N = 4 moles of Ar2, P = 1.90 atm, V = ?
T = 50C = 273 + 50K = 323K
PV = nRT --> V = nRT/P
V = (4)(.0821)(323)/1.90
V = 106.07/ 1.9
V = 55.8 L
Notice q=3/2, is half of the original q = 3(<span>1/2</span>)<span>t/28.8
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An ionization suppressor is an alkali metal capable of preventing ionization, which can be used in atomic spectroscopy to determine matter composition.
<h3>What is ionization?</h3>
Ionization refers to the phenomena capable of converting neutral atoms/molecules to electrically charged atoms/ions.
Ionization is a process by which radiation (e.g., alpha, beta, gamma rays) can pass energy to inert matter.
Some examples of ionization suppressors include salts of alkali metals (for example, potassium), which can be used in atomic spectroscopy to determine matter composition.
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The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL
<h3>Balanced equation </h3>
2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O
From the balanced equation above,
- The mole ratio of the acid, HCl (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 1
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
- Volume of acid, HCl (Va) = 36 mL
- Molarity of acid, HCl (Ma) = 1.63 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(1.63 × 36) / (1.48 × Vb) = 2
58.68 / (1.48 × Vb) = 2
Cross multiply
2 × 1.48 × Vb = 58.68
2.96 × Vb = 58.68
Divide both side by 2.96
Vb = 58.68 / 2.96
Vb = 19.8 mL
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