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ivann1987 [24]
3 years ago
8

Write the appropriate symbol for each of the following isotopes: (a) Z 11, A 23; (b) Z= 28, A= 64; (c) Z= 50, A =115; (d) Z= 20,

A= 42.
Chemistry
2 answers:
prohojiy [21]3 years ago
6 0

Explanation:  

Z = atomic mass of the element and  , A = atomic mass of the element .

a) Z = 11, A =  23

Element = Sodium

  symbol: ²³₁₁Na  .

b) Z = 28, A =  64

Element = Nickel

  symbol: ⁶⁴₂₈Ni  .

c) Z = 50, A = 115

Element = tin

  symbol: ¹¹⁵₅₀Sn  .

d) Z = 20, A = 42

Element = Calcium

  symbol: ⁴²₂₀Ca .

Elden [556K]3 years ago
4 0

Answer:

(a) ^{23}_{11}Na

(b) ^{64}_{28}Ni

(c) ^{115}_{50}Sn

(d) ^{42}_{20}Ca

Explanation:

Hello,

In this case, Z stands for the isotope's atomic number and A for the isotope's atomic mass, in such a way, looking for each isotope one concludes that the proper symbol, in the form of ^{A}_{Z}Element turn out being:

(a) ^{23}_{11}Na

(b) ^{64}_{28}Ni

(c) ^{115}_{50}Sn

(d) ^{42}_{20}Ca

Best regards.

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A sample weighing 3.110 g is a mixture of Fe 2 O 3 (molar mass = 159.69 g/mol) and Al 2 O 3 (molar mass = 101.96 g/mol). When he
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Answer:

The mass fraction of ferric oxide in the original sample :\frac{723}{3110}

Explanation:

Mass of the mixture = 3.110 g

Mass of Fe_2O_3=x

Mass of Al_2O_3=y

After heating the mixture it allowed to react with hydrogen gas in which all the ferric oxide reacted to form metallic iron and water vapors where as aluminum oxide did not react.

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

Mass of mixture left after all the ferric oxide has reacted = 2.387 g

Mass of mixture left after all the ferric oxide has reacted = y

x=3.110 g- y=3.110 g - 2.387 g = 0.723 g

The mass fraction of ferric oxide in the original sample :

\frac{0.723 g}{3.110 g}=\frac{723}{3110}

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The size (radius) of an oxygen molecule is about 2.0 ×10−10m. Make a rough estimate of the pressure at which the finite volume o
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Answer:

Explanation:

We can calculate the volume  of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.

We know that volume=4/3×πr³

volume =4/3×π(2.0×10⁻¹⁰m)³

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Volume of oxygen molecule=33.40×10⁻³⁰m³

we know the ideal gas equation as:

PV=nRT

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R=k×Na

PV=n×k×Na×T

n×Na=N

PV=Nkt

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v is volume  of gas

T is temperature of gas

N is numbetr of molecules

Na is avagadros number

k is boltzmann constant =1.38×10⁻²³J/K

R is real gas constant

So to calculate pressure using the  formula;

PV=NkT

P=NkT/V

Since there is only one molecule of oxygen so N=1

P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³

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