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makvit [3.9K]
2 years ago
15

What is the volume in mL of a 1.11 carat diamond, given the density of diamond is 3.51 g/mL? (1 carat = 200 mg). Use significant

figures
Chemistry
1 answer:
kifflom [539]2 years ago
7 0
We will have to have first in mind the formula of density:
ρ= \frac{m}{V} 
 , where ρ is the density, m is the mass and V is the volume. 
We will next calculate the mass of the diamond:
1 carat ............................................... 200 mg
1.1 carat ............................................ x mg.
 =\ \textgreater \   \frac{200 * 1.1}{1} = 220 mg

We will next find out the volume:
ρ= \frac{m}{V} => V = \frac{m}{p} = \frac{0.220}{3.51} = 0,062 mL
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4 0
2 years ago
You have a mixture that contains 0.380 moles of Ne(g), 0.250 moles of He(g), and 0.500 moles CH4(g) at 400 K and 7.25 atm. What
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2 years ago
Summarize the process a scientist goes through to come up with a<br> satisfactory solution.
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Guess and check, test, trial and error, completion.
6 0
2 years ago
The heat of fusion for ice is 334 joules per gram. Adding 334 joules of heat to one gram of ice at STP will
yulyashka [42]

C) change to water at the same temperature

Explanation:

Adding 334Joules of heat to one gram of ice at STP will cause ice to change to water at the same temperature.

  • The heat of fusion is the amount of energy needed to melt a given mass of a solid
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3 0
2 years ago
Consider the Williamson ether synthesis between 2-naphthol and 1-bromobutane in strong base. A reaction was performed in which 0
Lera25 [3.4K]

Answer:

0.70 g

41 %

Explanation:

We can write the Williamson ether synthesis in a general form as:

R-OH + R´-Br ⇒  R-O-R´

where R-OH is an alcohol and R´-Br is an alkyl bromide.

We then see that the reaction occurs in a 1:1 mole ratio to produce 1 mol product.

Therefore what we need to calculate the theoretical yield and percent yield is to compute the theoretical number of moles of   2-butoxynaphthalene produced from 0.51 g 2-naphthol, and from there we can calculate the percent yield.

molar mass  2-naphthol = 144.17 g/mol

moles 2-naphthol  = 0.51 g / 144.17 g/mol = 0.0035 mol 2-naphthol

The number of moles of  produced:

= 0.0035 mol 2-naphthol  x ( 1 mol 2-butoxynaphthalene /mol 2-naphthol )

= 0.0035 mol  2-butoxynaphthalene

The theoretical yield will be

= 0.0035 mol 2-butoxynaphthalene  x  molar mass 2-butoxynaphthalene

= 0.0035 mol x  200.28 g/ mol = 0.70 g

percent yield=  ( 0.29 g / 0.70 ) g  x 100 = 41 %

7 0
2 years ago
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