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-Dominant- [34]

1 year ago

5

A particular reaction has an enthalpy and entropy of reaction of ∆H = +33 kJ/mol and ∆S = +0.15 kJ/mol⋅K. At the three indicated

temperatures, determine if either reactants or products are favored?

1 answer:

Thepotemich [5.8K]1 year ago

4 0

Answer:

Explanation: check my paage

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What do the superscripts and subscripts in the notation ^40 K represent?

raketka [301]

Explanation:

subscript is K

superscript is ^

subscript K means a unit of temperature like F or C

superscript ^ means to the power of

So all together it means to the power of 40 K

3 0

2 years ago

The molar solubility of magnesium carbonate is 1.8 Ã— 10â€“4 mol/l. What is ksp for this compound?

Serhud [2]

Answer:

3.2 × 10⁻⁸

Explanation:

Let's consider the solution of magnesium carbonate.

MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)

We can relate the molar solubility (S) with the solubility product (Ksp) using an ICE chart.

MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)

I 0 0

C +S +S

E S S

The Ksp is:

Ksp = [Mg²⁺] × [CO₃²⁻] = S × S = S² = (1.8 × 10⁻⁴)² = 3.2 × 10⁻⁸

4 0

3 years ago

Read 2 more answers

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

3 years ago

Mg(OH)2 + 2 HBr à MgBr2 + 2 H2O

AnnyKZ [126]

Explanation:

The balanced equation of the reaction is given as;

Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)

1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?

From the reaction;

2 mol of HBr produces 1 mol of MgBr2

Converting to masses using;

Mass = Number of moles * Molar mass

Molar mass of HBr = 80.91 g/mol

Molar mass of MgBr2 = 184.113 g/mol

This means;

(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2

18.3g would produce x

161.82 = 184.113

18.3 = x

x = (184.113 * 18.3 ) / 161.82 = 20.8 g

2. How many moles of H2O will be produced from 18.3 grams of HBr?

Converting the mass to mol;

Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol

From the reaction;

2 mol of HBr produces 2 mol of H2O

0.226 mol would produce x

2 =2

0.226 = x

x = 0.226 * 2 / 2 = 0.226 mol

3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?

From the reaction;

2 mol of HBr reacts with 1 mol of Mg(OH)2

18.3g of HBr = 0.226 mol

2 = 1

0.226 = x

x = 0.226 * 1 /2

x = 0.113 mol

5 0

2 years ago

At an ACME plant in Scottsdale Arizonia methanol is produced commercially

patriot [66]

Answer:

thats nice to know i will send Wile E. Coyote there for his supplys

Explanation:

4 0

2 years ago