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4 years ago

What did Dalton's atomic theory contribute to science?

Chemistry

1 answer:

Serga [27]4 years ago

4 0

Answer:

Explanation:

Pasted below is his 5 theory's and all of them are the building blocks of chemistry today.

1. Matter is made up of atoms that are indivisible and indestructible.

2. All atoms of an element are identical.

3. Atoms of different elements have different weights and different chemical properties.

4. Atoms of different elements combine in simple whole numbers to form compounds.

5. Atoms cannot be created or destroyed. When a compound decomposes, the atoms are recovered unchanged.

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vampirchik [111]

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3 years ago

What are Free Radicals ?

DENIUS [597]

Answer:

an uncharged molecule (typically highly reactive and short-lived) having an unpaired valency electron.

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3 years ago

Read 2 more answers

An object is traveling at a speed of 7500.0 centimeters per second. How many kilometers per day

ivann1987 [24]

Answer:

6,480 kilometers

Explanation:

1 day = 86,400 seconds

1 kilometer = 100,000 centimeters

Equation:

86,400 x 7,500 = 648,000,000

648,000,000 ÷ 100,000 = 6,480

Hope this helped : )

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3 years ago

How are white blood cells linked to the lymphatic system?

Anon25 [30]

Answer:

The lymphatic system produces white blood cells, known as lymphocytes. There are two types of lymphocyte, T cells and B cells. They both travel through the lymphatic system. As they reach the lymph nodes, they are filtered and become activated by contact with viruses, bacteria, foreign particles, and so on in the lymph fluid.

Explanation:

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3 years ago

Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica

Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = $d_{ideal} = \dfrac{P\times M}{RT}$

here:

P = pressure of the O₂ gas = 310 bar

= $310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}$

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂ gas = 32 g/mol

∴

$d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}$

$d_{ideal}$ = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol² &

b = 0.0319 L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

$P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}$

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

$310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}$

$310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}$

$310V^3 -44.389V^2+1.382V-0.044=0$

After solving;

V = 0.1152 L

∴

$d_{Van \ der \ Waal} = \dfrac{32}{0.1152}$

$d_{Van \ der \ Waal}$ = 277.77 g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

5 0

3 years ago