Water and carbon dioxide are waste products released as a direct result of materials
We have gravity here on earth while in space there’s none
Answer: The empirical formula for the given compound is 
Explanation : Given,
Percentage of C = 38.8 %
Percentage of H = 16.2 %
Percentage of N = 45.1 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 38.8 g
Mass of H = 16.2 g
Mass of N = 45.4 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Moles of Nitrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.
For Carbon = 
For Hydrogen = 
For Oxygen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N = 1 : 5 : 1
Hence, the empirical formula for the given compound is 
Answer:
The answer to your question is 1.11 M
Explanation:
Data
volume 1 = 287 ml
concentration 1 = 1.6 M
volume 2= 412 ml
concentration 2 = ?
Formula
Volume 1 x concentration 1 = Volume 2 x concentration 2
Solve for concentration 2
concentration 2 = (volume 1 x concentration 1) / volume 2
Substitution
concentration 2 = (287 x 1.6) / 412
Simplification
concentration 2 = 459.2 / 412
Result
concentration 2 = 1.11 M
Answer:
Fe
Explanation:
The cell potential is:
ΔE°cell = E°red(red) - E°red(oxid)
Where, E°red(red) is the reduction potential of the substance that is reducing, and E°red(oxid) is the reduction potential of the substance that is oxidizing. For the reaction be spontaneous and happen, ΔE°cell > 0.
The reduction takes place in the cathode, which is the negative pole, and the oxidation in the anode, which is the positive pole. So, the electrons flow from the positive pole to the negative pole (anode to cathode).
Then, if the voltmeter measured a negative potential, it means that is was attached incorrectly. So, the anode is Fe.
