What statement is FALSE giving brainliest

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

The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2

Gre4nikov [31]

Answer:

No the substance is not water.

Explanation:

The balance chemical equation for the decomposition of water is as follow;

2 H₂O = 2 H₂ + O₂

Step 1: Calculate moles of H₂O;

Moles = Mass / M.Mass

Moles = 5.0 g / 18.01 g/mol

Moles = 0.277 moles of H₂O

Step 2: Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:

According to equation,

2 moles of H₂O produced = 1 mole of O₂

So,

0.277 moles of H₂O will produce = X moles of O₂

Solving for X,

X = 0.277 mol × 1 mol / 2 mol

X = 0.138 moles of O₂

Also,

According to equation,

2 moles of H₂O produced = 2 mole of H₂

So,

0.277 moles of H₂O will produce = X moles of H₂

Solving for X,

X = 0.277 mol × 2 mol / 2 mol

X = 0.227 moles of H₂

Step 3: Calculate Mass of O₂ and H₂ as;

For O₂:

Mass = Moles × M.Mass

Mass = 0.138 mol × 31.99 g/mol

Mass = 4.44 g of O₂

For H₂:

Mass = Moles × M.Mass

Mass = 0.227 mol × 2.01 g/mol

Mass = 0.559 g of H₂

Conclusion:

From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

6 0

3 years ago

Write the balanced chemical equation for the following: The destruction of a marble statuary by acid rain: aqueous nitric acid r

Mashcka [7]

Answer: The coefficient in front of nitric acid is 2

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas. Liquids are represented by (l) and gases are represented by (g).

The balanced chemical reaction is :

$CaCO_3(s)+2HNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+CO_2(g)+H_2O(l)$