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Drupady [299]
3 years ago
5

How many moles is in 2.52*10^24 molecules of water?

Chemistry
1 answer:
nataly862011 [7]3 years ago
3 0
Hi friend
--------------
Your answer
-------------------

Water = H2O

Number of molecules in one mole of water = 6.022 × 10²³ [Avogadro's constant]

Given number of molecules = 2.52 × 10²³

So,
------

Number of moles =
\frac{2.52 \times 10 {}^{24} }{6.022 \times 10 {}^{23} }  \\  \\  = 4.184 \: (approximately)

HOPE IT HELPS
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Answer:

Take E(alpha particle energy) = 5.5 MeV (5.5x106x1.6x10-19)

If the charge on the lead nucleus is +82e(atomic number of lead is 82) = +82x1.6x10-19 C and the charge on the alpha particle is +2e = 2x1.6x10-19 C

Using dc = (1/4πεo)qQ/Eα we have

dc = [9x10^9x(2x1.6x10-19x82x1.6x10-19)]/5.5x10-13 = 6.67x10^-13m. = 6.67 x 10^-13 x 10^15 = 6.67 x 10^2fm

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Which statement describes a scientific theory?
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C) is the best description, although a scientific theory is not necessarily a "fact". They are, however, based on evidence.

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D) is a scientific "hypothesis"- an explanation for an observed phenomena that is yet to be proven.
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A neutral atom in the ground state of Sulfur has its outer most valence electron in which orbital? f d
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<u>Answer:</u> The outermost valence electron enters the p orbital.

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Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti
gayaneshka [121]

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Explanation :

The given chemical reaction is:

CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)

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\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]

where,

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Now put all the given values in this expression, we get:

\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]

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R = gas constant  = 8.314 J/L.atm

T = temperature  = 30.0^oC=273+30.0=303K

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-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K

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