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Lynna [10]
3 years ago
5

Naturally occurring copper exists in two isotopic forms: 63Cu and 65Cu. The atomic mass of copper is 63.55 amu. What is the appr

oximate natural abundance of 63Cu?
Chemistry
1 answer:
Mariulka [41]3 years ago
3 0
To calculate the average atomic weight, each exact atomic weight is multiplied by its percent abundance, then, add the results together.  If the natural abundance of 63Cu is assigned x, the natural abundance of 65Cu is 1-x (the two abundance always add up to 1). So the solution is: (63)(x)+(65)(1-x) = 63.55, 63x+65-65x=63.55, x=0.725=72.5%. The natural abundance of 63Cu is 72.5%, that of 65Cu is 1-72.5%=27.5%.
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What is the molecular formula of a compound with the empirical formula
ipn [44]

Answer:  C_2H_2O_2

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

The empirical formula is CHO

The empirical weight of CHO = 1(12)+1(1)+1(16)= 29 g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula:

n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{60}{29}=2

The molecular formula will be=2\times CHO=C_2H_2O_2

Thus molecular formula will be C_2H_2O_2

5 0
3 years ago
What does AS> O mean?
Vlad1618 [11]

Answer:

ΔS> 0 means Letter A

Explanation:

Processes that involve an increase in entropy of the system (ΔS > 0) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:

\displaystyle \Delta {S}_{\text{univ}}=\Delta {S}_{\text{sys}}+\Delta {S}_{\text{surr}}

5 0
3 years ago
Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir
guajiro [1.7K]

Answer:

y_{O2} =4.3%

Explanation:

The ethanol combustion reaction is:

C_{2}H_{5} OH+3O_{2}→2CO_{2}+3H_{2}O

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}

Dividing the previous equation by x:

1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles

Calculate the number of moles of CO2 and water considering the same:

0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)

0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)

The total number of moles at the reactor output would be:

N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)

So, the oxygen mole fraction would be:

y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3%

6 0
3 years ago
How would an element on the left side of row 2 of the periodic table differ from an element in the middle of the same row?
maw [93]

Explanation:

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7 0
3 years ago
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How many moless is 4.91x10^22 molecules of H3PO4?
Lynna [10]
1 mol = 6.023x10^23 number of molecules (Avogadro's number)

1 : 6.023x10^23
X : 4.91x10^22

(6.023x10^23)X = 4.91x10^22

X = 4.91x10^22/6.023x10^23

X = 0.082 Moles
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