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SVETLANKA909090 [29]
3 years ago
14

What is the smallest part of a compound that retains the properties of that compound?

Chemistry
1 answer:
snow_lady [41]3 years ago
5 0
<span>"A molecule represents the </span>smallest part<span> into which a </span>compound<span> can be divided and still </span>retain<span> its chemical and physical </span><span>properties" (socratic.org)</span>
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A mineral forms in water heated by magma inside Earth.
allsm [11]

Answer:

the mineral formed when a hot water solution cooled

4 0
3 years ago
Read 2 more answers
Iron has many isotopes but only 4 are found in significant amounts in naturally found mixtures. The amounts by mass percent are:
AnnZ [28]

Answer:

The average mass of iron to be is 54.76 amu.

Explanation:

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

1)

Mass of  Fe-54 isotope = 53.9396 amu

Percentage abundance of Fe-54 = 5.845%

Fractional abundance of Fe-54 = 0.05845

2)

Mass of  Fe-56 isotope = 55.9349 amu

Percentage abundance of Fe-56 = 91.754 %

Fractional abundance of Fe-56= 0.91754

3)

Mass of  Fe-57 isotope = 56.9354 amu

Percentage abundance of Fe-57 = 2.119%

Fractional abundance of Fe-57 = 0.002119

4)

Mass of  Fe-58 isotope = 57.9333 amu

Percentage abundance of Fe-58 = 0.282%

Fractional abundance of Fe-58 = 0.00282

Average atomic mass of iron :

=53.9396 amu\times 0.05845+55.9349 amu\times 0.91754 +56.9354 amu\times 0.002119 + 57.9333 amu\times 0.00282=54.76 amu

The average mass of iron to be is 54.76 amu.

5 0
3 years ago
The free energy of formation of nitric oxide, NO, at 1000 K (roughly the temperature in an automobile engine during ignition) is
Assoli18 [71]

<u>Answer:</u> The value of K_p for the chemical equation is 8.341\times 10^{-5}

<u>Explanation:</u>

For the given chemical equation:

N_2(g)+O_2(g)\rightarrow 2NO(g)

To calculate the K_p for given value of Gibbs free energy, we use the relation:

\Delta G=-RT\ln K_p

where,

\Delta G = Gibbs free energy = 78 kJ/mol = 78000 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = temperature = 1000 K

K_p = equilibrium constant in terms of partial pressure = ?

Putting values in above equation, we get:

78000J/mol=-(8.314J/Kmol)\times 1000K\times \ln K_p\\\\Kp=8.341\times 10^{-5}

Hence, the value of K_p for the chemical equation is 8.341\times 10^{-5}

4 0
4 years ago
10 points get it right ADVPH
Sav [38]

Answer:

Advanced Pharmaceutics Inc

Explanation:

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8 0
3 years ago
Scientific notation steps
FrozenT [24]
1identify the #
2place a decimal in a location which would make a #between 1and9
3 count the # of space values that the decimal moved
4 plug the numbers into this formula number from step 2×10 place value
example:
1) 330,000
2) 3.3 0000
3) count the places moved 5
4)plug the # in 3.30×10 to the 5th power.
7 0
3 years ago
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