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Vadim26 [7]
2 years ago
12

Write the name of the molecular compound C2Br4

Chemistry
2 answers:
cupoosta [38]2 years ago
6 0

Answer:

Tetrabromoethylene

Explanation:

Tetrabromoethylene I guess

8_murik_8 [283]2 years ago
4 0
The name of the molecular compound is Tetrabromoethylene
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Give an<br> example of when a plant or animal might<br> use energy they have stored.
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Water is a ______ molecule and the bond between the h-o is a __ ______ bond.
omeli [17]

Answer: polar molocule

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3 years ago
Considering the chart shown, at what temperature does the substance boil? 200ºC 1,200ºC 2,200ºC 2,400ºC
kodGreya [7K]

Answer:

The answer is 2,200ºC

Explanation:
I took the assignment for Edge, I don't think I can send the image because it might pick up on that and get reported, sorry!

The only other rationale that I have is that it's boiling because the graph shows that it's at a constant temperature/rate at 2,200ºC for quite a while. Typically when something boils, it stays at that constant rate of boiling, unless you turn the temperature up or it's finally able to peak..?

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3 years ago
What are three ways fluid flow is important in the food industry?
yarga [219]
Viscosity measurements are used in the food industry to maximize production efficiency and cost effectiveness. It affects the rate at which a product travels through a pipe, how long it takes to set or dry, and the time it takes to dispense the fluid into packaging.
8 0
3 years ago
Compute the atomic density (the number of atoms per cm3 ... rather than the mass density g/cm3) for a perfect crystal of silicon
serious [3.7K]

Answer:

        \large\boxed{\large\boxed{5.00\times 10^{22}atoms/cm^3}}

Explanation:

You can convert the <em>density</em> into <em>atomic density</em> using the <em>atomic weight </em>and Avogadro's number

A dimensional analysis is very helpful:

           \dfrac{g}{cm^3}\times \dfrac{mol}{g}\times \dfrac{atoms}{mol}=\dfrac{atoms}{cm^3}

Follow the chain: g cancels with g, mol cancels with mol; at the end, what remains is atoms/cm³, which is what you want.

Use that with your data:

         \dfrac{2.33g}{cm^3}\times \dfrac{1mol}{28.09g}\times \dfrac{6.022\times 10^{23}atoms}{mol}=\approx 5.00\times10^{22}atoms/cm^3

3 0
3 years ago
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