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Dima020 [189]
3 years ago
13

How will you solve this?

Chemistry
1 answer:
11111nata11111 [884]3 years ago
8 0
Rock a is a sedimentary and rock B is a sediment rock also known as sandstone
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What is the result of a neutralization reaction between nitric acid (HNO3) and potassium hydroxide (KOH)?
scZoUnD [109]
HNO3+KOH = H2O+KNO3 . When nitric acid react with pottasuim hydroxide, the reaction will produce water (H20) and pottasuim trioxonitrate
3 0
3 years ago
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What would be the major product if 1,4-dibromo-4-methylpentane was allowed to react with:
Levart [38]

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.

7 0
3 years ago
How is it that when a salt sample dissolves in water, the delta S for the process is positive?
Bess [88]

Answer:-

Water is highly ordered. In water each oxygen atom is connected to others around it through hydrogen bonding via bridging hydrogen atoms. When a salt like NaCl is dissolved, some of these Hydrogen bonds break.

When a salt like NaCl dissolves in water, the NaCl breaks in to ions Na+ and Cl-.

The water molecules now surround these ions.

The slightly negative oxygen end of water molecule gets near the Na+, while the slightly positive Hydrogen of water molecule gets near the Cl-.

So before salt sample dissolve, the water molecules were highly ordered due to hydrogen bonding. Now after salt dissolve there is a decrease in order and thus an increase in disorder of the water molecules.

Due to increase in disorder, entropy which is a measure of disorder increases. Since entropy increases, delta S for the process is positive.

4 0
3 years ago
Just as one dozen eggs always has 12 eggs in it, one mole of a
zzz [600]

Answer:

6.022x10^{23}atoms \ Al

Explanation:

Hello,

In this case, given the described concept regarding the Avogadro's number, we can easily notice that 27.0 g of aluminium foil has 6.022x10²³ atoms as shown below based on the mass-mole-particles relationship:

27.0gAl*\frac{1molAl}{27.0gAl} *\frac{6.022x10^{23}atoms \ Al}{1molAl} \\\\=6.022x10^{23}atoms \ Al

Notice this is backed up by the fact that aluminium molar mass if 27.0 g/mol.

Best regards.

8 0
3 years ago
A company is designing a new range of saucepans. The base of the saucepan should
earnstyle [38]
D)plastic copper
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