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3 years ago

Which of the following is a mixture? a. Orange soda, b. Alcohol, c. Mercury, d. Water

Chemistry

2 answers:

yaroslaw [1]3 years ago

8 0

It will be a letter c

Maurinko [17]3 years ago

5 0

I think it is c mercury

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Find the mass of a table that has a density of 2.35 g/cm3 and a volume of 12cm3.

Hunter-Best [27]

Answer:

<h2>The answer is 28.2 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of table = 12 cm³

density = 2.35 g/cm³

The mass of the table is

mass = 2.35 × 12

We have the final answer as

Hope this helps you

8 0

3 years ago

A concentrated binary solution containing mostly species 2 (but x2 ≠ 1) is in equilib- rium with a vapor phase containing both s

marusya05 [52]

Answer:

x1= 4.5 × 10^-3, y1= 0.9

Explanation:

A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2

Take the basis as the pressure of the 2 phase system is 1 bar. The assumption are as follows:

1. The vapor phase is ideal at pressure of 1 bar

2. Henry's law apply to dilute solution only.

3. Raoult's law apply to concentrated solution only.

Where,

Henry's constant for species 1 H= 200bar

Saturation vapor pressure of species 2, P2sat= 0.10bar

Temperature = 25°C= 298.15k

Apply Henry's law for species 1

y1P= H1x1...... equation 1

y1= mole fraction of species 1 in vapor phase.

P= Total pressure of the system

x1= mole fraction of species 1 in liquid phase.

Apply Raoult's law for species 2

y2P= P2satx2...... equation 2

From the 2 equations above

P=H1x1 + P2satx2

200bar= H1

0.10= P2sat

1 bar= P

Hence,

P=H1x1 + (1 - x1) P2sat

1bar= 200bar × x1 + (1 - x1) 0.10bar

x1= 4.5 × 10^-3

The mole fraction of species 1 in liquid phase is 4.5 × 10^-3

To get y, substitute x1=4.5 × 10^-3 in equation 1

y × 1 bar = 200bar × 4.5 × 10^-3

y1= 0.9

The mole fraction of species 1 in vapor phase is 0.9

4 0

3 years ago

Read 2 more answers

When a 5.00-g metal piece, A, was immersed in 38.0 mL of water, the water level rose to 50.0 mL. Similarly, when a 5.00-g metal

Mnenie [13.5K]

Answer:

A is denser than B as it's volume for the same mass is smaller.

Explanation:

Hello.

In this case, we first need to take into account that the density of each metal A and B is computed by dividing the mass over the volume of each metal which is actually computed by substracting the volume of water from the volume of the water and the solid:

$V_A=50.0mL-38.0mL=12.0mL\\\\V_B=60.0mL-38.0mL=22.0mL$

Next, we compute the densities as shown below:

$\rho_A=\frac{m_A}{V_A}=\frac{5.00g}{12.0mL}=0.42g/mL\\ \\\rho_B=\frac{m_B}{V_B}=\frac{5.00g}{22.0mL}=0.23g/mL$

In such a way, A is denser is B as it's volume for the same mass is smaller.

Best regards.

4 0

3 years ago

Which of the following elements has the largest atomic radius?

Pachacha [2.7K]

Answer:

nitrogen

Explanation:

as the group number increases, the atomic radius decreases

3 0

3 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

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

3 years ago