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4 years ago

What is the chemical formula of the ionic compound that forms from potassium and bromine?

Chemistry

1 answer:

vekshin14 years ago

6 0

solution:

This is an example of a binary salt. A binary salt is composed of one cation (+ charged ion) and one anion (- charged ion). Cations are usually elements of first group of the periodic table and anions are usually elements of 17th group of the periodic table. All the cations of the elements of the first group of the periodic table take the charge of 1+. Remember that a molecule has no charge, so if K(potassium) is the cation and has a charge of +1, the anion must have a charge of -1. Thus, the ionic compound formed when potassium and bromine combine is KBr.

KBr is an example of this. In order for K to lose all of its outer shell electrons, you must trace your finger from K, and go left. Nothing there! K, Potassium only has one electron in its outer shell to give away. After giving the electron away, K looks externally (in terms of electrons) like Argon (a noble gas).

Now, let's look at Br. We go right, because that is the closest way to get to the noble gases, and so have a complete outer shell with no extras (called valence electrons) to react with. Br needs exactly one electron.

Therefore, pairing K and Br results in a stable state where K donates and completely gives Br an electron, making K now positively charged (lost an electron) and Br negatively charged (gained one). This electron transfer forms an ionic bond because opposite charges attract.

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Estimate ΔH for the reaction: C2H6(g) + Cl2(g)--> C2H5Cl(g) + HCl(g) given the following average bond energies (in kJ/mol): C

Leno4ka [110]

Explanation:

The reaction equation will be as follows.

$C_{2}H_{6}(g) + Cl_{2}(g) \rightarrow C_{2}H_{5}Cl(g) + HCl(g)$

Using bond energies, expression for calculating the value of $\Delta H$ is as follows.

$\Delta H = \sum B.E_{reactants} - \sum B.E_{products}$

On reactant side, from $C_{2}H_{6}$ number of bonds are as follows.

C-C bonds = 1

C-H bonds = 6

From $Cl_{2}$ ; Cl-Cl bonds = 1

On product side, from $C_{2}H_{5}Cl$ number of bonds are as follows.

C-C bonds = 1

C-H bonds = 5

C-Cl bonds = 1

From HCl; H-Cl bonds = 1

Hence, using the bond energies we will calculate the enthalpy of reaction as follows.

$\Delta H = \sum B.E_{reactants} - \sum B.E_{products}$

= $[(1 \times 348 kJ/mol) + (6 \times 414 kJ/mol) + (1 \times 242 kJ/mol)] - [(1 \times 348 kJ/mol) + (5 \times 414 kJ/mol) + (1 \times 327 kJ/mol) + (1 \times 431 kJ/mol)]$ = -102 kJ/mol

Thus, we can conclude that change in enthalpy for the given reaction is -102 kJ/mol.

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3 years ago

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True, the hardness of all materials in relative

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The crane dropped the 75 kg mass with velocity of 3.2m/s. Calculate the kinetic energy of the mass

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3 years ago

If 150 g of phenacetin were dissolved in 100 ml of water, how much ether would be required to extract 90% of phenacetin in a sin

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Distribution coefficient in ether to water Kew = (1/90) ether /(1/1310) water = 14.56 (b) If 50 mg of phenacetin were dissolved in 100 mL of water.

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3 years ago

Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib

KengaRu [80]

Answer:

For methanol: Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol: Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol: Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

For methanol:

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol:

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol:

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

Hence energy in kJ/mol:

Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

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3 years ago