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wel
3 years ago
7

Strontium crystallizes in a face-centered cubic unit cell having an edge length of 77.43 pm. What is the atomic radius of stront

ium (in picometers) based on this structure
Chemistry
1 answer:
kiruha [24]3 years ago
5 0

Answer:

Atomic radius of Strontium is 27.38pm

Explanation:

In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:

a² + a² = b² = (4r)²

2a² = 16r²

a = √8 r

As edge length of Strontium is 77.43pm:

77.43pm / √8 = r

27.38pm = r

<h3>Atomic radius of Strontium is 27.38pm</h3>

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A sample of nitrogen occupies a volume of 250 mL at 25 oC. What volume will it occupy at 95 oC?
Volgvan

Answer:

308.7 mL

Explanation:

V1/T1 =V2/T2           T must be in K units

V2 = V1T2/T1

     =  (.250)(95 + 273.15) / (25 + 273.15)  = .3087 LITER

4 0
2 years ago
An iron object alloyed with cobalt rusts more quickly than a pure iron object. However, an iron object alloyed with manganese ru
HACTEHA [7]

Rusting is an electrochemical reaction. Iron rusts faster when alloyed with cobalt than when alloyed with manganese because, in the iron-manganese alloy, manganese is rendered the anode and iron is rendered the cathode

An alloy is a combination of two metals. There are various reasons for producing alloys such as greater tensile strength, corrosion resistance and improved aesthetic appearance.

When iron is alloyed with cobalt, the iron rusts faster than pure iron because iron is rendered the anode and cobalt is rendered the cathode. When the iron is alloyed with manganese, it rusts more slowly than pure iron because in the iron-manganese alloy, manganese is rendered the anode and iron is rendered the cathode.

Missing parts;

An iron object alloyed with cobalt rusts more quickly than a pure iron object. However, an iron object alloyed with manganese rusts less quickly than a pure iron object under the same conditions. This is true because

(1) cobalt is a stronger reducing agent than iron

(2) iron is a stronger reducing agent than manganese

(3) cobalt exhibits more metallic character than either iron or manganese

(4) in the iron-manganese alloy, manganese is rendered the anode and iron is rendered the cathode

(5) in the iron-cobalt alloy, cobalt is rendered the anode and iron is rendered the cathode

Learn more: brainly.com/question/14281129

5 0
3 years ago
Materials
nikitadnepr [17]

Answer:

Limestone caves are formed when acid rain occurs.

Explanation:

When chalk pieces which is made of calcium carbonate react with acetic acid, it produces carbon dioxide, calcium acetate, and water. When calcium carbonate react with Carbonated Water produces calcium bicarbonate. Limestone cave forms when the dissolution of limestone occur when rainwater which is acidic due to the addition of carbondioxide in the air react with limestone. This dissolution removes the lime from the rock and caves are formed.

8 0
4 years ago
In reading a line drawing, how do you know where atoms of these elements are in the structure if they are missing from the drawi
Tanya [424]

Answer:

The atoms of an element are represented in a chemical line drawing with its chemical formula.

Explanation:

chemical structural drawing helps to represent the pattern for which an element is formed. Chemical elements are made up of atoms that represent their single state.

The line drawing is made up of lines (representing the chemical bond between atoms) and the atoms or various atoms that make up the element.

7 0
3 years ago
A horizontal pipe 15 cm in diameter and 4 m long is buried in the earth at a depth of 20 cm. The pipe-wall temperature is 75 deg
Basile [38]

Explanation:

The given data for case (1) is as follows.

  h = 20 cm = 0.2 m

Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.

              Q = kA \frac{\Delta T}{L}

  where,    A = area

                  L = length

                  k = thermal conductivity = 0.8 W/m

             \Delta T = change in temperature.

Therefore, putting the given values into the above formula as follows.

            Q = kA \frac{\Delta T}{L}

                = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{0.2}

                = 168 W

For case (2), h = 180 cm = 1.8 m

Therefore, heat lost will be calculated as follows.

                       Q = kA \frac{\Delta T}{L}

                           = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{1.8}

                           = 18.67 W

Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.

8 0
3 years ago
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