Answer:
Water is the solvent
Both the ethanol and the hydrogen peroxide are the solute
Explanation:
Both the hydrogen peroxide and ethanol are sisobable in water.
There are 0.05 moles of ethanol.
1 litreof water contains 55.55 moles of water.
0.2 g of hydrogen peroxide contains 0.2/34 = 0.0059 moles of hydrogen peroxide (the 34 is the molar mass of hydrogen peroxide).
Since there are more moles of water, water becomes the solvent and the other two liquids dissolve in it.
Answer : Amoxicillin Suspension 125 mg/ 5 ml is 125 mg of Amoxicillin per 5 ml of suspension is an example of weight to volume.
Explanation :
Weight by volume (w/v) means that the mass of solute present in 100 mL volume of solution.
Weight by weight (w/w) means that the mass of solute present in 100 gram of solution.
Volume by volume (v/v) means that the volume of solute present in 100 mL volume of solution.
As per question, amoxicillin suspension is, 125 mg/ 5 ml that means 125 mg of Amoxicillin present in 5 mL of suspension. So, it is an example of weight to volume.
Hence, it is an example of weight to volume.
Mitochondria are rod-shaped organelles are basically considered the power producers of the cell, it converts oxygen and nutrients into adenosine triphosphate or ATP, which is the chemical energy,also known as "currency" of the cell which powers the metabolic actions of the cell. This process is called aerobic respiration and it is the reason animals breathe oxygen. Cellular repiration happens in the mitochodrion. The 3 phases of cellular respiration are Krebs Cycle, Electron Transport and Glycolysis (Fermentation). Glycolysis takes place in the cytoplasm while the Krebs cycle and electron transport take place in the mitochondria.
Answer:
= 13.0 moles O2
Explanation:
1] Given the equation: 2C8H18 + 25 O2 ----> 16CO2 + 18H2O
a. How many moles of oxygen gas are required to make 8.33 moles of carbon dioxide?
8.33 moles CO2 X
25mol O2
16mol CO2
= 13.0 moles O2
Answer:
The correct answer is because the molecular structure.
Explanation:
The difficulty of ammonia and methane to be represented on paper is due to the molecular structure. These compounds have a three-dimensional projection with defined angles. Ammonia presents angles of 109.5º between the atom of Nitrogen and those of Oxygen. The ammonia presents 107.8º between the oxygen atoms.
In the methane molecule, there is 109.5º between the hydrogen molecules and the carbon atom. This results in the need for a 3D representation of the molecule.
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