MEASUREMENT AND MATTER Interconverting temperatures in Celsius and Kelvins The metal osmium becomes superconducting at temperatures below 655.mk Calculate the temperature at which osmium becomes superconducting in degrees Celsius. Be sure your answer has the correct number of significant digits. Ac

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Answer:

A. 2-keto acid decarboxylase and alcohol dehydrogenase

Explanation:

2-keto acid decarboxylase and alcohol dehydrogenase are used to produce many higher alcohols. These enzymes also display a high degree of specificity on their substrate

Answer:

226.2 kJ/mol

Explanation:

Let's consider the following thermochemical equation for the combustion of acetylene.

C₂H₂(g) + (5/2) O₂(g) → 2 CO₂(g) + H₂O(l), ΔH°rxn = –1299 kJ/mol.

We can also calculate the enthalpy of the reaction per mole of acetylene from the enthalpies of formation.

ΔH°rxn = 2 mol × ΔH°f(CO₂(g)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(C₂H₂(g)) - 1 mol × ΔH°f(O₂(g))

1 mol × ΔH°f(C₂H₂(g)) = 2 mol × ΔH°f(CO₂(g)) + 1 mol × ΔH°f(H₂O(l)) - ΔH°rxn - 1 mol × ΔH°f(O₂(g))

1 mol × ΔH°f(C₂H₂(g)) = 2 mol × (-393.5 kJ/mol) + 1 mol × (-285.8 kJ/mol) - (-1299 kJ) - 1 mol × (0 kJ/mol)

ΔH°f(C₂H₂(g)) = 226.2 kJ/mol

Answer:

The enthalpy of formation of acetylene is 226.2 kJ/mol

Explanation:

Step 1: Data given

C2H2 (g) + (5/2)O2 (g)  → 2CO2 (g) + H2O (l)  Heat of Reaction (Rxn) = -1299kJ/mol

Standard formation [CO2 (g)]= -393.5 kJ/mol

Standard formation [H2O (l)] = -285.8 kj/mol

Step 2: The balanced equation

The formation of acetylene is:

2C(s) + H2(g)   → C2H2(g)

Step 3: Calculate the enthalpy of formation of acetylene

It doesn't matter if the process will happen in 1 step or in more steps. What matters is the final result. This is Hess' law of heat summation.

To have the reaction of the formation of acetylene we have to take:

⇒ the reverse equation of the combustion of acetylene

   2CO2 (g) + H2O (l) → C2H2 (g) + (5/2)O2 (g)

⇒  The equation of formation of CO2 (multiplied by 2)

2C(s) + 2O2(g) → 2CO2(g)

⇒ the equation of formation of H2O

H2(g) + 1/2 O2(g) → H2O(l)

2CO2 (g) + H2O (l) + 2C(s) + 2O2(g) + H2(g) + 1/2 O2(g  → C2H2 (g) + (5/2)O2 (g) + 2CO2(g) + H2O(l)

Final reaction = 2C(s) + H2(g)   → C2H2(g)

Calculate the enthalpy of formation of acetylene =

ΔHf = 1299 kJ/mol + (2*-393.5) kJ/mol + (-285.8) kJ/mol

ΔHf = 226.2 kJ/mol

The enthalpy of formation of acetylene is 226.2 kJ/mol

Answer:

Polarity

Cohesion  

Adhesion

High Specific Heat

Explanation:

Answer:

The gases will expand 8.2 L against the constant pressure of 710 torr.

Explanation:

Given that:

the original volume V₁ = 35 cm³ = 35 × 10⁻⁶ m³

Since the combustion of the mixture releases energy then :

the work W = - 775 J

Pressure = 710 torr

Since 1 torr = 133.322 Pa

710 torr = 94658.62 Pa

We all know that:

W = -PdV

-775 = - 94658.62 Pa ( V₂ - V₁ )

-775 = - 94658.62 ( V₂ - 35 × 10⁻⁶)

-775/ - 94658.62 =  V₂ - 35 × 10⁻⁶

0.008187 = V₂ - 35 × 10⁻⁶

V₂ = 0.008187 + 35 × 10⁻⁶

V₂ = 0.008222 m³

The change in volume dV = V₂ - V₁

The change in volume dV = 0.008222 m³ - 35 × 10⁻⁶ m³

The change in volume dV = 0.008187   m³

To litres

The change in volume dV = 8.2 L

Thus, the gases will expand 8.2 L against the constant pressure of 710 torr.

Answer:

12 mol CO₂

General Formulas and Concepts:

Atomic Structure

• Compounds
• Moles
• Mole Ratio

Stoichiometry

• Analyzing reactions rxn
• Using Dimensional Analysis

Explanation:

Step 1: Define

Identify

[rxn] 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

[Given] 2 mol C₆H₁₂O₆

[Solve] mol CO₂

Step 2: Identify Conversions

[rxn] 6CO₂ → C₆H₁₂O₆

Step 3: Convert

1. [DA] Set up:                                                                                                  
2. [DA] Multiply [Cancel out units]: