Classical bonding in NaH2PO4
1 answer:
, a crystal structure with a short symmetrical hydrogen bond.
Classical models of the chemical bond. By classical, we mean models that do not take into account the quantum behaviour of small particles, notably the electron. These models generally assume that electrons and ions behave as point charges which attract and repel according to the laws of electrostatics.
Sodium dihydrogen phosphate is a derivative composed of glycerol derivatives formed by reacting mono and diglycerides that are derived from edible sources with phosphorus pentoxide followed by neutralization with sodium carbonate.
Bonding in
, a crystal structure with a short symmetrical hydrogen bond. Sodium dihydrogen phosphate (
) is monoclinic, space group P2,/c, with a= 6.808 (2), b= 13.491 (3), c=7.331 (2)/~, fl=92.88 (3) ; Z=8.
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Answer:
See below.
Step-by-step explanation:
Ethers react with HI at high temperature to produce an alky halide and an alcohol.
R-OR' + HI ⟶ R-I + H-OR'
<em>Benzylic ethers</em> react by an Sₙ1 mechanism by forming the stable benzyl cation.
- PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻ Protonation of the ether
- PhCH₂-O⁺(H)R ⟶ PhCH₂⁺ + HOR Sₙ1 ionization of oxonium ion
- PhCH₂⁺ + I⁻ ⟶ PhCH₂-I Nucleophilic attack by I⁻
If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:
ROH +HI ⟶ R-I + H-OH
Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).
The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).
Answer:
28g remain after 13.5 hours
Explanation:
Element decayment follows first order kinetics law:
ln[Pa-234] = -kt + ln [Pa-234]₀ <em>(1)</em>
<em>Where [Pa-234] is concentration after t time, k is rate constant in time, and [Pa-234]₀ is initial concentration</em>
Half-life formula is:
6.75 = ln2 / k
<em>k = 0.1027hours⁻¹</em>
Using rate constant in (1):
ln[Pa-234] = -0.1027hours⁻¹×13.5hours + ln [112.0g]
ln[Pa-234] = 3.332
[Pa-234] = <em>28g after 13.5 hours</em>
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