<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
Answer:0.178 moles
Explanation: carbon trihydride seems to be an unusual name for the methyl group CH3–
ionic wt 15
moles = 2.67/15 = 0.178
Answer: 0.176 atm
Explanation: Solution attached:
Use Boyle's Law to find the new pressure of the gas.
P1V1 = P2V2
Derive for P2
P2 = P1V1 / V2
= 5.5 atm ( 4.8 L ) / 150 L
= 0.176 atm
Answer: The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster. Effusion rate is inversely proportional to molar mass.
Explanation:
Warm is less heaver then cold air so warm air rise and cold air sinks
