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vivado [14]
3 years ago
14

What is the average isotopic mass of an element that has the following abundances in nature? Isotopic mass: X20 X22 X23 Abundanc

es in natures: 90.5 % 8.0 % 1.5 %
Chemistry
1 answer:
Paladinen [302]3 years ago
8 0

Answer: 20.2

Explanation:

Mass of isotope 1 = 20

% abundance of isotope 1 = 90.5% = \frac{90.5}{100}=0.905

Mass of isotope 2 = 22

% abundance of isotope 2 = 8.0% = \frac{8}{100}=0.08

Mass of isotope 3 = 23

% abundance of isotope 3 = 1.5% = \frac{1.5}{100}=0.015

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(20\times 0.905+(22\times 0.08)+(23\times 0.015]

A=20.2

Therefore, the average atomic mass of the element is 20.2.

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In order to derive a simplified version of the Nernst equation by assuming standard temperature we can substitute which of the f
Nikitich [7]

Answer:

c. 298 K

Explanation:

Nernst equation is an equation used in electrochemistry that relates the reduction potential of a reaction with the standard potential, temperature and concentrations of the reactants in that are been reducted and oxidized. The formula is:

E = E° - RT / nF ln [Red] / [Ox]

<em>Where R is gas constant (8.314J/molK), T is absolute temperature (In Kelvin), n are moles of electrons and F is faraday constant (K/Volt*mol)</em>

<em />

In electrochemistry, standard temperature is taken as 298K. That means by assuming standard temperature we can substitute T as:

<h3>c. 298 K</h3>
5 0
3 years ago
Which of the following is an example of radiation?
Lapatulllka [165]
I believe the best answer to that question wud be D. I cud b wrong 
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3 years ago
How many grams of PbBr2 will precipitate when excess CuBr2 solution is added to 77.0 mL of 0.595 M Pb(NO3)2 solution?Pb(NO3)2(aq
Alexxx [7]

Answer:

16.89g of PbBr2

Explanation:

First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:

Molarity of Pb(NO3)2 = 0.595M

Volume = 77mL = 77/1000 = 0.077L

Mole =?

Molarity = mole/Volume

Mole = Molarity x Volume

Mole of Pb(NO3)2 = 0.595x0.077

Mole of Pb(NO3)2 = 0.046mol

Convert 0.046mol of Pb(NO3)2 to grams as shown below:

Molar Mass of Pb(NO3)2 =

207 + 2[ 14 + (16x3)]

= 207 + 2[14 + 48]

= 207 + 2[62] = 207 +124 = 331g/mol

Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g

Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol

Equation for the reaction is given below:

Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2

From the equation above,

331g of Pb(NO3)2 precipitated 367g of PbBr2

Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2

8 0
3 years ago
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol

b) To calculate the constant we have the following expression:

lnK=-\frac{deltaG_{rxn} }{RT}

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066

c) The equilibrium pressure of O₂ over M is:

K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm

3 0
3 years ago
Which of the following statements is true?
zalisa [80]
It's A, t<span>The figure is a molecule and an element.</span>
6 0
3 years ago
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