Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
Answer:
11
Explanation:
Moles of KOH = 
Volume of water = 10 liters
Concentration of KOH is given by
![[KOH]=\dfrac{10^{-2}}{10}\\\Rightarrow [KOH]=10^{-3}\ \text{M}](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5Cdfrac%7B10%5E%7B-2%7D%7D%7B10%7D%5C%5C%5CRightarrow%20%5BKOH%5D%3D10%5E%7B-3%7D%5C%20%5Ctext%7BM%7D)
is strong base so we have the following relation
![[KOH]=[OH^{-}]=10^{-3}\ \text{M}](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E%7B-%7D%5D%3D10%5E%7B-3%7D%5C%20%5Ctext%7BM%7D)
![pOH=-\log [OH^{-}]=-\log10^{-3}](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E%7B-%7D%5D%3D-%5Clog10%5E%7B-3%7D)

So, pH of the solution is 11
Answer:
192g
Explanation:
so for this find out there moles of CH4 by moles=mass/mr
48/16=3 then use molar ratio 1:2 so times it by 2 which is 6 moles. then uses mass =moles*mr so 6*32=192g is the answer hope this helps to understand.
Because their atoms have the same number of electrons in the highest occupied energy level