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2 years ago

Differences between allotropy and isotopy

Chemistry

1 answer:

raketka [301]2 years ago

3 0

Answer:

Property of an element by virtue of which it exists in two or more forms which differ only in their physical properties is known as allotropy. Allotropes are the different physical forms in which the element can exist. Allotropes are different physical forms of the same element.

Also-

Allotropes are different forms of the same element in the molecular level. Isotopes are different forms of atoms of the same chemical element. The key difference between allotropes and isotopes is that allotropes are considered at the molecular level, whereas isotopes are considered at the atomic leve

Explanation:

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Marizza181 [45]

Answer:

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Explanation:

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2 years ago

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D. is the right answer

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3 years ago

Whats the formula for selenium pentaoxide, iodine trichloride, zinc (1) nitride, chromium (III) bicarbonate, and there molar mas

Phoenix [80]

Answer:

1) Se2O5

2) I2O6

3)Zn3n2

4) Cr(HCO3)3

Explanation:

selenium pentaoxide (= also called diselenium pentoxide)

= Se2O5

⇒ Se = 78.97 g/mol

⇒ O = 16 g/mol

⇒ 2*78.97 + 5*16 = 237.94 g/mol

iodine trichloride

= I2O6

⇒ I = 126.9 g/mol

⇒ Cl = 35.45 g/mol

⇒ 2* 126.9 + 6 * 35.45 = 466.5 g/mol

zinc (1) nitride does not exist (it's Zinc(ii)nitride

The oxidation number for zinc is always 2

Zn3n2

⇒ Zn = 65.38 g/mol

⇒ N = 14 g/mol

⇒3*65.38 + 2* 14 = 224.14 g/mol

chromium (III) bicarbonate

Cr(HCO3)3

⇒ Cr = 52 g/mol

⇒ H = 1.01 g/mol

⇒ C = 12 g/mol

⇒ O = 16 g/mol

52 + 3*1.01 + 3*12 + 6*16 = 235.03 g/mol

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2 years ago

The oldest rock on the bottom and newest on the top is called ____________

AysviL [449]

Answer:

Law of position

Explanation:

6 0

3 years ago

Read 2 more answers

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

3 years ago

Differences between allotropy and isotopy​

Differences between allotropy and isotopy