13.26 g has<br> significant figures<br> 1.<br> 0<br> 1<br> 2<br> 3<br> 4.

HELP!!! name two properties of metals

Olenka [21]

Answer:

1) Ductile

2) Malleable

please mark brainlest if right

Explanation:

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

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

3 years ago

What is the smallest halogen atom<br>

ad-work [718]

The smallest halogen atom is fluorine

4 0

3 years ago

Read 2 more answers

Convert from moles to particles 6.02x1023 particles 1 mole Question 2. 2.7 moles of lithium Question 3. 1.8 moles of sodium chlo

spayn [35]

Considering the definition of Avogadro's number:

the number of molecules of lithium is 1.62621×10²⁴ molecules.
the number of molecules of sodium chloride is 1.08414×10²⁴ molecules.

<h3>Definition of Avogadro's number</h3>

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

<h3>Amount of molecules in this case</h3>

You can apply the following rule of three, considering the Avogadro's number: If 1 mole of lithium contains 6.023×10²³ molecules, 2.7 moles of lithium contains how many molecules?

amount of molecules of lithium= (6.023×10²³ molecules × 2.7 moles)÷1 mole

<u><em>amount of molecules of lithium=1.62621×10²⁴ molecules</em></u>

Finally, the number of molecules of lithium is 1.62621×10²⁴ molecules.

On the other hand you can apply the following rule of three, considering the Avogadro's number: If 1 mole of sodium chloride contains 6.023×10²³ molecules, 1.8 moles of sodium chloride contains how many molecules?

amount of molecules of sodium chloride= (6.023×10²³ molecules × 1.8 moles)÷1 mole

<u><em>amount of molecules of sodium chloride=1.08414 ×10²⁴ molecules</em></u>

Finally, the number of molecules of sodium chloride is 1.08414×10²⁴ molecules.

Learn more about Avogadro's Number:

brainly.com/question/11907018

brainly.com/question/1445383

brainly.com/question/1528951

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4 0

2 years ago

A student decides to conduct an investigation to determine the boiling points of water and ethanol. The student will heat sample

lilavasa [31]