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stiks02 [169]
2 years ago
8

*

Chemistry
1 answer:
Natasha_Volkova [10]2 years ago
4 0

to be honestl id ldm

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Consider the structures of vinegar and triglyceride and draw in any bond dipoles that exist in the molecules above using a dipol
bagirrra123 [75]

Answer:

Given molecules are vinegar and triglycerides.

Explanation:

The dipole is a vector quantity and it is heading from less electronegative atom to more electronegative atom in a polar covalent bond.

The structures and the bond dipoles in the given molecules are shown below:

5 0
3 years ago
What is another common name for the lanthanoids on the periodic table?
RUDIKE [14]
Lanthanide Lanthanoid, also called Lanthanide
7 0
3 years ago
Please help me with this chemistry
Elena-2011 [213]

Answer: gasoline, water, sea water, chloroform and mercury so B

Explanation:

6 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
2 years ago
The atoms of some elements can be made
natima [27]

Answer:

I think 4 I got it right on edg but i duno if its the same

Explanation:

8 0
2 years ago
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