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Pachacha [2.7K]
3 years ago
10

Combustion of ethylene glycol leads to a change in internal energy (\DeltaU) of –1189 kJ at 298 K. What is the corresponding cha

nge in enthalpy in kJ\DeltaH? HOCH_2CH_2OH(l) + 5/2 O_2(g) → 2 CO_2(g) + 3 H_2O(l) \DeltaU = –1189 kJ
Chemistry
1 answer:
ddd [48]3 years ago
7 0

Answer:

-1190.24 kJ

Explanation:

The enthalpy change in a chemical reaction that produces or consumes  gases is given by the expression:

ΔH = ΔU + Δngas RT

where  Δn gas is the change of moles of gas, R is the gas constant,and T is temperature.

Now from the given   balanced chemical reaction, the change in number of mol gas is equal to:

Δn gas = mole gas products - mole gas reactants =  2 - 5/2 = -1/2 mol

Sionce we know ΔU and the temperature (298 K), we are in position to calculate the change in enthalpy.

ΔH = -1189 x 10³ J + (-0.5 mol ) 8.314 J/Kmol x 298 K

ΔH = -1.190 x 10⁶ J = -1.190 x 10⁶ J x 1 kJ/1000 J = -1.190 x 10³ J

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Extend your thinking: The slow decay of radioactive materials can be used to find the age of rocks, fossils, and archaeological
lisov135 [29]

The age in years of the Egyptian papyrus, the Aboriginal charcoal, the Mayan headdress, and the Neanderthal skull are 4000 years, 13106.5years , 2040 years, and 30353 years respectively.

<h3>What is the half-life of a radioactive material?</h3>

The half-life of a radioactive material is the time taken for half the atoms present in the material to decay or disintegrate.

The half-life, t_\frac{1}{2}, the age, t, and amount remaining, A_{r}, of a radioactive material are related by the formula below:

  • t = \frac{t_\frac{1}{2}*A_{r}}{-ln2}

Half-life of carbon-14 = 6000 years

For the Egyptian papyrus with 63% of its original carbon-14 atoms:t = \frac{6000*0.63}{-0.63} = 4000\:years

For the Aboriginal charcoal with 22% of its original carbon-14 atoms:

t = \frac{6000*0.22}{-0.63} = 13106.5\:years

For the Mayan headdress with 79% of its original carbon-14 atoms:

t = \frac{6000*0.79}{-0.63} = 2040\:years

Neanderthal skull with 3% of its original carbon-14 atoms:

t = \frac{6000*0.03}{-0.63} = 30353\:years

Therefore, the age in years of the Egyptian papyrus, the Aboriginal charcoal, the Mayan headdress, and the Neanderthal skull are 4000 years, 13106.5years , 2040 years, and 30353 years respectively.

Learn more about half-life at: brainly.com/question/26689704

#SPJ1

6 0
2 years ago
Ik i already asked this but i need diff point of views
salantis [7]

Answer:

Is it prescribe to you?If so than yes if not then no need to

Explanation:

5 0
3 years ago
Read 2 more answers
Calculate the vapor pressure (in torr) of a 35°C solution of ethanol and propanol where the mole fraction of propanol is .75.
astraxan [27]

Answer:

i wish i know

Explanation:

4 0
3 years ago
What is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa
Nezavi [6.7K]

60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.

Explanation:

Data given:

V1 = 75 ml

T1 = 30 Degrees or 273.15 + 30 = 303.15 K

P1 = 91 KPa

V2  =?

P2 = 1 atm or 101.3 KPa

T2 = 273.15 K

At STP the pressure is 1 atm and the temperature is 273.15 K

applying Gas Law:

\frac{P1VI}{T1}= \frac{P2V2}{T2}

putting the values in the equation of Gas Law:

V2 = \frac{P1V1T2}{P2T1}

V2 = \frac{91 X 75 X 273.15}{303.15 X 101.3}

V2 = 60.7 ml

at STP the volume of carbon dioxide gas is 60.7 ml.

4 0
3 years ago
A phosphate buffer solution (25.00 mL sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The compone
AysviL [449]

Answer:

0,07448M of phosphate buffer

Explanation:

sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:

Na₂HP + HCl ⇄ NaH₂P + NaCl <em>(1)</em>

NaH₂P + HCl ⇄ H₃P + NaCl <em>(2)</em>

The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:

0,01862L of HCl×\frac{0,1000mol}{L}= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.

The concentration is:

\frac{1,862x10^{-3}moles}{0,02500L} = <em>0,07448M of phosphate buffer</em>

<em></em>

I hope it helps!

7 0
3 years ago
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