Answer: Remove the watch and watchband immediately, and rinse his wrist for at least 15 minutes to be sure all hazardous material has been washed away. Let the TA inspect his wrist to see if it is okay.
Explanation:
The question is incomplete, the choices are:
a. Take him to the safety shower, and make sure he stands under it for at least 15 minutes to be sure all hazardous material has been washed away. b. Remove the watch and watchband immediately, and rinse his wrist for at least 15 minutes to be sure all hazardous material has been washed away. c. Let the TA inspect his wrist to see if it is okay.
d. Rinse the watch band before allowing him to put it back on.
An acid is traditionally considered to be any chemical compound that, when dissolved in water, produces a solution with greater hydronium cation activity than pure water, i.e., a pH less than 7. Both corrosive acids (and also corrosive bases) <u>are capable of destroying skin</u> by catalyzing the hydrolysis of fats, which are chemically esters. Also, when acids attack metals, hydrogen gas is often released.
This is why your lab partner should remove the watch and watchband immediately, and rinse his wrist for at least 15 minutes to be sure all hazardous material has been washed away. This ensures that the acid is can not damage the skin. And also the person should make sure that someone qualified checks it to see if it is okay.
Answer : The molarity of solution is, 1.73 mole/L
Explanation :
The relation between the molarity, molality and the density of the solution is,
where,
![d=M[\frac{1}{m}+\frac{M_b}{1000}]](https://tex.z-dn.net/?f=d%3DM%5B%5Cfrac%7B1%7D%7Bm%7D%2B%5Cfrac%7BM_b%7D%7B1000%7D%5D)
d = density of solution = 
m = molality of solution = 2.41 mol/kg
M = molarity of solution = ?
= molar mass of solute (toluene) = 92 g/mole
Now put all the given values in the above formula, we get the molality of the solution.
![0.876g/ml=M\times [\frac{1}{2.41mol/kg}+\frac{92g/mole}{1000}]](https://tex.z-dn.net/?f=0.876g%2Fml%3DM%5Ctimes%20%5B%5Cfrac%7B1%7D%7B2.41mol%2Fkg%7D%2B%5Cfrac%7B92g%2Fmole%7D%7B1000%7D%5D)

Therefore, the molarity of solution is, 1.73 mole/L
TRUE i hope this helps let me know if you need more help
Not sure, just setting up account
(125 mg Na) x (1 g/1000mg) x (1 mol of Na/22.99 g) = 5.43 E -3 mols of Na
5.43 E -3 mols x 6.022 E 23 = 3.27 E 21 Na atoms