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just olya [345]
3 years ago
8

An element has two naturally occurring isotopes, X-85 with a mass of 84.9118 amu and a natural abundance of 72.17%, and X-87 wit

h a mass of 86.9092 amu and a natural abundance of 27.83%. Calculate the atomic mass of this element.
Chemistry
1 answer:
QveST [7]3 years ago
5 0

Answer:

This element is Rubidium (Rb) and has an average atomic mass of 85.468 u

Explanation:

The average mass of an element is calculated by taking the average of the atomic masses of its stable isotopes.

The enitre atomic mass = 100 % or 1

⇒ this consists of X-85 with 72.17 % abundance  with atomic massof  84.9118 g/mol

72.17 % = 0.7217

⇒ this consists of X-87 with  27.83 % abundance with atomic mass of 86.9092 g/mol

27.83 % = 0.2783

To calculate the mass of this isotope we use the following:

0.7271 * 84.9118 + 0.2783 * 86.9092 =85.468 g/mol

This element is Rubidium(Rb) and has an average atomic mass of 85.468 u

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7 0
3 years ago
What is the pH of a substance with a hydrogen concentration of 1.0 x 10-13?
Nikolay [14]

Answer:

Favorite Answer

1.0 x10^-14 = (1.0 x 10^-13) (x)

x = 1.0 x 10^-1 = 0.1 M (this is the [OH-])

pOH = -log 0.1 = 1.0

Explanation:

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4 0
3 years ago
What is the molar mass of 81.50 g of gas
garik1379 [7]

The molar mass  of gas = 238.29 g/mol

<h3>Further explanation</h3>

Given

mass = 81.5 g

P=1.75 atm

V=4.92 L

T=307 K

Required

molar mass

Solution

The gas equation can be written  

\large{\boxed{\bold{PV=nRT}}

\tt n=\dfrac{mass}{molar~mass}

So the equation becomes :

\tt Molar~mass=\dfrac{mRT}{PV}

Input the value :

\tt M=\dfrac{81.5\times 0.082\times 307}{1.75\times 4.92}\\\\M=238.29~g/mol

5 0
3 years ago
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8 0
2 years ago
Assume the molality of isoborneol in your product is 0.275 mol/kg. What is the melting point of your impure sample given that th
aliina [53]

Answer:

168°C is the melting point of your impure sample.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = ?

Depression in freezing point = \Delta T_f

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

K_f = The freezing point depression constant

m = molality of the sample  = 0.275 mol/kg

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  non electrolyte)

\Delta T_f=1\times 40^oC kg/mol\times 0.275 mol/kg

\Delta T_f=11^oC

\Delta T_f=T- T_f

T_f=T- \Delta T_f=179^oC-11^oC=168^oC

168°C is the melting point of your impure sample.

4 0
3 years ago
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