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astraxan [27]
3 years ago
13

The molar mass of Beryllium (Be)

Chemistry
1 answer:
e-lub [12.9K]3 years ago
8 0
The molar mass of Beryllium is 9.012182 u (symbol I can't put down) 0.000003 U
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How many nitrogen atoms are in 2.79 mo.of NH3
alexandr1967 [171]
1 mol of any particles has 6.02 * 10 ²³ particles.

If we look at 1 NH3 (1 mol NH3 or 1 molecule NH3), we can see that 1 molecule NH3 has 1 atom of N and 3 atoms of H; also 1 mole of NH3 has 1 mole of N atoms and 3 moles of H atoms.

So, 1 mol of NH3 has 1 mol of N atoms,
 and 2.79 mol NH3 have 2.79 mol of N atoms.

2.79 mol of N atoms* 6.02 * 10 ²³ N atoms/ 1 mol N atoms = 1.68*10²⁴ N-atoms 

 Answer is 1.68*10²⁴ N-atoms. 
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Nuclear energy originates from splitting of uranium atoms a process called fission. This generates heat to produce steam.

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Elements with atomic numbers 7 and 83 have how many valence electrons? *
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2 years ago
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]

Answer : The value of K for this reaction is, 2.6\times 10^{15}

Explanation :

The given chemical reaction is:

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

Now we have to calculate value of (\Delta G^o).

\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,

\Delta G^o = Gibbs free energy of reaction = ?

n = number of moles

\Delta G^0_{(HCH_3CO_2(g))} = -389.8 kJ/mol

\Delta G^0_{(CH_3OH(g))} = -161.96 kJ/mol

\Delta G^0_{(CO(g))} = -137.2 kJ/mol

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)]

\Delta G^o=-89.4kJ/mol

The relation between the equilibrium constant and standard Gibbs, free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs, free energy  = -89.4 kJ/mol = -89400 J/mol

R = gas constant  = 8.314 J/L.atm

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

K = equilibrium constant = ?

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

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

K=2.6\times 10^{15}

Thus, the value of K for this reaction is, 2.6\times 10^{15}

4 0
3 years ago
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