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UkoKoshka [18]
2 years ago
7

Suppose you are measuring the mass of a sample using a balance that employs a standard mass of density dw = 8.0 g/ml. what is th

e buoyancy correction (m/m\'), if the sample density is less than 8.0 g/ml?
Chemistry
1 answer:
sleet_krkn [62]2 years ago
5 0

The buoyant force formula is:

B = \rho Vg

where B is Buoyant force, \rho is density, V is displaced body volume and g is 9.806 ms^{-2} (standard gravity).

The standard mass density of the sample, \rho _{1} = 8 g/ml  

If the dansity of the sample, \rho _{2} < 8 g/ml

The buoyancy correction is determined by the ratio:

\frac{B_{1}}{B_{2}} = \frac{\rho_{1} Vg}{\rho_{2} Vg}

\frac{B_{1}}{B_{2}} = \frac{\rho_{1}}{\rho_{2}}

\frac{B_{1}}{B_{2}} = \frac{\rho_{1}}{\rho_{2}}>1

Hence, the the buoyancy correction is positive.

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<span>So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.


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Answer is "B - 700,000".<span>

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Answer:

X=600 degrees celsius

Explanation:

We can evaluate by using the equation ==> Q=m(c)(delta T)

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-Therefore, we need to solve for delta T (which will be represented by x)

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The unit of measurement (g-grams) can be crossed out on the left side of our expression.

66,300 J=0.3(390 J 0C) (t)

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------------       =        -----------

117 J/0C                 117 J/0C

-Joules on both sides can be crossed off

*117 J/0C came from the product in terms of 0.3(390 J 0C)

*If we find the quotient of 66,300 J and 117 J/0C, we get 566.6 repeating

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