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pogonyaev
3 years ago
7

Sodium metal is sometimes used as a cooling agent in heat exchange units because of its relatively high molar heat capacity of 2

8.2 J/mol°C. What is the specific heat of sodium in J/g°C?
Chemistry
1 answer:
IgorC [24]3 years ago
4 0

Answer:

The specific heat of sodium is 1,23J/g°C

Explanation:

Using the atomic weight of sodium (23g/mol) and the atomic weight definition, we have that each mole of the substance has 23 grams of sodium.

starting from this, we use the atomic weight of sodium to convert the units from J / mol ° C to J / g ° C

28,2 \frac{J}{mol C} x \frac{1mol}{23g} = 1,23 J/g C

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What is the total number of protons in an atom with the electron configuration 2-8-18-32-18-1.
svp [43]

The total number of protons in this atom is 79 because an atomic number of an element is equal to the number of protons. Here, the atomic number is 79 after adding the given electronic configuration.

What are Protons?

Every atom has a proton, a subatomic particle, in its nucleus. The particle possesses an electrical charge that is positive and opposite to the electrons.

What is Atom?

A nucleus plus one or more electrons bound to the nucleus make up an atom. The quantity of protons or electrons in an element's atom determines how different it is from other similar elements. The atomic number of an element, which serves as its primary identification, is the sum of its protons or electrons.

What is Electronic configuration?

The arrangement of electrons in atomic or molecular orbitals within an atom, molecule, or other physical structure is known as the electron configuration.

Hence, the total number of protons in this atom is 79, after adding 2 + 8 + 18 + 32 + 18 + 1.

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5 0
1 year ago
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

p = 0.0453 lbs/in³

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

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

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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3 years ago
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Answer:

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Today, as part of the series of posts on soils, we are going to look at ‘soil texture’. Soil forms the basis for all life but it’s important to know about its mineral constitution as well as its biological profile.

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