How many grams of h2 gas can be produced by the reaction of 63.0 grams of al(s) with an excess of dilute hydrochloric acid in the reaction shown below? 2 al(s) + 6 hcl(aq) → 2 alcl3(aq) + 3 h2(g)?
MNaHCO₃: 23+1+12+(48×3) = 84g
mCH₃COOH: 12+(1×3)+12+(16×2)+1 = 60g
.................
84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with...........
84g ----- 60g
83g ----- X
X = 59,29g CH₃COOH
We used 70g CH₃COOH, it' too much.
So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.
_______________________________
B) Amount of CH3COOH is in excess.
:•)
Answer:
B. PhCHO
Explanation:
Every organic group shows a characteristic IR absorption at certain wavelength . With the help of these absorption spectra we can identify the group present on organic molecules .
The wave number of 2710 cm⁻¹ is absorbed by aldehyde bond stretching .
The wave number of 1705 cm⁻¹ is shown by conjugated aldehyde . So the most likely compound among given compounds is PhCHO .
Answer:
The correct answer is 28.2 %.
Explanation:
Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,
Ptotal = PO2 + PN2 + PHe
= 55 + 90 + 50
= 195 atm
The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100
Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,
%VO2 = (55/195) * 100 = 28.2%
%VN2 = (90/195) * 100 = 46%
%VHe = (50/195) * 100 = 25.64%
Hence, the percent oxygen by volume present in the blend is 28.2 %.
Answer:
Cu(NO3)2 --> MM187.5558
NiNO3 *COEF2* --> 120.6983