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Bas_tet [7]
3 years ago
14

the term diffusion comes from a latin word meaning to spread apart. how is the term diffusion related to its latin word of origi

n
Chemistry
2 answers:
Lena [83]3 years ago
5 0

Answer:

It is related to certain common meanings.

Explanation:

The word diffusion comes from the Latin diffusio (wide, dilated, scattered, something that is not appreciated correctly). The term diffusion is related to its Latin word because it is used to refer to the idea that something is scattered, scattered or spread. For example, the word diffusion is used to refer to the idea of the dissemination of news, culture or knowledge.

Have a nice day!

Nitella [24]3 years ago
3 0

it means basically the same thing but can be used in different context. like if someone says, "lets diffuse the situation." it means lets separate it and find out what's going on with it. Like if you don't understand a sentence when you were younger your teacher would tell you to break down the sentence, and try to use context clues. What you are doing is diffusing the sentence so that your not looking at it as one big confusing sentence, your "breaking it down" and that helps you a lot. I hope I helped you, if not let me know and I can diffuse the topic a little more to make it more approachable and easy to understand. :)

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Convert 10g to milligrams
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What mass, in grams, of sodium bicarbonate, nahco3, is required to neutralize 1000.0 l of 0.350 m h2so4?
oksano4ka [1.4K]
Hello!

The net chemical equation between Sodium Bicarbonate and H₂SO₄ is the following one:

2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2H₂O + 2CO₂

To calculate the amount of Sodium Bicarbonate needed to neutralize 1000 L of 0,350 M H₂SO₄ we'll need to use the following conversion factor, to go from the volume of H₂SO₄ to grams of Sodium Bicarbonate:

 1000 L H_2SO_4* \frac{0,350 moles H_2SO_4}{1 L}* \frac{2 moles NaHCO_3}{1 mol H_2SO_4}* \frac{84,007gNaHCO_3}{1molNaHCO_3} \\ \\ =58804,9 g NaHCO_3

So, to neutralize 1000 L  of 0,350 moles of H₂SO₄ you'll need 58804,9 grams of NaHCO₃
3 0
3 years ago
Experiment Initial [A], mol/L Initial [B], mol/L Initial [C], mol/L Initial rate, mol/L.s 1 0.0500 0.0500 0.0100 6.25 x 10-3 2 0
icang [17]

Answer:

k[A][B][C]

Explanation:

Initial [A], mol/L Initial [B], mol/L Initial [C], mol/L Initial rate, mol/L.s

1      0.0500       0.0500        0.0100       6.25 x 10-3

2      0.100         0.0500        0.0100        1.25 x 10-2

3      0.100         0.100            0.0100       2.50 x 10-2

4      0.0500      0.0500         0.0200      1.25 x 10-2

Comparing equations 1 and 2, the reaction is first order with respect to A. This is because the concentration of A doubles, while the concentration of B and C remained constant leading to a double of the rate of reaction.

Comparing equations 2 and 3, the reaction is first order with respect to B. This is because the concentration of B doubles, while the concentration of A and C remained constant leading to a double of the rate of reaction.

Comparing equations 1 and 4, the reaction is first order with respect to C. This is because the concentration of C doubles, while the concentration of A and B remained constant leading to a double of the rate of reaction.

This means our rate law = k[A][B][C];

That is first order with respect to A, B and C

4 0
2 years ago
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