Mole<span>: the amount of a substance that contains 6.02 x </span>10<span>. 23 respective particles of that substance. Avogadro's number: 6.02 x </span>10<span>. 23. Molar Mass: the mass of one </span>mole<span> of an element. CONVERSION FACTORS: 1 </span>mole<span> = 6.02 x </span>10<span>. 23 </span>atoms<span> 1 </span>mole<span> = </span>atomic<span> mass (g). Try: 1. How </span>many atoms<span> are in 6.5</span>moles<span> of zinc</span>
A B and C are all chemical changes. paper tearing is not.
Answer:
Energy can neither be created nor destroyed, but it does change its form. And not all forms of energy are usable and it gets dissipated as heat energy and sound energy. The energy that is wasted cannot be recovered.
Using the ideal gas law equation, we can find the number of H₂ moles produced.
PV = nRT
Where P - pressure - 0.811 atm x 101 325 Pa/atm = 82 175 Pa
V - volume - 58.0 x 10⁻³ m³
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 32 °C + 273 = 305 K
substituting these values in the equation,
82 175 Pa x 58.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 305 K
n = 1.88 mol
The balanced equation for the reaction is as follows;
CaH₂(s) + 2H₂O(l) --> Ca(OH)₂(aq) + 2H₂(g)
stoichiometry of CaH₂ to H₂ is 1:2
When 1.88 mol of H₂ is formed , number of CaH₂ moles reacted = 1.88/2 mol
therefore number of CaH₂ moles reacted = 0.94 mol
Mass of CaH₂ reacted - 0.94 mol x 42 g/mol = 39.48 g of CaH₂ are needed
Answer:
The 99.68% of the aspirin is present in the neutral form
Explanation:
Aspirin, Acetylsalicylic acid, is a weak acid with pKa = 3.5
Using Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
<em>Where [A⁻] is the ionized form and HA the neutral form of the acid</em>
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Replacing with a pH of stomach of 1.0:
1.0 = 3.5 + log [A⁻] / [HA]
-2.5 = log [A⁻] / [HA]
3.16x10⁻³ = [A⁻] / [HA] <em>(1)</em>
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A 100% of aspirin is = [A⁻] + [HA]
100 = [A⁻] + [HA] <em>(2)</em>
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Replacing (2) in (1)
3.16x10⁻³ = 100 - [HA] / [HA]
3.16x10⁻³[HA] = 100 - [HA]
1.00316 [HA] = 100
[HA] = 99.68%
<h3>The 99.68% of the aspirin is present in the neutral form</h3>
