Which metallic crystal structure has a coordination number of 8

Describe three different scenarios where a chemical process or reaction shows an increase in entropy.

Jet001 [13]

The request is characterized as knowing where things are and having the capacity to discover and utilize the things.
In a compound procedure, there is more issue, more entropy when the particles
1. warm up, increment in temperature. The atoms are more disorganized
2. get stirred up and must be isolated with exertion. Bedlam.
3. state changes, dissolves, vaporizes. The atoms are more turbulent
4. respond to frame a pack of various particles. More disorder

4 0

3 years ago

Determine the empirical formula of a compound having the following percent composition by mass: K: 24.74%; Mn: 34.76%; O: 40.50%

Dovator [93]

Answer: The empirical formula of the compound is $KMnO_4$

Explanation:

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Let the mass of the compound be 100 g

Given values:

% of K = 24.7%

% of Mn = 34.76%

% of O = 40.50%

Mass of K = 24.7 g

Mass of Mn = 34.76 g

Mass of O = 40.50 g

To calculate the empirical formula of a compound, few steps need to be followed:

Step 1: Calculating the number of moles of each element

We know:

Molar mass of K = 39.10 g/mol

Molar mass of Mn = 54.94 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of K}=\frac{24.7g}{39.10g/mol}=0.632 mol$

$\text{Moles of Mn}=\frac{34.76g}{54.94g/mol}=0.633 mol$

$\text{Moles of O}=\frac{40.50g}{16g/mol}=2.53 mol$

Step 2: Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.632 moles

$\text{Mole fraction of K}=\frac{0.632}{0.632}=1$

$\text{Mole fraction of Mn}=\frac{0.633}{0.632}=1$

$\text{Mole fraction of O}=\frac{2.53}{0.632}=4$

Step 3: Writing the mole fraction as the subscripts of each of the element

The empirical formula of the compound becomes $K_1Mn_1O_4=KMnO_4$

Hence, the empirical formula of the compound is $KMnO_4$

4 0

3 years ago

plz help me asap and find the surface area here are the numbers 8 10 10 5 5 10 10. here is the other one 10 10 12 8 12 10 10 8

statuscvo [17]

I'm pretty sure if it was a cube, this would be impossible to find since each side of a cube is a square and their all equal. I use a calculator called mathsoup and symbolab. Maybe those can help

4 0

3 years ago

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The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F.

exis [7]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. $Na_{2} CO_{3}$

Major Species at B:

1. $Na_{2} CO_{3}$

2. $NaHCO_{3}$

Major Species at C:

1. $NaHCO_{3}$

Major Species at D:

1. $NaHCO_{3}$

2. $H_{2}CO_{3}$

Major Species at E:

1. $H_{2}CO_{3}$

Major Species at F:

1. $H_{2}CO_{3}$

b) pH calculation:

At Halfway point B:

pH = pK $a_{1}$ + log[ $CO_{3}$ $.^{-2}$ ]/[H $CO_{3}$ $.^{-1}$ ]

pH = pK $a_{1}$ = 6.35

Similarly, at halfway point D.

At point D,

pH = pK $a_{2}$ + log [H $CO_{3}$ $.^{-1}$ ]/[H2 $CO_{3}$ ]

pH = pK $a_{2}$ = 10.33

8 0

3 years ago

The molar solubility of ba3(po4)2 is 8.89 x 10-9 m in pure water. calculate the ksp for ba3(po4)2. the molar solubility of ba3(p

Flura [38]

Answer is: The molar solubility of ba3(po4)2 is 6.00 x 10-39.
Balanced chemical reaction: Ba₃(PO₄)₂(s) → 3Ba²⁺(aq) + 2PO₄³⁻(aq).
s(Ba₃(PO₄)₂) = 8.89·10⁻⁹ M.
[Ba²⁺] = 3s(Ba₃(PO₄)₂) = 3s.
[PO₄³⁻] = 2s.
Ksp = [Ba²⁺]³ · [PO₄³⁻]².
Ksp = (3s)³ · (2s)².
Ksp = 108s⁵.
Ksp = 108·(8.89·10⁻⁹ M)⁵.
Ksp = 108 · 5.55·10⁻⁴¹ = 6·10⁻³⁹.

3 0

3 years ago

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