Saturated fats<span> are solid at room temperature, while </span>unsaturated fats<span> are liquid at room temperature. This is because </span>saturated and unsaturated fats<span> differ in their chemical structures. </span>Saturated fats<span> have no double bond </span>between<span> molecules, which means there are no gaps and the </span>fat<span> is </span>saturated<span> with hydrogen molecules.</span>
Answer: 500L
Explanation:
No of moles= volume× molarity/1000
No of moles =0.5moles
Volume=?
Molarity of a gas at stp = 1M
Stp means standard temperature and pressure
No of moles = volume ×molarity/1000
Substitute the values
0.5=volume×1/1000
Cross multiply
Volume = 1000×0.5
Volume = 500L
The volume is 500L
Answer:
2. Isotope, one of two or more species of atoms of a chemical element with the same atomic number and position in the periodic table and nearly identical chemical behavior but with different atomic masses and physical properties. Every chemical element has one or more isotopes.
3. Since the vast majority of an atom's mass is found its protons and neutrons, subtracting the number of protons (i.e. the atomic number) from the atomic mass will give you the calculated number of neutrons in the atom. In our example, this is: 14 (atomic mass) – 6 (number of protons) = 8 (number of neutrons).
Explanation: Try rewording the questions when looking it up. Hope this helps
Answer:
75 mg
Explanation:
We can write the extraction formula as
x = m/[1 + (1/K)(Vaq/Vo)], where
x = mass extracted
m = total mass of solute
K = distribution coefficient
Vo = volume of organic layer
Vaq = volume of aqueous layer
Data:
m = 75 mg
K = 1.8
Vo = 0.90 mL
Vaq = 1.00 mL
Calculations:
For each extraction,
1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62
x = m/1.62 = 0.618m
So, 61.8 % of the solute is extracted in each step.
In other words, 38.2 % of the solute remains.
Let r = the amount remaining after n extractions. Then
r = m(0.382)^n.
If n = 7,
r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg
m = 75 - 0.088 = 75 mg
After seven extractions, 75 mg (99.999 %) of the solute will be extracted.
