Answer:
xy (-b+c+q) is the answer to this
<h3>Answer:</h3>
a) Moles of Caffeine = 1.0 × 10⁻⁴ mol
b) Moles of Ethanol = 4.5 × 10⁻³ mol
<h3>Solution:</h3>
Data Given:
Mass of Caffeine = 20 mg = 0.02 g
M.Mass of Caffeine = 194.19 g.mol⁻¹
Molecules of Ethanol = 2.72 × 10²¹
Calculate Moles of Caffeine as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 0.02 g ÷ 194.19 g.mol⁻¹
Moles = 1.0 × 10⁻⁴ mol
Calculate Moles of Ethanol as,
As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.
The relation between Moles, Number of Particles and Avogadro's Number is given as,
Number of Moles = Number of Molecules ÷ 6.022 × 10²³
Putting values,
Number of Moles = 2.72 × 10²¹ Molecules ÷ 6.022 × 10²³
Number of Moles = 4.5 × 10⁻³ Moles
<em><u>A molecule </u></em><em><u>can </u></em><em><u>possess polar bonds and still be nonpolar.</u></em>
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answer:
in plants
Transport manufactured food from the leaves to others parts of the plant
Facilitates gaseous exchange through the stomata in the leaves to other parts of the plant
in animal
Exchange of respiratory gases across respiratory services
Excretion of nitrogenous waste in some unicellular organisms
Explanation:
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Answer:
Explanation:
Hello!
In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:
![[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}](https://tex.z-dn.net/?f=%5BMgCl_2%5D%3D1.672%5Cfrac%7BmolMgCl_2%7D%7BL%7D%2A%5Cfrac%7B95.211gMgCl_2%7D%7B1molMgCl_2%7D%3D159.2%5Cfrac%7BgMgCl_2%7D%7BL%7D)
Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:
![[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14](https://tex.z-dn.net/?f=%5BMgCl_2%5D%3D159.2%5Cfrac%7BgMgCl_2%7D%7BL%7D%2A%5Cfrac%7B1L%7D%7B1000mL%7D%2A%5Cfrac%7B1mL%7D%7B1.137g%7D%3D0.14)
Which is also the by-mass fraction and in percent it turns out:
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