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3 years ago

Why do we use ammonium formate as a hydrogen surrogate instead of using hydrogen gas in this experiment?

Chemistry

1 answer:

NemiM [27]3 years ago

7 0

Answer:

Ammonium formate can also be used in palladium on carbon (Pd/C) reduction of functional groups. In the presence of Pd/C, ammonium formate decomposes to hydrogen, carbon dioxide, and ammonia.

Explanation:

This hydrogen gas is adsorbed onto the surface of the palladium metal, where it can react with various functional groups.

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statuscvo [17]

Answer:

2E-24

Explanation:

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3 years ago

Avagadros Law..<br>Definition please..

Nikitich [7]

Answer:

"Avogadro's law is an experimental gas law relating the volume of a gas to the amount of substance of gas present. The law is a specific case of the ideal gas law. A modern statement is: Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

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2 years ago

Read 2 more answers

The formula for Gibbs free energy is

UNO [17]

At constant temperature and pressure, the change in Gibbs free energy is defined as DG= DH-TDS hope this helps bb ♡

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2 years ago

Read 2 more answers

How many moles of water are there in 1.000 l at stp? assume the density of water is 1000 kg/m3?

EastWind [94]

Assume 1 liter = 1 kilogram of water = 1000 grams of water.

Part A)
MW of hydrogen is 1.008g/mol, and oxygen is 16.00g/mol.

Find the MW of water by
2*(1.008) + (16.00) = 18.016g/mol.

Convert 1000g H2O to moles :

(1000g H2O)*(1mol H2O / 18.016g H2O) = 55.51 mol

Part B)
Using the answer from part A and Avogadro's number:

(55.51mol)*(6.022*10^23) =

3.343*10^25 molecules.
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7 0

3 years ago

Calculate Delta H in KJ for the following reactions using heats of formation:

lozanna [386]

Answer:

$\Delta H\textdegree = -2856.8\;\text{kJ}$ per mole reaction.

$\Delta H\textdegree = -22.3\;\text{kJ}$ per mole reaction.

Explanation:

What is the standard enthalpy of formation $\Delta H_f\textdegree{}$ of a substance? $\Delta H_f\textdegree{}$ the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, $\Delta H_f\textdegree{} = 0$ for the most stable allotrope of each element under standard conditions. For example, oxygen $\text{O}_2$ (not ozone $\text{O}_3$ ) is the most stable allotrope of oxygen. Also, under STP $\text{O}_2$ is a gas. Forming $\text{O}_2\;(g)$ from itself does not involve any chemical or physical change. As a result, $\Delta H_f\textdegree{} = 0$ for $\text{O}_2\;(g)$ .

Look up standard enthalpy of formation $\Delta H_f\textdegree{}$ data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam $\text{H}_2\text{O}$ in particular) and the sign of the enthalpy changes.

$\text{C}_2\text{H}_6\;(g)$ : $-84.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{CO}_2\;(g)$ : $-393.5\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{H}_2\text{O}\;{\bf (g)}$ : $-241.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}\;(s)$ : $-217.9\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}_2\;(s)$ : $-276.6\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{Pb}_3\text{O}_4\;(s)$ : $-734.7\;\text{kJ}\cdot\text{mol}^{-1}$

How to calculate the enthalpy change of a reaction $\Delta H_\text{rxn}$ (or simply $\Delta H$ from enthalpies of formation?

Multiply the enthalpy of formation of each product by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Reactants}))$ to show that this value takes the coefficients into account.
Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Products}))$ to show that this value takes the coefficient into account.
Change = Final - Initial. So is the case with enthalpy changes. $\Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))$ .

For the first reaction:

$\Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}$ .

Try these steps for the second reaction:

$\Delta H_\text{rxn} = -22.3\;\text{kJ}\cdot\text{mol}^{-1}$ .

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3 years ago