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ValentinkaMS [17]

2 years ago

6

At a certain temperature K = 1.1 X 10^3 L/mol for the reaction given below. Calculate the concentrations of Fe3+, SCN?, and FeSC

N2+ at equilibrium if 0.020 mol of Fe(NO3)3 is added to 1.0 L of 0.10 M KSCN. (Neglect any volume change.)Fe3+(aq) + SCN ?(aq) FeSCN2+(aq)Fe3+____________MSCN- ______________MFeSCN2+ ____________M

1 answer:

lianna [129]2 years ago

6 0

Answer:

Explanation:

Fe⁺³ + SCN⁻ = FeSCN²⁺

x x x

If x mole of Fe⁺³ reacted , unreacted Fe⁺³ = .02 - x

Remaining SCN⁻ unreacted = .1 - x

K_c = [ FeSCN²⁺] / [Fe⁺³] [ SCN⁻]

1.1 x 10³ = x / ( .02 - x ) (.1 - x )

1100 = x / .002 - .12x ( neglecting x² )

x = .0165

Concentration of Fe⁺³ = .02 - x = .0035 M

Concentration of SCN⁻ = .1 - x = .0835 M

Concentration of FeSCN²⁺ = x = .0165M

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A reaction container holds 5.77 g of P4 and 5.77 g of O2.

Dvinal [7]

Answer:

a) O2 is the limiting reactant

b) 5.75 grams P4O10

c) 5.79 grams P4O6

Explanation:

Step 1: Data given

Mass of P4 = 5.77 grams

Mass of O2 = 5.77 grams

Molar mass of P4 = 123.90 g/mol

Molar mass O2 = 32.0 g/mol

Step 2: The balanced equation

P4 + 3O2 → P4O6

Step 3: Calculate moles of P4

Moles P4 = mass P4 / molar mass P4

Moles P4 = 5.77 grams / 123.90 g/mol

Moles P4 = 0.0466 moles

Step 4: Calculate moles O2

Moles O2 = mass O2 / molar mass O2

Moles O2 = 5.77 grams / 32.0 g/mol

Moles O2 = 0.1803 moles

Step 5: Calculate limiting reactant

P4 is the limiting reactant in this reaction. It will completely be consumed (0.0466 moles). O2 is in excess, there will react 3*0.0466 = 0.1398 moles

There will remain 0.1803 - 0.1398 = 0.0405 moles O2

Step 6: Calculate the amount of P4O6

For 1 mol P4 we'll have 1 mol P4O6

For 0.0466 moles P4 we'll have 0.0466 moles P4O6

Step 7: The balanced equatio

P4O6 + 2O2 → P4O10

We have 0.0466 moles P4O6 and 0.0405 moles O2

Step 8: Calculate the limiting reactant

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

O2 is the limiting reactant. It will completely be consumed (0.0405 moles)

P4O6 is in excess. There will react 0.0405/2 = 0.02025 moles

There will remain 0.0466 - 0.02025 = 0.02635 moles P4O6

This is 0.02635 * 219.88 g/mol = 5.79 grams P4O6

Step 9: Calculate moles and mass of P4O10

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

For 0.0405 moles O2 we'll have 0.02025 moles P4O10

This is 0.02025 * 283.89 g/mol = 5.75 grams P4O10

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2 years ago

In which orbitals would the valence electrons for carbon <br> c. be placed?

katrin2010 [14]

Answer: 2s and 2p

Explanation: Carbon is an element with atomic number of 6 and thus contains 6 electrons. The electrons are filled in order of increasing energies and follows Afbau's rule.The electrons are singly filled first in each orbital having same spin, then only pairing occurs. This rule was known as Hund's Rule.

The valence electrons are the electrons which are present in last shell. Thus valence electrons are 4, two in s and 2 in p orbitals.

$C: 6:1s^22s^22p_x^12p_y^1$

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2 years ago

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jan is holding an ice cube. what causes the ice to melt? thermal energy from the ice is transferred to the air. thermal energy f

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Answer: Ice is melting due to the transfer of thermal energy from Jan's hand to ice.

Explanation: The melting of ice is a physical change and is happening when the thermal energy from Jan's hand is transferred to ice. Due to this energy transfer, the particles of ice starts to move faster and hence, making the ice melt.

In this, the physical state of ice is changing from solid to liquid state.

$H_2O(s)\rightleftharpoons H_2O(l)$

8 0

2 years ago

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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

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

2 years ago

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The correct answer is option d, that is, atoms of the element.

As the atoms are neither destroyed nor created in a chemical reaction, the sum of the mass of the products in a reaction must be equivalent to the sum of the mass of the reactants.

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2 years ago

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