Answer: 0.8M
Explanation:
Given that,
Amount of moles of NaCl (n) = ?
Mass of NaCl in grams = 1.40 g
For molar mass of NaCl, use the molar masses:
Sodium, Na = 23g;
Chlorine, Cl = 35.5g
NaCl = (23g + 35.5g)
= 58.5g/mol
Since, amount of moles = mass in grams / molar mass
n = 1.40g / 58.5g/mol
n = 0.024 mole
Now, given that:
Amount of moles of NaCl (n) = 0.024
Volume of NaCl solution (v) = 30.0mL
[Convert 30.0mL to liters
If 1000 mL = 1L
30.0mL = 30.0/1000 = 0.03L]
Concentration of NaCl solution (c) = ?
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
c = 0.024 mole / 0.03 L
c = 0.8 M (0.8M means concentration is in moles per litres)
Thus, the concentration of the solution is 0.8M
Answer:
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The state in which all of the external forces acting upon an object are balanced; there is no acceleration. friction ..... quadrupling. doubling distance and quadrupling mass has the overall effect of the force
Answer: d) -705.55 kJ
Explanation:
Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

Reversing the reaction, changes the sign of 


On multiplying the reaction by
, enthalpy gets half:


Thus the enthalpy change for the given reaction is -705.55kJ
Answer : The correct answer for a) 4-bromo-2-iodo-4-methyl pentane and b)5-bromo-2-ethoxy-2-methyl pentane.
A) Reaction with NaI :
Reaction of alkyl halide with NaI is known as Finkelstein Reaction . The acetone is used as solvent . It involves bimolecular nucleophillic substitution rmechanism (SN²) . There is replecement of one halogen with other occurs .
The incoming Nucleophile(Nu⁻) (halide) attacks on carbon from back side , while the leaving group (halide) leaves the compound from front side , simultaneously. The product so formed have is inverted .(Image)
NaI releases I⁻ ion which act as nucelophile and attacks on C1 carbon and Br⁻ from C1 carbon is released . Out of two bromines at C1 and C4 carbons , C1 is primary carbon which is less sterically hindered while C-4 is tertiary carbon and sterically hindered . So it is easy for incoming Nu⁻ to attack on C1 carbon .So Br⁻ is repleaced by I⁻.
1,4-dibromo-4-methylpentane + NaI → 4-bromo-1-iodo-4-methylpentane
The product formed from reaction between 1,4-dibromo-4-methylpentane and NaI is 4-bromo-1-iodo-4-methylpentane . (Image)
B) Reaction with AgNO3 :
Reaction of alkyl halide with AgNO3 in ethanol takes place via SN¹ ( unimolecular nucleophilic substitution ) mechanism . In this leaving group(halide) leaves from alkyl halide forming an intermediate carbocation species . The incoming Nu⁻ attack on this carbocation.
AgNO3 reacts releases Ag⁺ion which abstract Br⁻ of C-4 carbon from 1,4-dibromo-4-methylpentane. THis forms tertiary carbocation which is more stable than carbocation formed by removal of Br from C-1 . The ethanol being more Nucleophilic than NO₃⁻ (from AgNO₃), attacks on this carbocation .(Image )
The product formed as a result is 5-bromo-2-ethoxy-2-methyl pentane.