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2 years ago

What mass of solid lead would displace exactly 234.6 liters of water?

2 answers:

6 0

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The mass of solid lead would displace exactly 234.6 liters of water should be 2,674,440

BARSIC [14]2 years ago

3 0

The answer is 2660.6 kg

Exactly 234.6 liters of water is displaced by lead hence giving us the volume of the metal, according to Archimedes' principle of displacement.

Lead has a density of 11341 kg/m3 and we know that 1000L = 1m3. Therefore;

234.6 liters = 0.2346m3

0.2346 * 11341

= 2660.6 kg

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mario62 [17]

A net worth: $79.0 billion.
Value of stock : $3.20 billion.
New net worth:
$79.0 + $3.20 = $82.20 billion = $82.20 * 10^9 = $8.20 * 10^10

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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

calculation

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

If the kinetic energy of a particle is equal to twice its rest mass, what is the velocity of the particle? Determine if relativi

ivann1987 [24]

Answer:

The velocity of the particle is 2 m/s,

Explanation:

Kinetic energy is defined as energy of the body due to its motion. It is given by :

$K.E=\frac{1}{2}mv^2$

Where :

m = mass of the object

v = velocity of the object

We have , particle with mass m and its kinetic energy is twice its mass.

$K.E=2m$

$2m=\frac{1}{2}mv^2$

$v^2=\frac{4}{1}$

$v=2$

And unit of velocity are m/s , so the velocity of the particle is 2 m/s.

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

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

Which of the following statements concerning a metal crystallized in a face-centered cubic cell is/are CORRECT? 1. One metal ato

AleksAgata [21]

Answer:

Only 3 is correct.

Explanation:

The crystal of a metal or an ionic compound is called a cell, and there are 7 types of unit cells: cubic, tetragonal, orthorhombic, monoclinic, hexagonal, rhombohedral, and triclinic.

In a face-centered cubic cell (FCC) all angles are 90º and all lengths are equal. Each cubic cell has 8 atoms in each corner of the cube, and that atom is shared with 8 neighboring cells. So for a metal crystal, the atom is located at each of the eight lattice points, where it is shared equally between eight unit cells.

5 0

2 years ago