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3 years ago

7

When producing a soluble salt in a reaction between an acid and an alkali, how can you prepare dry solid crystals from the solut

ion?

1 answer:

igor_vitrenko [27]3 years ago

7 0

Answer:

aqueous solution

Explanation:

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dsp73

Answer:

1384 kJ/mol

Explanation:

The heat absorbed by the calorimeter is equal to the heat released due to the combustion of the organic compound. C is the total heat capacity of the calorimeter and Δt is the change in temperature from intial to final:

Q = CΔt = (3576 J°C⁻¹)(30.589°C - 25.000°C) = 19986.264 J

Extra significant figures are kept to avoid round-off errors.

We then calculate the moles of the organic compound:

(0.6654 g)(mol/46.07) = 0.0144432 mol

We then calculate the heat released per mole and convert to the proper units. (The conversion between kJ and J is infinitely precise and is not involved in the consideration of significant figures)

(19986.264 J)(1kJ/1000J) / (0.0144432 mol) = 1384 kJ/mol

8 0

3 years ago

Which element in each of the following sets would you expect to have the lowest IE₃?

AysviL [449]

Answer:

(a) AL

(b) Sc

(c)Al

Explanation:

Ionization Energy is the energy required to remove electrons from the outer most orbitals of atom.

The higher the electron is on energy level the farther its from nucleus and more loosely bonded thus need lesser energy.

By looking at electron configuration we can figure out which electron will need more energy.

<h3>(a)Na, Mg, Al</h3>

1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Na₁₁ ⇒ 1s², 2s², 2p⁶, 3s¹

Mg₁₂ ⇒ 1s², 2s², 2p⁶, 3s²

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

<h3>(b) K, Ca, Sc</h3>

K₁₉⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Ca₂₀⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Sc₂₁⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Sc will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

<h3>(c) Li, Al, B</h3>

Li₃ ⇒ 1s², 2s¹

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

B₅ ⇒ 1s², 2s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

3 0

3 years ago

Read 2 more answers

List the substances Ar, Cl2, CH4, and CH3COOH, in order of increasing strength of intermolecular attractions. List the substance

Viefleur [7K]

Answer:

Order of increasing strength of intermolecular attraction:

$CH_3COOH$ > $Cl_2$ > $CH_4$ > Ar

Explanation:

$CH_3COOH$ can form hydrogen bond as H atom is attached with electronegative atom O.

Rest three, $Cl_2$ , $CH_4$ , Ar are non-polar molecules.

In non-polar molecules, van der Waal's intermolecular forces of attractions exist. Hydrogen bonding is stronger intermolecular attraction then van der Waal's intermolecular forces of attraction, hence, $CH_3COOH$ has strongest intermolecular attractions.

Ar will have least intermolecular attraction, as it behaves almost as ideal gas and there is no intermolecular attraction exist between molecules of ideal gases.

Molecular size and mass of $Cl_2$ is high as compared to $CH_4$ .

van der Waals intermolecular forces of attraction increases with increase in size.

Therefore,

Order of increasing strength of intermolecular attraction will be:

$CH_3COOH$ > $Cl_2$ > $CH_4$ > Ar

8 0

3 years ago

Read 2 more answers

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

<u> </u>

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

<u> </u>

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

How did mendeleev orginally organize the element in his periodic table

horrorfan [7]

From increasing atomic mass. He also made groups and periods.

4 0

3 years ago

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