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irinina [24]
3 years ago
6

In 1961, the reference for the atomic mass unit (amu or u) was changed from naturally occurring oxygen with a value of 16.000 u

to one based on a single isotope of carbon (carbon-12). Why was it changed
Chemistry
1 answer:
vovangra [49]3 years ago
3 0

To determine the mass of an atom in terms of atomic mass unit (a.m.u.) the standard should have the fixed mass otherwise the calculated mass of the other atoms will get error. Initially oxygen was used as the reference standard for the calculation of other atoms mass. But after the invention of different isotopes of oxygen i.e. O^{17} and O^{18} which ratios with respect to O^{16} is 1.002 the mass of standard oxygen became unstable. Thus the mass of the new atom have error. Thus in 1961 the standard reference is canged to carbon-12. After that onward the mass of the other atoms are measured with respect to carbon.

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Conjugated dienes routinely undergo 1,2 and 1,4 addition reactions with a variety of electrophilic reagents; this suggests that electrophilic reagents are likely intermediates during these reactions.

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Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): 2H2O2
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Answer:

23.0733 L

Explanation:

The mass of hydrogen peroxide present in 125 g of 50% of hydrogen peroxide solution:

Mass=\frac {50}{100}\times 125\ g

Mass = 62.5 g

Molar mass of H_2O_2 = 34 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus, moles are:

moles= \frac{62.5\ g}{34\ g/mol}

moles= 1.8382\ mol

Consider the given reaction as:

2H_2O_2_{(aq)}\rightarrow2H_2O_{(l)}+O_2_{(g)}

2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen gas.

Also,

1 mole of hydrogen peroxide decomposes to give 1/2 mole of oxygen gas.

So,

1.8382 moles of hydrogen peroxide decomposes to give \frac {1}{2}\times 1.8382 mole of oxygen gas. Moles of oxygen gas produced = 0.9191 molGiven: Pressure = 746 torr&#10;The conversion of P(torr) to P(atm) is shown below:&#10;[tex]P(torr)=\frac {1}{760}\times P(atm)

So,

Pressure = 746 / 760 atm = 0.9816 atm

Temperature = 27 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (27 + 273.15) K = 300.15 K

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9816 atm × V = 0.9191 mol × 0.0821 L.atm/K.mol × 300.15 K

<u>⇒V = 23.0733 L</u>

8 0
3 years ago
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