjugate base conjugate acid
HCO3^- ion is an intermediate molecule of CO2 and CO3^2-. When we add OH- to HCO3^-, we produce CO3^2-. And when we add H+ to HCO3, we produce CO2. </span>
Given:
Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL
The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:
45ml * 0.10 M = C analyte * 20 ml
C analyte = 0.225 M
