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Furkat [3]
1 year ago
15

Sixty-five percent of the mass of bone is a compound called hydroxyapatite. sixty-five percent of the mass of bone is a compound

called hydroxyapatite.
a. True
b. False
Chemistry
1 answer:
mafiozo [28]1 year ago
5 0
True. Its simple how i remeberd was the bone has 650,000 cells and the bone has 65% hydroxyapatite
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What element speed of sound is 323 m/s?​
Pepsi [2]

Answer:

Speed of Sound

Explanation:

Speed of sound, fluid phases

 

m/s

notes

WEL 206  

use 323 27 °C

CRC 323 27 °C

WEL 319

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2 years ago
What is the name for molecular compound N4O8?
zmey [24]
Tetranitrogen octoxide
8 0
2 years ago
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A food worker had safely cooled a large pot of soup to 70 F (21 C) with two hours. What temperature must the soup Reach in the f
padilas [110]

The temperature reached after four hours of cooling is 140 F (or) 42 C.

Explanation:

A food worker cooled a cup of soup for two hours,

step 1: The temperature reached in <u>two hours</u>= 70 F (or) 21 C

step 2: Then for <u>four hours</u>, <u>it is twice the value</u>

∴ The temperature reached in four hours= 2(70)= 140 F (or) 2(21)= 42 C

4 0
2 years ago
By which process is a precipitate most easily separated from the liquid in which it is suspended
Gelneren [198K]
<span>Filtration, if its a precipitate that means its insoluble. </span>
7 0
2 years ago
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A hypothetical AX type of ceramic material is known to have a density of 3.55 g/cm3 and a unit cell of cubic symmetry with a cel
postnew [5]

Explanation:

For AX type ceramic material, the number of formula per unit cells is as follows.

         \rho = \frac{n'(A_{c} + A_{A})}{V_{C}N_{A}}

or,     n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}

where,   n' = no. of formula units per cell

           A_{c} = molecular weight of cation = 90.5 g/mol

           A_{a} = molecular weight of anion = 37.3 g/mol

          V_{c} = volume of cubic cell = 3.55 g/cm^{3}

           a = edge length of unit cell = 3.9 \times 10^{-8} cm

        N_{A} = Avogadro's number = 6.023 \times 10^{23}

          \rho = density = 3.55 g/cm^{3}

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

           n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}

                      = \frac{3.55 g/cm^{3} \times (3.9 \times 10^{-8})^{3} \times 6.023 \times 10^{23}}{(90.5 + 37.3)}

                     = 0.9

                    = 1 (approx)

Therefore, we can conclude that out of the given options crystal structure of cesium chloride is possible for this material.

3 0
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