What type of model did Rutherford, Thompson and Bohr develop?

matter that is composed of two or more different elements chemically combined in a fixed proportion is classified as

Helga [31]

Answer: The answer for this is Compound.

Explanation: A matter is something which occupies space and has some mass.

Elements are defined as the smallest substances which cannot be broken down into another substance by any chemical means.

A compound is a defined as a substance composed of two or more different elements when they combine chemically in definite proportion. For example: $CO_2$ is named as carbon dioxide and is a compound made by combining carbon and oxygen in a fixed ratio of 1 : 2

So, the answer to the given question is compound.

5 0

3 years ago

A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path

Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

$V_{2}$ = 8.34 L, $V_{1}$ = 2.67 L

Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

= -2818.68 J

Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

= -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

= -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

W = -574.54 J

Hence, work for path B is -574.54 J.

3 0

4 years ago

1. Propane gas (C3H8) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide gas (CO2) and water vapor (H2O

vovikov84 [41]

1) Balanced equation

C3H8 + 5O2 -> 3 CO2 + 4 H2O

2) 0.700 L C3H8

Given the pressure and temperature do not change, the molar ratio is equivalent to volume ratio

1molC3H8 / 5 mol O2 => 1 L C3H8 / 5 L O2

0.700 L C3H8 / x L O2 = 1 L C3H8 / 5 L O2 => x = 0.700 L C3H8 * 5 L O2 / 1 L C3H8

x = 3.500 L O2

3) CO2 produced

1 L C3H8 / 3 L CO2 = 0.700 L C3H8 / x L CO2 =>

x = 0.700 L C3H8 * 3 L CO2 / 1 L C3H8 = 2.100 L CO2

4) Water vapor produced

1) 1 L C3H8 / 4 L H2O = 0.700 LC3H8 / x L H2O =>

x = 0.700 L C3H8 * 4 L H20 / 1 L C3H8 = 2.800 L H2O

3 0

4 years ago

B) A company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you

Yuri [45]

Answer:

43.1atm is the pressure using gas law and 27.2atm using Van der Waals Law.

Explanation:

Ideal gas law is:

PV = nRT

Where P is pressure in atm

V is volume = 4.00L

n are moles of the gas (For chlorine Molar Mass: 70.90g/mol):