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STatiana [176]
4 years ago
6

In medical imaging, radioactive substances are sometimes used to help "see" parts of the body.

Chemistry
1 answer:
Kaylis [27]4 years ago
4 0
This is true as some parts cannot be checked but the radioactive substances can be used in x-rays
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The nature of reactants, temperature, concentration, surface area and catalysts.
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Objects that can bew seen only under magnification
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The answer is (D) microscopic. You can remember this, because the name is very close to "microscope," an instrument used to greatly magnify and observe tiny organisms and objects.
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Como se representan los elementos en la tabla¿?
Shalnov [3]

Answer:

Na sodiops los elementos químicos se representan como el átomo, el núcleo, donde se necuentran los protones y neutrones van dentro, los electrones afuera, girando de forma elíptica alrededor del núcleo. ... por ejemplo el oxígeno numero atómico 8, tiene 8 protones y 8 electrones. el número de neutrones es diferente.

Explanation:

4 0
3 years ago
Write a rap about fossils please
givi [52]

Answer:

CHORUS 1:

Fossil rocks are in the ground tonight

They’ve been down there for a really long time

They contain the history of life

They’ve been down there for a really long time

Fossil rocks are in the ground tonight

They’ve been down there for a really long time (They’re the schist*)

They are literally on the grind

Take your hammer out and... break that

VERSE 1:

Look at the ground. Partly rock

Plus plants and animals stuck in their spot. Huh

Each new layer stackin’ up

So the oldest on the bottom and the newest on top

“So how old is your fossil, bro?”

Well it was pretty far down so it’s pretty darn old

And I dated this one, so I gots to know

It’s from 3.5 billion years ago

Yo. I got that lava flow, Volcano

Air full of particles & smoke

There goes life as they know

Now that’s a blast! But most rocks ain’t movin’ that fast

The sediment’s evidence. Like a present tellin’ us of the past

CHORUS 2:

Take your chisel out and… flake that

BREAK:

Everyday I’m Shovelin’…

Brushin’ ‘em, Brushin’ ‘em

VERSE 2:

Brush it off and hear the story of life as told through rock

The story of the earth as told through life. Now stop... do the trilobite

Continental plates (are driftin’)

Mountains in the states (upliftin’)

Tend to isolate (species)

By affecting who you mate with (believe me)

Slow moves. Long time. Push up mountains to the sky

Slow moves. Long time. Open up a great divide

Slow moves. Long time. Fossils found on either side

Fossils found they tell us why... tell the history of life....

Slow moves. Long time. Slow moves. Long time

Slow moves. Long time. Slow moves. Long time

Plates are driftin’ at ya boy. Plates are driftin’ at ya boy

Plates are driftin’ at ya boy. Now boom. There’s an asteroid

   

Explanation:

4 0
3 years ago
Read 2 more answers
Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
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