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3 years ago

explain why the crystal structure of barium fluoride, BaF2, differs from that of magnesium fluoride, MgF2

Chemistry

1 answer:

Leto [7]3 years ago

8 0

The crystal structure of barium fluoride BaF₂ and magnesium fluoride MgF₂ is cubic and tetragonal lattice respectively.

There are difference between the cubic and tetragonal lattice of any crystal. For cubic lattice a = b = c and α = β = γ = 90⁰ and for tetragonal lattice a = b ≠ c and α = β = γ = 90⁰ (where a, b and c are the edge length of the lattice and α, β, γ angle of the crystal lattice).

The difference of the structure depends on the size of the ions present in the compound. In BaF₂, the ions are Ba²⁺ and F⁻. In MgF₂, the ions are Mg²⁺ and F⁻.

Although the size of the anion in both the crystal is same, there is a huge difference between the size of the cation, which affect the crystal lattice of the compound.

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Ainat [17]

Answer:

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Explanation:

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As protein molecules in the given experiment is radio labelled that"s why the pellet fraction will be radio labeling part because the heavy protein molecules will be present in the pellet fraction.

5 0

3 years ago

A gas is contained in a cylinder with a volume of 2.9 L at a temperature of 32.7oC and a pressure of 645.3 torr. The gas is then

stealth61 [152]

Answer: 41 atm

Explanation:

Given that:

Original Volume of gas V1 = 2.9L

Temperature T1 = 32.7°C

Convert Celsius to Kelvin

(32.7°C + 273 = 305.7K)

Pressure P1 = 645.3 torr

New Volume V2 = 0.23 L

New temperature T2 = 894.7°C

Convert Celsius to Kelvin

(894.7°C + 273 = 1167.7K)

New pressure = ?

Then, apply the combined gas equation

(P1V1)/T1 = (P2V2)/T2

(645.3 torr x 2.9L)/305.7K = (P2 x 0.23L)/1167.7K

1871.37 / 305.7 = 0.23P2 / 1167.7

To get P2, Cross multiply

1871.37 x 1167.7 = 305.7 x 0.23P2

2185198.749 = 70.311P2

Divide both sides by 70.311

2185198.749/70.311 = 70.311P2/70.311

31079.045 torr = P2

Now, convert pressure in torr to atmosphere

Since 760 torr = 1 atm

31079.045 torr = Z

cross multiply

760 torr x Z = 31079.045 torr x 1 atm

Z = 31079.045 torr / 760 torr

Z = 40.89 atm (Round to the nearest whole number as 41 atm)

Thus, new pressure of gas is 41 atm

3 0

3 years ago

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2

IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at $75^{o}F$ and 14.6 psia.

$\varepsilon$ = 0.00015 ft, Flow rate, (Q) = 48000 $ft^{3}/m$

(a) Formula to calculate hydraulic radius $(r_{H})$ is as follows.

$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

= $\frac{2 \times 1}{2(1) + 2(2)}$

= $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

$D_{eq} = 4 \times r_{H}$

= $4 \times \frac{1}{3} ft$

= $\frac{4}{3}$ ft

(b) Formula for velocity floe is as follows.

Q = VA

V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

5 0

3 years ago

There are 70 fisherman and 5 sea bass in the lake. Are sea bass considered a limiting factor?

Olin [163]

Answer:

yes

Explanation:

8 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

EleoNora [17]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$

4 0

3 years ago