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3 years ago

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Chemistry

2 answers:

MArishka [77]3 years ago

7 0

Answer:

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Ne4ueva [31]3 years ago

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Answer:

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Explanation:

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The Ideal Gas Law is as follows: PV = nRT.

Dima020 [189]

Answer:

p=nRT/V

v=nRT/P

n=PV/RT

t=PV/nR

5 0

3 years ago

Which of the following is an example of the conversion of electrical energy to electromagnetic energy?

GuDViN [60]

The ccorrect answer is C a generator tubrine

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4 years ago

Read 2 more answers

2. (2 pts) How would you prepare 1.5 liters of 2 M KCI (MW=74.55 g/mol)

ra1l [238]

Answer:

Dissolve 226 g of KCl in enough water to make 1.5 L of solution

Explanation:

1. Calculate the moles of KCl needed

$n = \text{1.5 L} \times \dfrac{\text{2 mol}}{\text{1 L}}= \text{3.0 mol}$

2. Calculate the mass of KCl

$m = \text{3.0 mol} \times \dfrac{\text{74.55 g}}{\text{1 mol}}= \text{224 g}$

3. Prepare the solution

Measure out 224 g of KCl.
Dissolve the KCl in a few hundred millilitres of distilled water.
Add enough water to make 1.5 L of solution. Mix thoroughly to get a uniform solution.

8 0

3 years ago

The two naturally occurring isotopes of antimony, 121Sb (57.21 percent) and 123Sb (42.79 percent), have masses of 120.904 and 12

Alex

Answer:

The correct answer is option c.

Explanation:

Formula used to determine an average atomic mass :

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

Mass of isotope Sb-121 = 120.904 amu

Fractional abundance of Sb-121 = 57.21% = 0.5721

Mass of isotope Sb-123 = 122.904 amu

Fractional abundance of Sb-123 = 42.79% = 0.4279

Average atomic mass of Sb:

$120.904 amu\times 0.5721+ 122.904 amu\times 0.4279=121.7598 amu \approx 121.76 amu$

7 0

3 years ago

The radius of an indium atom is 0.163 nm. If indium crystallizes in a face-centered unit cell, what is the length of an edge of

adoni [48]

<u>Answer:</u> The edge length of the unit cell is 0.461 nm

<u>Explanation:</u>

We are given:

Atomic radius of iridium = 0.163 nm

To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:

$a=2\sqrt{2}R$

Putting values in above equation, we get:

$a=2\sqrt{2}\times 0.163=0.461nm$

Hence, the edge length of the unit cell is 0.461 nm

8 0

3 years ago