Determine ΔT if T1 = 5oC and T2 = 123oC

How long will it take 10.0 mL of Ne gas to effuse through a porous barrier if it has been observed that 125 minutes are required

cupoosta [38]

Answer:

88.8 minutes

Explanation:

Graham's law of diffusion relates rate of difusion by the following formula

Rate1 / rate 2 = √( Mass of argon / Mass of Neon)

Where rate = volume divided by time

Rate 1 = 10 ml / t1

Rate 2 = 10 ml / t2

Rate 1/ rate 2 = 10 ml / t1 ÷ 10 ml/ t2 = t2/ t1

t2/t1 = √(Mass of argon / mass of Neon) = √( 39.984/20.179)

125 / t1 = 1.4026

t1 = 125 / 1.4026 = 88.8 minutes

7 0

2 years ago

Vinegar is an aqueous solution of acetic acid, ch3cooh. a 5.00 ml sample of a particular vinegar requires 26.90 ml of 0.175 m na

Fed [463]

The molarity of acetic acid in the vinegar is 0.94 M

<u><em> calculation</em></u>

Step 1: write the balanced equation between CH3COOH + NaOH

that is CH3COOH + NaOH → CH3COONa + H2O

step 2 : find the moles of NaOH

moles =molarity x volume in L

volume in liters = 26.90/1000=0.0269 l

moles = 0.175 mol /L x 0.0269 L =0.0047 moles of NaOH

Step 3: use the mole ratio to find moles of CH3COOH

that is the mole ratio of CH3COOH: NaOH is 1:1 therefore the moles of CH3COOH is =0.0047 moles

Step 4: find the molarity of CH3COOH

molarity = moles/volume in liters

volume in liter = 5.00/1000 =0.005 l

molarity is therefore=0.0047 moles/ 0.005 l = 0.94 M

4 0

2 years ago

When elemental sodium is added to water, the sodium atoms ionize spontaneously. uncharged na becomes na+. this means that the na

Temka [501]

Ionized, become charged, become a cation

hopefully that helps

5 0

2 years ago

A student made a graph to show the chemical equilibrium position of a reaction.

pogonyaev

Answer:

Concentration, because the amounts of reactants and products remain constant after equilibrium is reached.

Explanation:

The rate of reaction refers to the amount of reactants converted or products formed per unit time.

As the reaction progresses, reactions are converted into products. This continues until equilibrium is attained in a closed system.

When equilibrium is attained, the rate of forward reaction is equal to the rate of reverse reaction, hence the concentration of reactants and products in the system remain fairly constant over time.

When deducing the rate of reaction, concentration of the specie of interest is plotted on the y-axis against time on the x-axis.

8 0

2 years ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago