Well. Yeah. Basically.
1 mole of any gas is 22.4 L.
Not really sure if that answers your question but that's what i know.
Anyway if we are talking about the size then usually gases are filled with empty space.
Answer:
USe this Nernst equation and find your required answer
Explanation:
E = E0 - [0.0592 /n ] log [Cd+2]/[Co2+]
Wher Eo is +0.120 V
[Cd+2] =0.1
[Co2+] = 0.001
n is number of electons transfered int he process and in your case it is = 2
E= 0.120- 0.0296 * log 100
= ........ V
<u>Answer:</u> The correct option is 1 mole of acetic acid was required for 2 moles of sodium bicarbonate
<u>Explanation:</u>
We are given:
Average number of drops of sodium bicarbonate = 142
The chemical equation for the reaction of sodium bicarbonate and acetic acid in vinegar follows:
From the stoichiometry of the reaction:
1 mole of sodium carbonate reacts with 1 mole of acetic acid in vinegar.
Hence, the correct option is 1 mole of acetic acid was required for 2 moles of sodium bicarbonate
Answer:
Theoretical yield = 3.52 g
Percent yield =65.34%
Explanation:
Given data:
Mass of HgO = 46.8 g
Theoretical yield of O₂ = ?
Percent yield of O₂ = ?
Actual yield of O₂ = 2.30 g
Solution:
Chemical equation:
2HgO → 2Hg + O₂
Number of moles of HgO = mass/ molar mass
Number of moles of HgO = 46.8 g / 216.6 g/mol
Number of moles of HgO = 0.22 mol
Now we will compare the moles of HgO with oxygen.
HgO : O₂
2 : 1
0.22 : 1/2×0.22 = 0.11 mol
Theoretical yield:
Mass of oxygen = number of moles × molar mass
Mass of oxygen = 0.11 mol × 32 g/mol
Mass of oxygen = 3.52 g
Percent yield :
Percent yield = actual yield / theoretical yield × 100
Percent yield = 2.30 g/ 3.52 g × 100
Percent yield =65.34%
The gas production not recognized as nitrate reduction is because <span>organisms that can ferment also produce gas. Therefore even if gas is produced on the tube, you can't tell if it's by fermentation or by nitrate reduction. I hope the answer helps. </span>
