Answer:
A fire extinguisher
Explanation:
A lab station usually contains equipment for the people working at the station to use. However, most rooms will only have one fire extinguisher for the whole room meaning it would most likely be located somewhere that is easy to access by all and not just a singular lab station.
~Hope this Helps!~
*The molality of a solution is calculated by taking the moles of solute and dividing by
the kilograms of solvent* Basically if we had 1.00 mole of sucrose (it's about 342 3 grams) and
proceeded to mix it into exactly 1.00 liter water. It would dissolve and make sugar
water. We keep adding water, dissolving and stirring until all the solid was gone. We
then made sure everything was well-mixed.
What would be the molality of this solution? Notice that my one liter of water weighs
1000 grams (density of water = 1.00 g / mL and 1000 mL of water in a liter).
Answer:
For me
Explanation:
It's Cu because other compound contains negative radicals
<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234





So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left

Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.