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3 years ago

Ch3-ch=ch2-->br2/nacl

Chemistry

1 answer:

attashe74 [19]3 years ago

8 0

CH3-CH=CH2+Br2--->CH3-CH(Br)-CH2-Br

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The total volume of seawater is 1.5 x 10²¹ L. Seawater contains approximately 3.5% sodium chloride by mass. At that high of a co

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Answer:

There are $5.408\times 10^{22}$ grams contained in all the seawater in the world.

Explanation:

At first let is determinate the total mass of seawater ( $m_{sw}$ ), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

$m_{sw} = \rho_{sw}\cdot V_{sw}$

Where:

$\rho_{sw}$ - Density of seawater, measured in grams per liters.

$V_{sw}$ - Volume of seawater, measured in liters.

If $V_{sw} = 1.5\times 10^{21}\,L$ and $\rho_{sw} = 1030\,\frac{g}{L}$ , then:

$m_{sw}=\left(1030\,\frac{g}{L} \right)\cdot (1.5\times 10^{21}\,L)$

$m_{sw} = 1.545\times 10^{24}\,g$

The total mass of sodium chloride is determined by the following ratio:

$r = \frac{m_{NaCl}}{m_{sw}}$

$m_{NaCl} = r\cdot m_{sw}$

Given that $m_{sw} = 1.545\times 10^{24}\,g$ and $r = 0.035$ , the total mass of sodium chloride in all the seawater in the world is:

$m_{NaCl} = 0.035\cdot (1.545\times 10^{24}\,g)$

$m_{NaCl} = 5.408\times 10^{22}\,g$

There are $5.408\times 10^{22}$ grams contained in all the seawater in the world.

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