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avanturin [10]
4 years ago
8

The radius of a single atom of a generic element x is 197 picometers (pm and a crystal of x has a unit cell that is face-centere

d cubic. calculate the volume of the unit cell.
Chemistry
1 answer:
Vinvika [58]4 years ago
7 0
In a face centered cubic unit cell, the volume occupied by the particles of the substance is about 74% of the total unit cell. First, we find the volume of a single atom assuming it to be sphere:
Volume(atom) = 4/3 x π x r³
Volume(atom) = 4/3 x π x (197 x 10⁻¹²)³
Volume(atom) = 2.87 x 10⁻²⁹ m³

There are a total of 4 atoms in a FCC unit cell, so the total volume occupied by atoms is:
2.87 x 10⁻²⁹ x 4
= 1.15 x 10⁻²⁸ m³

Volume of cell = (1.15 x 10⁻²⁸ ) / 0.74
Volume of cell = 1.55 x 10⁻²⁸ m³
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Using the van der Waals equation, calculate the pressure for a 1.25 mol sample of xenon contained in a volume of 1.000L at 75°C
alina1380 [7]

Answer:

ABC

Explanation:

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5 0
3 years ago
On a balance, you have beakers of AgNO3AgNO3 solution and NaClNaCl solution. When mixed, they will form AgCl(s)AgCl(s). What wil
statuscvo [17]

Explanation:

According to the law of conservation of mass, mass can neither be created nor it can be destroyed. But it can be simply transformed from one form to another.

Therefore, when AgNO_{3} is added to NaCl then the compound formed will have same mass as that of reactants.

       AgNO_{3} + NaCl \rightarrow AgCl + NaNO_{3}

Total mass of reactants is (169.87 + 58.44) g/mol = 228.31 g/mol

Total mass of products is (143.32 + 84.99) g/mol = 228.31 g/mol

Thus, we can conclude that mass of the new mixture will stay the same.

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=I_%7B2%7D" id="TexFormula1" title="I_{2}" alt="I_{2}" align="absmiddle" class="latex-formula">
gogolik [260]

Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

Explanation:

An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.

A Reductant thus exactly the opposite.

Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:

I2 --> 2I-.

The oxidation number reduced from 0 to -1.

In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.

Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

7 0
3 years ago
As you move from left to right across a period, what happens to the atomic radil?
egoroff_w [7]

Answer:

Moving Across a Period

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4 0
4 years ago
Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p
Tanzania [10]

Answer :  The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)

The expression for rate of reaction :

\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}

\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}

\text{Rate of disappearance of }F=-\frac{d[F]}{dt}

\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}

\text{Rate of formation of }H=+\frac{d[H]}{dt}

\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}

Given:

-\frac{d[D]}{dt}=0.18mol/L.s

As,  

-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s

and,

+\frac{d[H]}{dt}=2\times 0.18mol/L.s

+\frac{d[H]}{dt}=0.36mol/L.s

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

5 0
3 years ago
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