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3 years ago

How are carbohydrates and lipids similar? how are they different?

1 answer:

6 0

"Lipids are like carbohydrates in way that the true fats contain only carbon, hydrogen, and oxygen. Both carbohydrates and lipids act as the main fuels and energy storage compounds of the human body. They are also called SACCHARIDES and grouped as: Monosaccharides, Disaccharides, Trisaccharides, Polysaccharides."

Source credit: https://www.quora.com/What-are-the-differences-and-similarities-between-carbohydrates-and-lipids

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Now suppose, instead, that 5.678 g of a volatile solute is dissolved in 150.0 g of water. This solute also does not react with w

n200080 [17]

Answer:

59.9 g/mol is the molar mass for the solute

Explanation:

Lowering vapor pressure → ΔP = P° . Xm

P° → Vapor pressure of pure solvent

ΔP = P° - Vapor pressure of solution

Xm = Mole fraction of solute

17.54 Torr - 17.344 Torr = 17.54 Torr . Xm

0.196 Torr / 17.54 Torr = Xm → 0.0112

These are the moles of solute / Total moles

Total moles = Moles of solute + Moles of solvent

We determine the moles of solvent → 150 g . 1mol/ 18 g = 8.33 moles

Now we can make this equation:

0.0112 = Moles of solute / Moles of solute + 8.33 mol

0.0112 Moles of solute + 0.0933 = Moles of solute

0.0933 = Moles of solute - 0.0112 Moles of solute

0.0933 = 0.9888 moles of solute → 0.0933 / 0.9888 = 0.0947 moles

Finally we can determine the molar mass (mol/g)

5.678 g / 0.0947 mol = 59.9 g/mol

4 0

3 years ago

Many trials are not needed before a hypothesis can be accepted

aalyn [17]

Your hypothesis is at the biggenning or the end of you summary

6 0

4 years ago

Given 1.00 grams of each of the following radioisotopes, Ca-37, U-238, P-32 and Rn-222,which would have the least remaining orig

erica [24]

1) Ca-37, with a half-life of 181.1(10) ms.

5 0

4 years ago

Consider the following unbalanced redox reactions. In each case, separate the whole reactions into half-reactions, balance the h

Marina86 [1]

Answer & Explanation:

(a)

$Fe^{2+} +NO_{3}^{-} => Fe{(OH)}_3 + N_2$

reducing agent = Fe²⁺

Oxidizing agent = NO₃⁻

oxidation

Fe²⁺ ⇒ Fe(OH)₃

reduction

NO₃⁻ ⇒ N₂

Oxidation Half Reaction

(redox reactions are balanced by adding appropriate H⁺ and H₂O atoms)

Fe²⁺ ⇒ Fe(OH)₃

Balance O atoms

Fe²⁺ + 3H₂O ⇒ Fe(OH)₃

Balance H atoms

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺

balance Charge

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺ + e⁻..............(1)

reduction Half Reaction

NO₃⁻ ⇒ N₂

Balance N atoms

2NO₃⁻ ⇒N₂

Balance O atoms by adding appropriate H₂O

2NO₃⁻ ⇒ N₂ + 6H₂O

Balance H atoms

2NO₃⁻ + 12H⁺ ⇒ N₂ + 6H₂O

Balance Charge

2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O.................(2)

Combine Equation (1) and (2)

(1) × 10: 10Fe² + 30H₂0 ⇒ 10Fe(OH)₃ + 30H⁺ + 10e⁻

(2) × 1: 2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O

(1) + (2): 10Fe² + 30H₂0 + 2NO₃⁻ + 12H⁺ + 10e⁻ ⇒10Fe(OH)₃ + 30H⁺ + 10e⁻ +

N₂ + 6H₂O

10Fe² + 24H₂0 + 2NO₃⁻ ⇒ 10Fe(OH)₃ + 18H⁺ + N₂

this is the balanced reaction

REDUCTION POTENTIAL

10Fe²⁺(aq) + 10e⁻ ⇒ 10Fe(OH)₃(aq) E°ox = 10(-0.44) = -4.4V

2NO₃⁻(aq) - 2e⁻ ⇌ N₂(g) + 18H⁺ E°red = 2(+0.80) = +1.6

10Fe² + 24H₂0 + 2NO₃⁻ ⇒ 10Fe(OH)₃ + 18H⁺ + N₂ E°cell = -2.8V

E°cell = E°red + E°ox

4 0

4 years ago

Why are deposits of pure sodium metal or pure potassium metal not seen in nature

Natasha2012 [34]

They are not seen in nature because they are always combined with something to make something else.

5 0

3 years ago