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2 years ago

The shape of an atomic orbital is associated with

Chemistry

1 answer:

Diano4ka-milaya [45]2 years ago

8 0

The properties of the atomic orbital are actually dependent on the quantum numbers.

size of atomic orbital: governed by the principal quantum number (n)

shape of atomic orbital: governed by the angular momentum quantum number (l)

orientation in space: governed by the magnetic quantum number (ml)

Since we are asked about the shape, hence the correct answer is:

angular momentum quantum number (l)

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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

Website

vredina [299]

Answer:

volume in L = 0.25 L

Explanation:

Given data:

Mass of Cu(NO₃)₂ = 2.43 g

Volume of KI = ?

Solution:

Balanced chemical equation:

2Cu(NO₃)₂ + 4KI → 2CuI + I₂ + 4KNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 2.43 g/ 187.56 g/mol

Number of moles = 0.013 mol

Now we will compare the moles of Cu(NO₃)₂ with KI.

Cu(NO₃)₂ : KI

2 : 4

0.013 : 4 × 0.013=0.052 mol

Volume of KI:

Molarity = moles of solute / volume in L

volume in L = moles of solute /Molarity

volume in L = 0.052 mol / 0.209 mol/L

volume in L = 0.25 L

6 0

2 years ago

A bottle containing 1,665 g of sulfuric acid (H2SO4, 98.08 g/mol) was spilled in a laboratory. The emergency spill kit contained

Tamiku [17]

Answer:

A. Yes, there is more than enough sodium carbonate.

Explanation:

Hello,

In this case, based on the given reaction which is:

$H_2SO_4(aq) + Na_2CO_3(s) \rightarrow Na_2SO_4(aq) + CO_2(g) + H_2O(l)$

By stoichiometry, one computes the grams of sodium carbonate that will neutralize 1,665 g of sulfuric acid as shown below:

$1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3$

Thus, the available mass is 2.0 kg so 0.2 kg are in excess, therefore: A. Yes, there is more than enough sodium carbonate.

Best regards.

5 0

2 years ago

During a lab experiment performed at STP conditions, you prepare HCl by reacting 100. ml of Cl2 gas with an excess of H2 gas.

Brrunno [24]

Answer: 19.4 mL Ba(OH)2

Explanation:

H2(g) + Cl2(g) --> 2HCl(aq) (make sure this equation is balanced first)

At STP, 1 mol gas = 22.4 L gas. Use this conversion factor to convert the 100. mL of Cl2 to moles.

0.100 L Cl2 • (1 mol / 22.4 L) = 0.00446 mol Cl2

Use the mole ratio of 2 mol HCl for every 1 mol Cl2 to find moles of HCl produced.

0.00446 mol Cl2 • (2 mol HCl / 1 mol Cl2) = 0.00892 mol HCl

HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.

0.00892 mol OH- • (1 mol Ba(OH)2 / 2 mol OH-) • (1 L Ba(OH)2 / 0.230 M Ba(OH)2) = 0.0194 L = 19.4 mL Ba(OH)2

3 0

2 years ago

The volume of hcl gas required to react with excess ca to produce 11.4 l of hydrogen gas at 1.62 atm and 62.0 °c is ________ l.

seropon [69]

Answer:

22.8 L

Step-by-step explanation:

We can use Gay-Lussac's Law of Combining Volumes to solve this problem:

Gases at the same temperature and pressure react in simple whole-number ratios.

1. Write the chemical equation.

Ratio: 2 L 1 L

Ca(s) + 2HCl(g) ⟶ CaCl₂(s) + H₂(g)

V/L: 11.4

2. Calculate the volume of HCl.

According to the law, 2 L of HCl form 1 L of H₂.

Then, the conversion factor is (2 L HCl/1 L H₂).

Volume of HCl = 11.4 L H₂ × (2 L HCl/1 L H₂)

= 22.8 L HCl

3 0

2 years ago