The quantity 44 liters expressed in cubic meters is ___

A 2.00-mol sample of hydrogen gas is heated at constant pressure from 294 K to 414 K. (a) Calculate the energy transferred to th

Furkat [3]

Answer:

a) The energy transferred is 6.91 kJ

b) The internal energy is 4.90 kJ

c) The work done on the gas is - 2.01 kJ

Explanation:

Step 1: Data given

Number of moles of hydrogen gas = 2.00 moles

Pressure = constant

Temperature is heated from 294 K to 414 K

Molar heat capacity of hydrogen gas = 28.8 J/mol*K

Step 2: Calculate the energy transferred to the gas by heat.

Q = n* Cp * ΔT

⇒with Q =the energy transferred

⇒with n = the number of moles = 2.00 moles

⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K

⇒ with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

Q = 2.00 * 28.8 * 120

Q = 6912 J = 6.91 kJ

Step 3: Calculate the increase in its internal energy.

ΔEint = n*Cv*ΔT

⇒with ΔEint = the increase in its internal energy.

⇒with n = the number of moles = 2.00 moles

⇒with Cv = The constant volume = 20.4 J/mol*K

⇒with ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

ΔEint = 2.00 * 20.4 * 120

ΔEint =4896 J = 4.90 kJ

Step 4: Calculate the work done on the gas.

Work done on the gas = -Q + ΔEint

W = -6.91 kJ + 4.90 kJ

W = -2.01 kJ

6 0

3 years ago

What is the correct answer?!????

ddd [48]

Answer:

B

Explanation:

Dalton worked with mainly about the chemistry of atoms.

how do atoms combine to form various molecules.

—rather than the details of the physical, internal structure of atoms, although he never denied the possibility of atoms' having a substructure.

4 0

3 years ago

GUYS I LITERALLY DONT DESERVE HIM I DO NOT DESERVE HIM IM DONE I CANT

postnew [5]

Just ignore such kind of people....

3 0

2 years ago

Read 2 more answers

In Rutherford's Gold foil experiment, were the alpha particles directed to different areas of the gold foil or only the same spo

MrRissso [65]

Answer:

See explanation

Explanation:

In the Rutherford experiment, alpha particles were directed at the same spot on a thin gold foil.

As the alpha particles hit the foil, most of the alpha particles went through the foil. In Rutherford's interpretation, most of the particles went through because the atom consisted largely of empty space.

However, some of the alpha particles were deflected through large angles, in Rutherford's interpretation, the deflected alpha particles had hit the dense positive core of the atom which he called the nucleus.

This accounted for their scattering through large angles throughout the foil in all directions.

8 0

3 years ago

Calculate the molality of a solution that contains 51.2 g of naphthalene, C10H8, in 500 mL of carbon tetrachloride. The density

PtichkaEL [24]

<u>Answer:</u> The molality of naphthalene solution is 0.499 m

<u>Explanation:</u>

Density is defined as the ratio of mass and volume of a substance.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ ......(1)

Given values:

Volume of carbon tetrachloride = 500 mL

Density of carbon tetrachloride = 1.60 g/mL

Putting values in equation 1, we get:

$\text{Mass of carbon tetrachloride}=(1.60g/mL\times 500mL)=800g$

Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molarity:

$\text{Molality of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent (in g)}}$ .....(2)

Given values:

Given mass of naphthalene = 51.2 g

Molar mass of naphthalene = 128.17 g/mol

Mass of solvent = 800 g

Putting values in equation 2, we get:

$\text{Molality of naphthalene}=\frac{51.2\times 1000}{128.17\times 800}\\\\\text{Molality of naphthalene}=0.499m$

Hence, the molality of naphthalene solution is 0.499 m

4 0

2 years ago

The quantity 44 liters expressed in cubic meters is ____.