1 mole ----------- 22.4 L ( at STP )
3.75 moles ----- ?
3.75 x 22.4 / 1 => 84.0 L
Answer:
2.9 grams.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:
no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.
no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.
<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>
<em></em>
<em><u>Using cross multiplication:</u></em>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.074 mol of Mg produce → 0.074 mol of MgO.
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>
So the ideal gas law is pv=nrt
The letter n stands for the number of moles. divide both side by rt to isolate.
pv/rt=n
Answer:
- <u>Alkaline or basic solution </u>(alkaline and basic means the same)
Explanation:
According to the <em>pH</em>, solutions may be classified as neutral, acidic, or alkaline (basic).
This table shows such classification:
pH classification
7 neutral
> 7 alkaline or basic
< 7 acidic
Thus, since the pH of the solution is 8.3, which is greater than 7, the solution is classified as basic (alkaline).
Additionally, you must learn that pH is a logarithmic scale for the concentration of hydronium ions in the solution.
You can calculate the concentration of hydronium ions using antilogarithm properties:
![pH=-log[H_3O^+]\\ \\ {[H_3O^+]}=10^{-pH}\\ \\ {[H_3O^+]}=10^{-8.3}=0.00000000501](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D10%5E%7B-pH%7D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D10%5E%7B-8.3%7D%3D0.00000000501)
NaOH solutions are alkaline solutions, bases, according to Arrhenius model, because they contain OH⁻ ions and release them when ionize in water.