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3 years ago

30 points for this question answer it plz .

Chemistry

1 answer:

kipiarov [429]3 years ago

4 0

Answer:

Low carbon steel is usually used for carburization because of its excellent toughness. For high duty applications alloy steels are also carburized for additional properties of both the case and the core. In alloy steels, the alloying elements remain in the form of various carbide and nitride precipitates.

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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:

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

6 0

4 years ago

. The pI is called ________________. The rule of calculating pI of an amino acid is that first, write the dissociation equation

umka21 [38]

Answer:

The isoelectric point is that the <u>pH </u>at which the compound is in an electronically neutral form.

For diss equations<u>, p</u>lease find them in the enclosed file.

The pIs of 2 amino acids:

Glutamate: pI = 3,2
Histidine: pI = 7,6

Explanation:

Formula for the pI calculation: pI = (pKa1 + pKa2)/2

Given 3 pKa :

Acid glutamic with an acid sidechain:

Use the lower 2 pKas (corresponding with 2 -COOH groups)

pKa1 = 2,19; pKa2 = 4,25; so pI = 3,2

Histidine with 2 amino groups:

Use the higher 2 pKas ( -COOH group and -NH= group)

pKa1 = 6; pKa2 = 9,17; so pI = 7,6

6 0

4 years ago

Arrange follwoing substances from lowest to highest lattice energy" MgS, KI, GAN, LiBr

sleet_krkn [62]

Answer: KI, LiBr, MgS,GaN

Explanation:

Ionic sizes are the deciding factor in lattice energy. The smaller the ion, the higher the lattice energy. Since K+> Li+ and I- is larger than Br-, LiBr will have a higher lattice energy. The size of the sulphide ion is much smaller than the nitride ion but the Gallium is much smaller due to lanthanide contraction hence the answer.

7 0

3 years ago

How does the energy in chemical bonds affect the energy in a chemical reaction?

zheka24 [161]

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8 0

3 years ago

How much heat is absorbed by 15.5g of steam to turn into liquid water when its temperature is decreased from 120.0°C to 100.0°C?

Natali [406]

Answer:

- 618.8 J.

Explanation:

To solve this problem, we can use the relation:

where, Q is the amount of heat released by steam (Q = ??? J).

m is the mass of the steam (m = 15.5 g).

c is the specific heat of the steam (c of steam = 1.996 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 120.0°C = - 20.0°C).

5 0

3 years ago