How does convection occur in the troposphere?

Chemistry

2 answers:

densk [106]3 years ago

5 0

Answer:

the warm air rises and cool air sinks

Explanation:

Tamiku [17]3 years ago

4 0

Answer:

the warm air rises and cool air sinks

Explanation:

The surface heated air expands as it warms, becomes less dense than surrounding cooler air and rises as buoyant and turbulent bubbles. This is convection and is the main process by which the troposphere mixes and heats. Although convection stirs and mixes the troposphere, the higher it is the colder it becomes.

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1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0%

bagirrra123 [75]

Answer:

$m_{PbI_2}=18.2gKI$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2$

Thus, we proceed to compute the reacting moles of Pb(NO3)2 and KI, by using the given concentrations and densities and molar masses which are 331.2 g/mol and 166 g/mol respectively:

$n_{Pb(NO_3)_2}=96.7mL*\frac{1.134g}{mL}*\frac{0.14gPb(NO_3)_2}{1g}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} =0.0464molPb(NO_3)_2\\\\n_{KI}=99.8mL*\frac{1.093g}{mL}*\frac{0.12gKI}{1g}*\frac{1molKI}{166gKI} =0.0789molKI$

Next, the 0.0464 moles of Pb(NO3)2 will consume the following moles of KI (consider their 1:2 molar ratio):

$n_{KI}^{consumed\ by\ Pb(NO_3)_2}=0.0464molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0928molKI$

Hence, as only 0.0789 moles of KI are available, KI is the limiting reactant, therefore the formed grams of PbI2, considering its molar mass of 461.01 g/mol and 2:1 molar ratio, are:

$m_{PbI_2}=0.0789molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gKI$