Answer:
a. NH₃ : base
CH₃COOH (acetic acid) : acid
NH₄⁺ : conjugate acid
CH₃COO⁻ : conjugate base
b. HClO₄ (perchloric acid) : acid
NH₃ : base
ClO₄⁻ : conjugate base
NH₄⁺ : conjugate acid
Hope this helps.
Answer:
my mass turned into gasses and other substances besides ash
Explanation:
burning a log products smoke and ash. the smoke takes some mass and the ash takes the rest.
Answer:
2192.64 PSI.
Explanation:
- From the general law of ideal gases:
<em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the container in L (V = 1650 L).
n is the no. of moles of the gas in mol (n = 9750 mol).
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature of the gas in (T = 35°C + 273 = 308 K).
∴ P = nRT/V = (9750 mol)(0.082 L.atm/mol.K)(308 K)/(1650 L) = 149.2 atm.
- <u><em>To convert from atm to PSI:</em></u>
1 atm = 14.696 PSI.
<em>∴ P = 149.2 atm x (14.696 PSI/1.0 atm) = 2192.64 PSI.</em>
Answer:
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Explanation:
but you follow me and give me brainliest ok by by by by
Answer:
Ka = 1.52 E-5
Explanation:
- CH3-(CH2)2-COOH ↔ CH3(CH2)2COO- + H3O+
⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]
mass balance:
⇒<em> C</em> CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M
charge balance:
⇒ [H3O+] = [CH3(CH2)2COO-]
⇒ Ka = [H3O+]²/(1 - [H3O+])
∴ pH = 2.41 = - Log [H3O+]
⇒ [H3O+] = 3.89 E-3 M
⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )
⇒ Ka = 1.519 E-5
