How many atoms of hydrogen are there in HCOOH2?

Sam lives in the city of Springdale and is interested in how the climate in his city compares to the climate in Ecuador. Which o

ratelena [41]

Both "Do high pressure systems prefer forming in Ecuador compared to in Springdale?" and "Are average yearly temperatures in Ecuador greater than in Springdale?" are acceptable questions. The others deal with subjectivity.

3 0

3 years ago

Why does Earthquakes occur?

Annette [7]

Shifting plates, underwater volcanoes. many different variables come in play

6 0

3 years ago

NAME THESE ALCOHOLS. CH3CH2CH2CH2OH AND CH3CH(OH)CH3

ale4655 [162]

"CH3CH2CH2CH2OH " is known by the name of "n-butanol" and "CH3CH(OH)CH3" is known by the name of "Isopropyl alcohol". These two given products are basically alcohols. I hope that this is the answer that you were looking for and the answer has actually come to your desired help. Thanks for joining brainly and getting your questions solved.

3 0

3 years ago

The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p

saul85 [17]

Answer : The vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

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 bromine at $10.0^oC$ = ?

$P_2$ = vapor pressure of propane at normal boiling point = 1 atm

$T_1$ = temperature of propane = $10.0^oC=273+10.0=283.0K$

$T_2$ = normal boiling point of bromine = $58.8^oC=273+58.8=331.8K$

$\Delta H_{vap}$ = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

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

$\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})$

$P_1=0.1448atm$

Hence, the vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

4 0

3 years ago

A mixture of helium and methane gases, at a total pressure of 821 mm Hg, contains 0.723 grams of helium and 3.43 grams of methan

Nookie1986 [14]

Answer:

For 1: The partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

For 2: The mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

Explanation:

For 1:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For helium:

Given mass of helium = 0.723 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$\text{Moles of helium}=\frac{0.723g}{4g/mol}=0.181mol$

For methane gas:

Given mass of methane gas = 3.43 g

Molar mass of methane gas = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane gas}=\frac{3.43g}{16g/mol}=0.214mol$

To calculate the mole fraction , we use the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$ .......(2)

To calculate the partial pressure of gas, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ......(3)

For Helium gas:

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{He}=\frac{0.181}{0.181+0.214}=0.458$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{He}=0.458$

Putting values in equation 3, we get:

$p_{He}=0.458\times 821mmHg=376mmHg$

For Methane gas:

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{CH_4}=\frac{0.214}{0.181+0.214}=0.542$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{CH_4}=0.542$

Putting values in equation 3, we get:

$p_{CH_4}=0.542\times 821mmHg=445mmHg$

Hence, the partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

For 2:

We are given:

Partial pressure of nitrogen gas = 363 mmHg

Partial pressure of carbon dioxide gas = 564 mmHg

Total pressure = (363 + 564) mmHg = 927 mmHg

Calculating the mole fraction of the gases by using equation 3:

For nitrogen gas:

$363=\chi_{N_2}\times 927\\\\\chi_{N_2}=\frac{363}{927}=0.392$

For carbon dioxide gas:

$564=\chi_{CO_2}\times 927\\\\\chi_{CO_2}=\frac{564}{927}=0.608$

Hence, the mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

6 0

3 years ago