Compounds with very high vapor pressures must have very minimal intermolecular forces true or false

HELP PLEASE!! ASAP What would the mass be in kg?

fgiga [73]

Answer:

i guess answer is 0.0600

Explanation:

Here,

Density=0.791g/cm^3

Volume=75.89ml

So,

Mass=Density*Volume

=0.791g/cm^3×75.89ml

=60.02g/cm^3.ml

Expressing them in kg

60.02/1000

=0.0600 kg

4 0

2 years ago

N2+3H2-2NH3

NemiM [27]

102 grams of ammonia is formed when 3 moles of nitrogen and 6.7 moles of hydrogen reacts.

Explanation:

The equation given is of Haeber's process in which the nitrogen is limiting factor in the ammonia formation and hydrogen if in excess gets delimited.

We know that 1 mole of Nitrogen gives 2 moles of ammonia.

We have 3 moles of nitrogen here,

So, 6 moles of ammonia will be form

so from the formula

no of moles=mass/atomic mass

mass= no. of moles*atomic mass

= 6*17

= 102 grams of ammonia will be formed.

So, 6 moles or 102 grams of ammonia is formed when 3 mole of nitrogen and 6.7 mole of hydrogen reacts.

7 0

3 years ago

The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.

adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

$\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}$

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

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

$\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol$

Now we have to calculate the energy required.

$Q=\frac{\Delta H}{n}$

where,

Q = energy required

$\Delta H$ = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

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

$Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ$

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

7 0

2 years ago

78.6 grams of O2 and 67.3 grams of F2 are placed in a container with a volume of 40.6 L. Find the total pressure if the gasses a

saul85 [17]

1) List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

2.1- List the known and unknown quantities.

Sample: O2.

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

2.2- Convert grams of O2 to moles of O2.

The molar mass of O2 is 31.9988 g/mol.

$mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2$

2.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

2.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}$ $P_{O_2}=1.57\text{ }atm$

The pressure of O2 is 1.57 atm.

3) Find the pressure of F2.

3.1- List the known and unknown quantities.

Sample: F2.

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- Convert grams of F2 to moles of F2.

The mmolar mass of F2 is 37.9968 g/mol.

$mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2$

3.3- Set the equation.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

$PV=nRT$

3.4- Plug in the known quantities and solve for P.

$(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)$

.

$P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}$ $P_{F_2}=1.13\text{ }atm$

The pressure of F2 is 1.13 atm.

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

4.1- Set the equation.

$P_T=P_A+P_B$

4.2- Plug in the known quantities.

$P_T=1.57\text{ }atm+1.13\text{ }atm$ $P_T=2.7\text{ }atm$

The total pressure in the container is 2.7 atm.

5 0

7 months ago

Which is NOT an example of plants demonstrating the characteristics of life?

Lena [83]

Answer:

D

Explanation:

trust me bro ive done this

3 0

3 years ago