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Anna11 [10]
3 years ago
13

Glacier National Park in Montana is approximately 4,100 ft above sea level with an atmospheric pressure of 681 torr. At what tem

perature does water (ΔHvap= 40.7 kJ/mol) boil at Glacier National Park?
Chemistry
2 answers:
777dan777 [17]3 years ago
8 0

Answer:

it is A

Explanation:

lapo4ka [179]3 years ago
6 0

Answer : The temperature of liquid is, 369.9 K

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of liquid at 373 K = 681 torr

P_2 = vapor pressure of liquid at normal boiling point = 760 torr

T_1 = temperature of liquid = ?

T_2 = normal boiling point of liquid = 373 K

\Delta H_{vap} = heat of vaporization = 40.7 kJ/mole = 40700 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760torr}{681torr})=\frac{40700J/mole}{8.314J/K.mole}\times (\frac{1}{T_1}-\frac{1}{373K})

T_1=369.907K\approx 369.9K

Hence, the temperature of liquid is, 369.9 K

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3 years ago
The nucleus of unstable _____ of an element will decay leading to emission of radiation.question 1 options:moleculescationsanion
Marta_Voda [28]

Isotopes of same element has different number of neutrons with different masses and having same number of protons and electrons.

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6 0
3 years ago
A standard solution contained 0.8 mg/mL. A student took 2 mL of the standard solution and added 10 mL of water. What is the new/
galben [10]
To get the concentration of the second solution let us use the following formulae

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7 0
3 years ago
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What is 1 third divided by 5<br>​
madreJ [45]
1/15 is the correct answer I think
5 0
3 years ago
A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g
Lynna [10]

Answer:

\%NaHCO_3=61.2\%

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

NaHCO_3+HA\rightarrow NaA+H_2O+CO_2

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%

Best regards.

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3 years ago
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