B. is ur answer I bieleve
Answer:
The gas that Dr. Brightguy added was O₂
Explanation:
Ideal Gases Law to solve this:
P . V = n . R . T
Firstly, let's convert 736 Torr in atm
736 Torr is atmospheric pressure = 1 atm
20°C = 273 + 20 = 293 T°K
125 mL = 0.125L
0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K
(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n
5.20x10⁻³ mol = n
mass / mol = molar mass
0.1727 g / 5.20x10⁻³ mol = 33.2 g/m
This molar mass corresponds nearly to O₂
A cold front is the leading edge of a cooler mass of air, replacing at ground level a warmer mass of air, which lies within a fairly sharp surface trough of low pressure.
Answer:
The resulting solution is basic.
Explanation:
The reaction that takes place is:
First we <u>calculate the added moles of HNO₃ and KOH</u>:
- HNO₃ ⇒ 12.5 mL * 0.280 M = 3.5 mmol HNO₃
- KOH ⇒ 5.0 mL * 0.920 M = 4.6 mmol KOH
As <em>there are more KOH moles than HNO₃,</em> the resulting solution is basic.
Answer : The mass of nitric acid is, 214.234 grams.
Solution : Given,
Moles of nitric acid = 3.4 moles
Molar mass of nitric acid = 63.01 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of nitric acid.
Therefore, the mass of nitric acid is, 214.234 grams.
