Sodium is very reactive but it’s a metal, and the problem asks specifically for a non-metal.
Silicone is technically reactive, but not super reactive.
Argon is a nonmetal, however it is an inert gas. It doesn’t react with anything.
We’re left with Chlorine, which is a non-metal in group 7, a highly reactive group, on the periodic table.
The molecular weight of water is <span>18.01528 g/mol.
So in 2.92 grams there are 2.92/</span>18.01528 = 0.1621 mol of particles.
1 mol contains 6,02214 × 10^<span>23 particles by definition.
So the nr of H2O molecules is </span>0.1621 * 6,02214 × 10^23 = 0,9761 × 10^23.
Every molecule has 2 H atoms, so you have to double that.
2* 0,9761 × 10^23 = 1.952 × 10^23.
The electromagnet and the permanent magnet -- interact with each other as any two magnets do. The positive end of the electromagnet is attracted to the negative pole of the permanent magnetic field, and the negative pole of the electromagnet is repelled by the permanent magnet's negative pole
Answer : The mass of sulfuric acid needed is 
.
Solution : Given,
pH = 8.94
Volume of solution = 380 ml = 
Molar mass of sulfuric acid = 98.079 g/mole
As we know,
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![5.06=-log[OH^-]](https://tex.z-dn.net/?f=5.06%3D-log%5BOH%5E-%5D)
![[OH^-]=0.00000871=8.71\times 10^{-6}mole/L](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.00000871%3D8.71%5Ctimes%2010%5E%7B-6%7Dmole%2FL)
Now we have to calculate the moles of 
.
Formula used : 
![\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%5Ctext%7B%20Concentration%20of%20%7D%5BOH%5E-%5D%5Ctimes%20Volume%5C%5C%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%288.71%5Ctimes%2010%5E%7B-6%7Dmole%2FL%29%5Ctimes%20%28380%5Ctimes%2010%5E%7B-3%7DL%29%3D3309.8%5Ctimes%2010%5E%7B-9%7Dmoles)
For neutralization, equal number of moles of 
ions will neutralize same number of ions.
![\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%5Ctext%7B%20Moles%20of%20%7D%5BH%5E%2B%5D%3D3309.8%5Ctimes%2010%5E%7B-9%7Dmoles)
As, 
From this reaction, we conclude that
2 moles of ion is given by the 1 mole of 
moles of ion is given by 
moles of
Now we have to calculate the mass of sulfuric acid.
Mass of sulfuric acid = Moles of × Molar mass of sulfuric acid
Mass of sulfuric acid = 
Therefore, the mass of sulfuric acid needed is .
Answer:
1.07 g
Explanation:
Half-life of Pu-234 = 4.98 hours
Initially present = 45 g
mass remains after 27 hours = ?
Solution:
Formula
mass remains = 1/ 2ⁿ (original mass) ……… (1)
Where “n” is the number of half lives
To find "n" for 27 hours
n = time passed / half-life . . . . . . . .(2)
put values in equation 2
n = 27 hr / 4.98 hr
n = 5.4
Mass after 27 hr
Put values in equation 1
mass remains = 1/ 2ⁿ (original mass)
mass remains = 1/ 2^5.4 (45 g)
mass remains = 1/ 42.2 (45 g)
mass remains = 0.0237 x 45 g
mass remains = 1.07 g
