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3 years ago

What is the percent by mass of oxygen in propanal, CH3CH2CHO?(1) 10.0% (3) 38.1%

2 answers:

8 0

17.6% oxygen
% mass = molar mass of part over molar mass of whole,
molar mass of oxygen is 16
molar mass of carbon is 12
molar mass of hydrogen is 1
so (16) / (16+12*6+3) = 16/91 = 17.58%

leonid [27]3 years ago

5 0

Answer : The percent by mass of oxygen in propanal is, 27.58 %

Solution : Given,

Molar mass of carbon = 12 g/mole

Molar mass of hydrogen = 1 g/mole

Molar mass of oxygen = 16 g/mole

First we have to calculate the molar mass of propanal, $CH_3CH_2CHO$

Molar mass of propanal, $CH_3CH_2CHO$ = $3(12)+6(1)+16=58g/mole$

Now we have to calculate the percent by mass of oxygen in propanal.

$\text{Percent by mass of oxygen}=\frac{\text{Mass of oxygen}}{\text{Mass of propanal}}\times 100=\frac{16}{58}\times 100=27.58\%$

Therefore, the percent by mass of oxygen in propanal is, 27.58 %

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3 years ago

Nina ran into a rough patch of pavement, but she thought she could ride right over it. Instead, the skateboard stopped suddenly

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Answer:

The rough pavement acted as an unbalanced force that was able to stop the skateboard.

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4 years ago

Which correctly lists the following in order of increasing entropy? i. 1 mol of HCl (g) at 50 ° C ii. 1 mol of NaCl (s) at 25 °

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Answer:

ii. < iv. < i. < iii.

Explanation:

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At first, the entropy is lower in solid-phase systems since their particles are molecularly closer. Next, by knowing that the higher the temperature, the higher the entropy, the gaseous HCl at 25 ºC occupies the second place. After that, it is clear 2 moles of HCl have more entropy than just 1 since the larger the amount of mater the higher the entropy.

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3 years ago

The molar volume of a certain solid is 142.0 cm3 mol−1 at 1.00 atm and 427.15 k, its melting temperature. the molar volume of th

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Answer:

$\Delta _{fus}H=3255.3J/mol$

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Explanation:

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Clausius Clapeyron equation is suitable in this case, since it allows us to relate the P,T,V behavior along the described melting process and the associated energy change. Such equation is:

$\frac{dp}{dT}=\frac{\Delta _{fus}H}{T\Delta _{fus}V}$

As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:

$p_2-p_1=\frac{\Delta _{fus}H}{\Delta _{fus}V}ln(\frac{T_2}{T_1} )$

Solving for the enthalpy of fusion we obtain:

$\Delta _{fus}H=\frac{(p_2-p_1)(V_2-V1)}{ln(\frac{T_2}{T_1})} =\frac{(11.84atm-1.00 atm)(156.6cm^3/mol-142.0cm^3/mol)}{ln(\frac{429.26K}{427.15K} )} \\\\\Delta _{fus}H=32127.3atm*cm^3/mol*\frac{101325Pa}{1atm}*(\frac{1m}{100cm} )^3\\\Delta _{fus}H=3255.3J/mol$

Finally the entropy of fusion is given by:

$\Delta _{fus}S=\frac{\Delta _{fus}H}{T_1} =\frac{3255.3J/mol}{427.15K}\\ \\\Delta _{fus}S=7.62\frac{J}{mol*K}$

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3 years ago