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hammer [34]
3 years ago
8

From where do the gems used in Jewlery come? A. they are mettallic ore B.they are nonmettallic silicates C. they are mettallic m

inerals D. they are non mettallic minerals.
Chemistry
2 answers:
mixas84 [53]3 years ago
6 0
The answer is D, because gems are usually not mettallic
UkoKoshka [18]3 years ago
5 0
Gems used in jewelry come are Non metallic minerals. So it would be D hope this helps!
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Answer:

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Explanation:

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Is barium nitrate more soluble in water than CH4 or the other way around?
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Barium nitrate and methane (CH4) are both soluble. They both will dissolve in water, however, barium nitrate will dissociate becoming barium 2+ ions and nitrate becoming NO3 1- ions. All nitrates are soluble and dissociate. CH4 is a weak base and does dissolves but doesn't dissociate. So in solubility terms.... they are both equally soluble just one happens to dissociate into its cations and anions. Hope this helps!
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3 years ago
A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat
Lady_Fox [76]

Explanation:

(a)  The given data is as follows.  

         Load applied (P) = 1000 kg

         Indentation produced (d) = 2.50 mm

         BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}    

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

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

                       = \frac{2000}{9.98}                          

                       = 200

Therefore, the Brinell HArdness is 200.

(b)     The given data is as follows.

               Brinell Hardness = 300

                Load (P) = 500 kg

               BHI diameter (D) = 10 mm

             Indentation produced (d) = ?

                      d = \sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}

                         = \sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}

                          = 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0
2 years ago
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Explanation:

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