You have to use the fact that percent mass is mass of salute over mass of solution and make the equation (x/250g)x100%=3.5% and solve for x. When done correctly you should get x=87.5g which is the mass of the solute needed.
I hope this helps.
4 3 4 111.6 3 r = 111.6 cm3<span> = (3.1416)r3 r= 3 = 2.987 cm V= 3 3 4(3.1416) 3A ... with it absent. </span>density<span> =</span>mass<span> 28.4 g rock = = 2.76 g/mL = 2.76 g/</span>cm3<span> volume 44.1 ... can be found using dimensional analysis. ethanol </span>mass<span> = </span>25<span> L gasohol 1000 mL ..... 100.00 g solution = 9.95 103 g solution 1 kg sucrose </span>10.05 g<span>sucrose 53.</span>
Regardless of how many protons this element has, it has 2 valence (outer) electrons. On the current periodic table, this element would go into the second group. If you want a more specific answer, try alkaline earth metals, or Group 2A.
In order to compute this, we must first take a couple of assumptions of:
1) The laboratory size so we can calculate its volume
2) The number of students working in the lab so we know the total gas produced
Let the lab be
11 m × 9 m × 6 m
The volume then computes to be:
594 m³
We know that
1 Liter is 1 dm³
1 m = 10 dm
1 m³ = 1000 dm³
Therefore, the room volume in liters is:
594,000 Liters
Let there be 30 students in the laboratory
Total gas being produced:
6 × 30
= 180 Liters
This works out to be:
0.03% of Hydrogen by volume
Therefore, there is no risk of explosion given our assumption of size and students.