Answer:
The molar mass of the liquid 62.89 g/mol
Explanation:
Step 1: Data given
Mass of the sample = 0.1 grams
Temperature = 70°C
Volume = 750 mL
Pressure = 0.05951 atm
Step 2: Calculate the number of moles
p*V = n*R*T
n = (p*V)/(R*T)
⇒ with n = the number of moles gas = TO BE DETERMINED
⇒ with p = The pressure = 0.05951 atm
⇒ with V = The volume of the flask = 750 mL = 0.750 L
⇒ with R = The gasconstant = 0.08206 L*atm/K*mol
⇒with T = the temperature = 70 °C = 343 Kelvin
n = (0.05951 *0.750)/(0.08206*343)
n = 0.00159 moles
Step 3: Calculate molar mass
Molar mass = mass / moles
Molar mass =0.1 gram / 0.00159 moles
Molar mass = 62.89 g/mol
The molar mass of the liquid 62.89 g/mol
D. Drop in barometric pressure, warm ocean water, humid air. The low pressure brings in a cool air mass causing collision of two different masses.
My father rode out a typhoon near Okinawa WWII, onboard the battleship USS Missouri BB-63.
Violent pitching, alarms going off for approaching capsize pitch. The captain came on loudspeaker “ don’t worry men, land is near... about a mile straight down”.
Answer:
0.263M of CH₃COOH is the concentration of the solution.
Explanation:
The reaction of acetic acid (CH₃COOH) with NaOH is:
CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O
<em>1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.</em>
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In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:
0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.
As the sample of acetic acid had a volume of 25.0mL = 0.025L:
6.56x10⁻³ moles of CH₃COOH / 0.0250L =
<em>0.263M of CH₃COOH is the concentration of the solution</em>
