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viktelen [127]
3 years ago
14

Help me I’ll do anything

Chemistry
2 answers:
seraphim [82]3 years ago
8 0

Answer:

A) something that causes physical pain

hope this helps

Explanation:

KIM [24]3 years ago
3 0

Answer:

its not A B or C soo choose D *This is what i think*

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D. Being cold temperatures can result in a cold nose. With prolonged exposure The body will start to lose heat faster than it can generate it, this is the result of hypothermia.
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Which statement explains why large atoms are more reactive than small atoms?
Lunna [17]

A. Large atoms have valence electrons farther from the nucleus and lose them more readily, so they are more reactive than small atoms.

For example, the valence electron of a small atom like Li is tightly held. <em>Lithium gently fizzes</em> on the surface as it reacts with the water to produce hydrogen.

In contrast, the valence electron of a large atom like Cs is so loosely held that <em>cesium exlodes </em>on contact with water.

8 0
3 years ago
How many moles are in 20 grams of O₂ gas?
Dima020 [189]
20 g O2 x 1 mol O2/32 g O = 0.625 mol O2
4 0
2 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
3 years ago
6.02 x <img src="https://tex.z-dn.net/?f=10%5E%7B2%7D" id="TexFormula1" title="10^{2}" alt="10^{2}" align="absmiddle" class="lat
VMariaS [17]
The answer is 602 but im not sure hehe
5 0
2 years ago
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