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Marrrta [24]
3 years ago
10

What should be the very last function performed in the lab before you leave?

Chemistry
1 answer:
lesya [120]3 years ago
6 0

Answer:

c) Wash your hands

Explanation:

A laboratory or most popularly called as lab is a place or facility which provides controlled conditions where some educational research, scientific studies, experiments, testings, can be performed. The place is equipped with sophisticated machines, test samples, hazardous chemicals, instruments  etc.

The university or college lab, there is a incharge of the lab called as the lab instructor who is responsible for all the belongings of the lab and the safety of the experimenter or the students.

After performing some experiments or testing with some chemicals, the last and the most important thing we need to do is to wash our hand with soap or hand wash. it is very important because we touch many test samples and hazardous chemical which might go inside our mouth or eyes accidentally, which might be harmful to us.

So, it is important to wash our hand at last before leaving the lab.

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1,05g+0,69g+1,82g=3,56g\\\\
3,56g \ \ \ \Rightarrow \ \ \ 100\%\\
1,82g \ \ \ \Rightarrow \ \ \ x\\\\
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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

Best regards.

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A removal of the predator populations from an ecosystem would most likely result in
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