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juin [17]
3 years ago
13

Calculate the average atomic mass of argon to two decimal places, given the following relative atomic masses and abundances of e

ach of the isotopes: argon-36 (35.97 u; 0.337%), argon-38 (37.96 u; 0.063%), and argon-40 (39.96 u; 99.600%)
Chemistry
1 answer:
user100 [1]3 years ago
4 0

Answer:

39.95 u

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})+(\frac {\%\ of\ the\ third\ isotope}{100}\times {Mass\ of\ the\ third\ isotope})

Given that:

<u>For first isotope, Argon-36: </u>

% = 0.337 %

Mass = 35.97 u

<u>For second isotope, Argon-38: </u>

% = 0.063 %

Mass = 37.96 u

<u>For third isotope, Argon-40: </u>

% = 99.600 %

Mass = 39.96 u

Thus,  

Average\ atomic\ mass=\frac{0.337}{100}\times {35.97}+\frac{0.063}{100}\times {37.96}+\frac{99.600}{100}\times {39.96}=0.1212189+0.0239148+39.80016=39.9452937

<u>Average atomic mass = 39.95 u</u>

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geniusboy [140]

Answer:

The change in temperature of a coffee cup calorimeter is 8.87°C.

Explanation:

Volume of the water = V = 150 g

Density of the water , d =1.0 g/mL

Mass of the water = M

M=d\times V=1.00 g/mL\times 150 ml =150.0 g

Mass of solution = m = M = 150.0 g

NaOH(s)\rightarrow NaOH(aq),\Delta H =-44.51 kJ/mol

Moles of NaOH = \frac{5.00 g}{40 g/mol}=0.125 mol

Energy released when 0.125 moles of NaOH added in water = Q

Q=0.125 moles\times (-44.51 kJ/mol)=-5.5638 kJ=-5,563.8 J

1 kJ = 1000 J

Heat gained by water = Q' = -Q ( conservation of energy)

Q'= 5,563.8 J

Specific heat of solution = c = 4.184 J/g°C

Change in temperature of the solution = \Delta T

Q'=mc\times \Delta T

5,563.8 J=150.0 g\times 4.184 J/g^oC\times \Delta T

\Delta T=\frac{5,563.8 J}{150.0 g\times 4.184 J/g^oC}=8.87^oC

The change in temperature of a coffee cup calorimeter is 8.87°C.

7 0
3 years ago
Make the equation equal<br> Fe(NO3)3+KSCN —&gt; Fe(SCN)3 + K(NO3)3
Advocard [28]

Answer:

Fe(NO₃)₃ + 3KSCN  →   Fe(SCN)₃ + 3KNO₃

Explanation:

Chemical equation:

Fe(NO₃)₃ + KSCN  →   Fe(SCN)₃ + KNO₃

Balanced Chemical equation:

Fe(NO₃)₃ + 3KSCN  →   Fe(SCN)₃ + 3KNO₃

Type of reaction:

It is double displacement reaction.

In this reaction the anion or cation of both reactants exchange with each other. In given reaction the cation Fe⁺³ exchange with cation K⁺.

The given reaction equation is balanced so there are equal number of atoms of each elements are present on both side of equation and completely hold the law of conservation of mass.

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

AB + CD → AC +BD

5 0
3 years ago
After the recycled plastic is released from the machine into the brick mold, explain what will happen to the kinetic energy in t
Rina8888 [55]

Answer: Depending on the data and the patterns, sometimes we can see that pattern in a simple tabular presentation of the data. Other times, it helps to visualize the data in a chart, like a time series, line graph, or scatter plot.

Explanation:

1960 5.91

1970 5.59

1980 4.83

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8 0
3 years ago
What is the atomic mass of neon to the nearest tenth<br><br><br>help me pls
motikmotik

Answer:

20.2

Explanation:

6 0
3 years ago
The density of a metal is 11.4 g/cm3. How much volume (in cm3) would a sample of 30.5 g have?
julia-pushkina [17]

<u>The Concept:</u>

We are given the density of a sample of the metal = 11.4 grams / cm³

and we need to find the volume occupied by a sample of 30.5 grams

For this solution, we will use dimensional analysis

from the given information, we can also say that the density of the metal is:

1 cm³ / 11.4 grams

If we multiply this value by 30.5 grams, the 'grams' in the numerator and the denominator will cancel out and we will be left with the volume occupied by 30.5 grams of the metal

<u>Solving for the volume:</u>

\frac{ 1 cm^{3} }{11.4 grams}  X  30.5 grams  = (30.5 / 11.4) cm³

Volume of 30.5 grams of the sample = 2.68 cm³

6 0
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