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soldier1979 [14.2K]
4 years ago
8

Identify the molecules or ions below as Lewis acids, Lewis bases, or neither. If there is more than one possible site in the mol

ecule/ion, focus on the central or the charged atom. a) _________ b) _________ c) _________ Submit AnswerRetry Entire Group2 more group attempts remaining Show HintPreviousNext

Chemistry
1 answer:
Varvara68 [4.7K]4 years ago
8 0

Answer:

This question is incomplete because of the absence of the molecules been referred to in the question. However, the molecules been referred to in the question is in the attachment below

Explanation:

First of all, we have to define lewis acid and lewis base.

A lewis acid is a substance that has the capacity to accept a pair of electrons. For example, H⁺. While a lewis base is a substance that is capable of donating a pair of electrons (to a lewis acid). For example, OH⁻.

Going by the definitions above, we can deduce the type of substance the molecules in the question (attachment) are

a. BF₃ (boron trifluoride) is a lewis acid because the central atom (as suggested to be focused on in the question) which is <u>boron, has the capacity to accept a pair of electrons</u>.

b. The central atom (carbon) in the compound (2-methyl propane) <u>also has the capacity to accept a pair of electrons</u> since it is positively charged (is electron deficient). Hence, it is a lewis acid.

c. The functional group in this compound (trimethyl borate) is the ether. Ethers are generally lewis bases because the oxygen atom in an ether can donate a pair of electrons from it's lone pair. However, the <u>presence of boron which is central to the compound</u> shows it is also a lewis acid, although weak (majorly due to the presence of the ethers).

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What is the reaction corresponding to the standard enthalpy of formation of K2SO3(s)?
nalin [4]

Answer:

  • 2K(s) + (1/8) S₈ (s) + (3/2) O₂(g)  →  K₂SO₃ (s)

Explanation:

The<em> standard enthalpy of formation </em>of a substance is the change in enthalpy that happens when one mole of the substance is formed from the elements in their standard states.

Thus, to calculate the standard state of formation of a compound you must:

  • 1. Identify the elements that form the compound
  • 2. Identify the standard form of each element
  • 3. Set the equation to form one mole of the compound, which may require to use fractional coefficients for some of the elements.

Applying that to our compound K₂SO₃

<u>1. Elements:</u>

  • potassium, K;
  • sulfur, S; and
  • oxygen, O.

<u />

<u>2. Standard forms of the elements:</u>

  • potassium: solid, K(s)
  • sulfur: solid, octatomic molecules, S₈ (s)
  • oxygen: diatomic gas, O₂(g)

<u>3. Reaction:</u>

  • K(s) + S₈ (s) + O₂(g)  →  K₂SO₃ (s)

Balance, keeping one mole of  K₂SO₃. You will need to use fractional coefficients for some elements:

  • 2K(s) + (1/8) S₈ (s) + (3/2) O₂(g)  →  K₂SO₃ (s) ← answer

8 0
3 years ago
Approximately how many ice cubes must melt to cool 650 milliliters of water from 29°C to 0°C? Assume that each ice cube contains
qwelly [4]

Answer : The number of ice cubes melt must be, 13

Explanation :

First we have to calculate the mass of water.

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}

Density of water = 1.00 g/mL

Volume of water = 650 mL

\text{Mass of water}=1.00g/mL\times 650mL=650g

Now we have to calculate the heat released on cooling.

Heat released on cooling = m\times c\times (T_2-T_1)

where,

m = mass of water = 650 g

c = specific heat capacity of water = 4.18J/g^oC

T_2 = final temperature = 29^oC

T_2 = initial temperature = 0^oC

Now put all the given values in the above expression, we get:

Heat released on cooling = 650g\times 4.18J/g^oC\times (29-0)^oC

Heat released on cooling = 78793 J = 78.793 kJ   (1 J = 0.001 kJ)

As, 1 ice cube contains 1 mole of water.

The heat required for 1 ice cube to melt = 6.02 kJ

Now we have to calculate the number of ice cubes melted.

Number of ice cubes melted = \frac{\text{Total heat}}{\text{Heat for 1 ice cube}}

Number of ice cubes melted = \frac{78.793kJ}{6.02kJ}

Number of ice cubes melted = 13.1 ≈ 13

Therefore, the number of ice cubes melt must be, 13

3 0
4 years ago
Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at 623 oC (this temperature is below the m
Degger [83]

Answer:

5.9 × 10^-6.

Explanation:

In the arrangements of crystal solids there is likely going to be an imperfection or defect and one of the defect or imperfections in the arrangements of solids is known as the Schottky defects. The Schottky defects is a kind of lattice arrangements imperfection that occurs when positively charged ions and negatively charged ions leave their position.

So, let us delve right into the solution of the question. We will be making use of the formula below;

Wb/ W = e^ - c/ 2kT.

Where Wb/ W= fraction of lattice sites, c= energy for defect formation = 1.86 eV, and T = temperature= 623° C= 896 k.

So, Wb/ W = e ^ -1.86/ (2 × 896 × 8.62 × 10^ -5).

Wb/ W= 0.000005896557435956372.

Wb/ W=5.9 × 10^-6.

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OlgaM077 [116]

Answer:

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Explanation:

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