<h2>DIFFERENCE BETWEEN CRYSTALLINE AND AMORPHOUS SOLIDS :</h2><h2><em><u> Amorphous solids do not have definite melting points but melt over a wide range of temperature because of the irregular shape. Crystalline solids, on the other hand, have a sharp melting point.</u></em></h2>
<span>PV=nRT Since P, V and R are constant, when T goes up, n must go down by the same factor </span>1:4
Explanation:
1.
Sodium, or Na, is 57.47% of the composition,
Hydrogen, or H, is 2.520% of the composition,
and Oxygen, or O, is 40.001% of the composition.
This is because mass% = mass/total mass x 100%.
2. For every 1 mole of C6H12O6, you need 6 moles of water. Multiply the 5.2 moles you are trying to make by the 6 moles of water you need, and you will need 31.2 moles.
3. x = 7.2 x 4 / 2 = 14.4 mol
Answer:
Neopentane (C(CH3)4) have a lower a lower boiling point than n-pentane (CH3CH2CH2CH2CH3) because of it branched chain.
Explanation:
Structural isomers are compounds that have the same molecular or chemical formula but have different arrangement of atoms in space. The connectivity of the atoms differs for this compounds but they possess the same molecular formula. An example of structural isomers are butane and methyl propane. Usually, as the number of carbon increases for alkane the number of isomers also increases.
Since Isomers are different compound they are bound to have different characteristic both in melting and boiling points. Generally, straight chained isomers have higher boiling points than branched chain isomers.
n-pentane (CH3CH2CH2CH2CH3) is a straight chained isomer while neopentane (C(CH3)4) is a branched chain isomer. Neopentane (C(CH3)4) have a lower a lower boiling point than n-pentane (CH3CH2CH2CH2CH3) because of it branched chain.
Answer: -
12.59
Explanation: -
Strength of NaOH = 0.0179 M
Volume of NaOH = 58.0 mL = 58.0/1000 = 0.058 L
Number of moles = 0.0179 M x 0.058 L
= 1.04 x 10⁻³ mol
Mol of [OH⁻] given by NaOH = 1.04 x 10⁻³ mol
Strength of Ba(OH)₂ = 0.0294 M
Volume of Ba(OH)₂ = 60.0 mL = 60.0/1000 = 0.060 L
Number of moles = 0.0294 M x 0.060 L
= 1.76 x 10⁻³ mol
Mol of [OH⁻] given by Ba(OH)₂ =2 x 1.76 x 10⁻³ mol
Total [OH⁻] = 1.04 x 10⁻³ mol + 2 x 1.76 x 10⁻³ mol
= 4.56 x 10⁻³ mol
Total volume of the mixture = 58.0 + 60.0
= 118.0 mL
118.0 mL of the solution has 4.56 x 10⁻³ mol [OH⁻]
1000 mL of the solution has 
= 0.0386 mol
Using the relation
pOH = - log [OH-]
= - log 0.0386
= 1.41
Using the relation
pH + pOH = 14
pH = 14 - 1.41
= 12.59
