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cricket20 [7]
3 years ago
9

Using the milligrams of ascorbic acid you entered above, the ratio of total sample volume to aliquot volume, and the total milli

grams of the Vitamin C tablet that you dissolved, calculate the mass of ascorbic acid in the Vitamin C tablet for each trial. Do this by scaling up to find the amount (mg) of ascorbic acid in your 250 mL flask Enter your calculated mass of ascorbic acid in the Vitamin C tablet, for each trial Be sure to enter your calculated mass in the corresponding order that you entered your milligrams of ascorbic acid The milligrams of ascorbic acid you entered for entry #1 previously should correspond to the mass of ascorbic acid that you enter for entry #1 here Ascorbic Acid (mg)
Entry # Volume (ml) Mass % of Ascorbic Acid
#1 9.650 3.84
#2: 7.500 2.99
#3: 9.500 3.78
Engineering
2 answers:
m_a_m_a [10]3 years ago
6 0
Using the milligrams of ascorbic acid you entered above, the ratio of total sample volume to aliquot volume, and the total milligrams of the Vitamin C tablet that you dissolved, calculate the mass of ascorbic acid in the Vitamin C tablet for each trial. Do this by scaling up to find the amount (mg) of ascorbic acid in your 250 mL flask Enter your calculated mass of ascorbic acid in the Vitamin C tablet, for each trial Be sure to enter your calculated mass in the corresponding order that you entered your milligrams of ascorbic acid The milligrams of ascorbic acid you entered for entry #1 previously should correspond to the mass of ascorbic acid that you enter for entry #1 here Ascorbic Acid (mg)
Entry # Volume (ml) Mass % of Ascorbic Acid
#1 9.650 3.84
#2: 7.500 2.99
#3: 9.500 3.782. 2 one is the answer
Dmitriy789 [7]3 years ago
3 0

Answer: 7.500 2.99

Explanation:

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Explanation:

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The train is traveling 26 meters A second .

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Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the
ra1l [238]

The evaporation rate of the n-Hexane is 7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}

<u>Explanation</u>:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

Where the molar flux is provided by,

N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}

\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}

P_{t}=1 a t m=101325 P a\\

\text { so, } P_{A 1}= the vapor pressure at hexane 25 \mathrm{C} =20158.2 \mathrm{Pa}

For wind, assume negligible hexane is present, hence P_{A 2}=0

Now,

\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}

\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}

P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\

=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}

=90873.57 \mathrm{Pa}

R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}

z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\

where T = 298 K

substituting all in the equation, we get

\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}

=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\

Now,Flux \times area  = Molar rate of evaporation

Evaporation rate = 0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)

Evaporation rate =7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}

6 0
3 years ago
A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr
Hoochie [10]

Answer:

\Delta m = 102.25\,kg

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

m_{o} = \rho_{air}\cdot V_{tank}

m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{o} = 31.25\,kg

The final mass inside the rigid tank is:

m_{f} = \rho \cdot V_{tank}

m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{f}= 133.5\,kg

The supplied air mass is:

\Delta m = m_{f}-m_{o}

\Delta m = 133.5\,kg-31.25\,kg

\Delta m = 102.25\,kg

4 0
3 years ago
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