The reason for this is because nonmetals, have close to fulfilling an octet and need to gain few more electrons to do this, not to lose more. Nonmetals, because of the fact they need only few more electrons to satisfy their octet they would receive or share electrons to do this.
The property that nonmetals have are that they are very electronegative, they possess a strong affinity to pull electron density closer, because they possess fewer electron shells and possess even protons this allows for this.
<u>Given:</u>
Mass of calcium nitrate (Ca(NO3)2) = 96.1 g
<u>To determine:</u>
Theoretical yield of calcium phosphate, Ca3(PO4)2
<u>Explanation:</u>
Balanced Chemical reaction-
3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2
Based on the reaction stoichiometry:
3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2
Now,
Given mass of Ca(NO3)2 = 96.1 g
Molar mass of Ca(NO3)2 = 164 g/mol
# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles
Therefore, # moles of Ca3(PO4)2 produced = 0.0589 * 1/3 = 0.0196 moles
Molar mass of Ca3(PO4)2 = 310 g/mol
Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g
Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g
Answer:
Enolate Alkylation
The anions from ketones, called enolates, can act as a nucleophile in SN2 type reactions. Overall an α hydrogen is replaced with an alkyl group and a new carbon-carbon bond is formed. These alkylations are affected by the same limitations as SN2 reactions previously discussed. A good leaving group, chloride, bromide, iodide or tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for preparing the enolate from the ketone. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylations occurring, and competing SN2 reactions with the base.
Explanation:
Design is illustrated in the attached document
Answer:
6.696 g/cm3
Explanation:
From the question;
Mass = 17.41g
Volume of water before = 46.3 cm3
Volume of water after = 48.9 cm3
Volume of antimony = Volume after - Volume before = 48.9 - 46.3 = 2.6 cm3
Density = Mass / Volume
Density = 17.41 / 2.6 = 6.696 g/cm3
