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Tju [1.3M]
3 years ago
12

3. A diamond contains 0.090 moles of carbon What is the mass of the diamond?

Chemistry
1 answer:
Nady [450]3 years ago
3 0

Answer:

1.0809

Explanation:

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The Farallon plate used to comprise what is now the Cocos plate of Mexico and Central America, and the Juan de Fuca plate in the region from N. Vancouver Island to the Cape Mendicino California, and a big sea floor tract in between. However, the middle portion of the Old Farallon plate disappeared underneath North America, it was subducted underneath California leaving the San Andreas fault system behind as the contact between the Pacific plates and North America.  

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Which type of reaction occurs in the following equation?
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The  type  of  reaction  which  occurs  is  referred to  as  redox  reaction.  This  kind  of  reaction  involve  both  oxidation  and  reduction.
Al(s)  is  oxidized  to  alluminium  ions,  while  cu2+  is reduced  copper metal.Reduction  occurs   at  the  cathode  while oxidation  occurs at  the anode.

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2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an
Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression d=\frac{m}{V} that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL

d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL

d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL

d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL

The problem says that all the samples have the same mass, so:

m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

V_{lithium}=\frac{m}{d_{lithium}}

V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m

V_{lithium}=1.88\frac{mL}{g}*m

V_{gold}=\frac{m}{d_{gold}}

V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m

V_{gold}=5.18*10^{-2}\frac{mL}{g}*m

V_{aluminum}=\frac{m}{d_{aluminum}}

V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m

V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m

V_{lead}=\frac{m}{d_{lead}}

V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m

V_{lead}=8.85*10^{-2}\frac{mL}{g}*m

If we assume m = 1g, we find that:

V_{lithium}=1.88mL

V_{gold}=5.18*10^{-2}mL

V_{aluminum}=3.70*10^{-1}mL

V_{lead}=8.85*10^{-2}mL

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

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Answer:

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Explanation:

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Covalent bond is the correct answer
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