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steposvetlana [31]
4 years ago
6

Using LeChâtelier’s principle, determine whether the reactants or products are favored and the direction of the shift in equilib

rium (left or right) in the following examples. 2CO(g) + O2(g)  2CO2(g) + heat a. Remove O2 b. Lower the temperature c. Add CO d. Remove CO2 e. Decrease pressure
Chemistry
1 answer:
AnnyKZ [126]4 years ago
6 0

Answer:

a. right

b. right

c. left

d. left

e. left

Explanation:

a. When you remove O2 that means that the reactant side is lacking in an O2 so the equilibrium shifts to the product side.

b. This is an exothermic reaction, because heat is on the products side. So since you are decreasing temperature, then it would shift toward your right.

c. WHen you add CO you are adding more to the reactants side therefore the equilibrium would shift toward the left.

d. When you remove CO2 that means that there is a lack of CO2, therefore the equilibrium would shift to the left.

e. When you decrease pressure, you are increasing in volume. So when you look at volume, if it is increasing you look at larger moles, and if it is decreasing you look at smaller moles. In this case, it is increasing therefore it would shift to the left.

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2. The empirical formula of a molecule is CH2O. In an experiment, the molar mass of the molecule was determined to be 360.3 g/mo
Vanyuwa [196]

Answer:

The answer to your question is:

2.- C₁₂ H₂₄ O₁₂

Explanation:

2.-

Data

CH2O

molar mass = 360.3 g/mol

Molar mass of CH2O = 12 + 2 + 16 = 30g

Divide molar mass given by molar mass obtain

                   

                           x = 360.3/30

                          x = 12

Finally

                      C₁₂ H₂₄ O₁₂

Molar mass = (12 x 12) + (24 x 1) + (16 x 12) = 144 + 24 + 192 = 360 g

3.- First we need to write the complete equation of the reaction and balanced it.

Then, we need to convert the mass given to moles of each compound.

After that, we need used rule of three calculate the amount of products based on the moles of reactants given.

Finally, convert the moles to grams.

4.-

a.- It is a relation between the mass of product obtain in an experiment and the mass of a product obtain theoretically times 100.

b.-

35 g of Mg reacted with excess O2

percent yield = 90%

Actual yield = ?

Formula

Percent yield = (actual yield/theoretical yield) x 100

Equation  

                       2Mg  + O2 ⇒ 2MgO

                      48.62 g of Mg ----------------- 80.62 g of MgO

                      35g                  ------------------  x

                     x = 58 g of MgO     (Theoretical yield)

Theoretical yield = 58 g of MgO

Actual yield = percent yield x theoretical yield / 100

                    = 90 x 58 / 100

                   = 52. 23 g

5 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
Imagine you are given a mystery element. It is, however, a discovered and known element. You may perform a maximum of two observ
miv72 [106K]

The classification of it being a metal, nonmetal, or metalliod will be useful in the process of elimination to determine what it is. Then for the second test, meauring the atomin radius will narrow it down quicker to the mystery elemet's name.

Since you determined what part of the periodic table it's on, then when measuring the atomic radius, you should be able to pinpoint what the element is more surely.

6 0
3 years ago
Read 2 more answers
What are the 3 principles of the Kinetic molecular theory?
Sever21 [200]

Answer:

u already know the answer

Explanation:answer

6 0
3 years ago
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Write the condensed electron configurations for the Ca atom. Express your answer in condensed form as a series of orbitals. For
umka21 [38]

Answer:

[Ar] 4s²

Explanation:

Ca is the symbol for Calcium. It is the 20th element and it has 20 electrons.

The full electronic configuration for calcium is given as;

1s²2s²2p⁶3s²3p⁶4s²

The condensed electronic configuration is given as;

[Ar] 4s²

6 0
3 years ago
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