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kvasek [131]
2 years ago
14

Isotopes are:

Chemistry
2 answers:
topjm [15]2 years ago
6 0

Hello and Good Morning/Afternoon

<u>Let's consider all the choices to determine the attributes of isotopes</u>

  • Choice (A) is wrong

        ⇒ isotopes are often found in nature as they are often the atoms

            that form the world around us

  • Choice (B) is wrong

        ⇒ isotopes describe a certain type of atom with a certain

             characterist that exists, thus it isn't theoretical

  • Choice (C) is wrong

        ⇒ as said before, isotopes are found in nature and form the world

  • Choice (D) is correct

        ⇒Isotopes are a type of atoms thus they are found in nature

<u>Answer: (D)</u>

<u></u>

Hope that helps!

#LearnwithBrainly

Oksana_A [137]2 years ago
4 0

Answer:

D. found in nature.

Explanation:

Isotope is any of two or more forms of a chemical element, having the same number of protons in the nucleus, or the same atomic number, but having different numbers of neutrons in the nucleus, or different atomic weights. There are 275 isotopes of the the 81 stable elements, in addition to over 800 radioactive isotopes, and every element has known isotopic forms. Isotopes of a single element possess almost identical properties.  An isotope of an element is just a version of that element with a particular number of neutrons. Some numbers are stable, some are not; the ones that aren't shoot particles out at extremely high speeds ("radiation").

Since radioisotopes generally have the same chemistry as their stable counterparts (since neutrons play almost no role in chemistry), they'll get wherever other atoms of that element would get. Iodine, for example, accumulates in your thyroid gland - so do radioactive forms of iodine, which can then cause thyroid cancer by irradiating it from the inside.

Also because they're chemically almost indistinguishable, they're almost impossible to separate out by any normal means. It's like giving someone a giant bin of golf balls, some of which are 1% heavier than the other golf balls but are otherwise exactly the same, and saying "okay, sort these". Except the golf balls are atoms and the bin is the size of a small country.

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Alexxandr [17]

Answer:

5.06 atm

Explanation:

Step 1:

Data obtained from the question. This includes:

Mass of S2O = 175g

Volume (V) = 16600 mL

Temperature (T) = 195°C

Pressure (P)

Step 2:

Determination of the number of mole of S2O in 175g of S2O.

Mass of S2O = 175g

Molar Mass of S2O = (32x2) + 16 = 64 + 16 = 80g/mol

Number of mole of S2O =.?

Number of mole = Mass/Molar Mass

Number of mole of S2O = 175/80

Number of mole of S2O = 2.1875 moles

Step 3:

Conversion to appropriate units.

It is essential to always express the various variables in the right units of measurement in order to obtain the desired answer in the right units.

For volume:

1000mL = 1L

Therefore, 16600mL = 16600/1000 = 16.6L

For temperature:

Temperature (Kelvin) = temperature (celsius) + 273

Temperature (celsius) = 195°C

Temperature (Kelvin) = 195°C + 273 = 468K

Step 4:

Determination of the pressure.

The pressure can be obtained by the application of the ideal gas equation. This is illustrated below:

Volume (V) = 16.6L

Temperature (T) = 468K

Number of mole (n) = 2.1875 moles

Gas constant (R) = 0.082atm.L/Kmol

Pressure (P) =

PV = nRT

P x 16.6 = 2.1875 x 0.082 x 468

Divide both side by 16.6

P = (2.1875 x 0.082 x 468) /16.6

P = 5.06 atm

Therefore, the pressure is 5.06 atm

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A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate
inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams  and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

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so the wet density of the CL sample is 0.0453 lb/in³

b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

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c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

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Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

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Answer:

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