Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.
<h3>What is volume?</h3>
Volume is the area occupied by the substance and is the ratio of the mass to the density.
At STP, 1 mole of gas occupies 22.4 L of volume
Given,
Volume of ammonia reacted = 0.720 L
The combustion reaction is shown as,
From the stoichiometry of the reaction, it can be said that,
L of ammonia reacts with L of oxygen gas.
So, 0.720 L of ammonia will react with:
Therefore, the volume of oxygen required is 900 mL.
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Answer:
turgor pressure can be done in a lab or a self test.
turgor pressure is key to the plant’s vital processes. It makes the plant cell stiff and rigid. Without it, the plant cell becomes flaccid. Prolonged flaccidity could lead to the wilting of plants.
Turgor pressure is also important in stomate formation. The turgid guard cells create an opening for gas exchange. Carbon dioxide could enter and be used for photosynthesis. Other functions are apical growth, nastic movement, and seed dispersal.
Explanation:
- salt is bad for turgor pressure.
- Turgidity helps the plant to stay upright. If the cell loses turgor pressure, the cell becomes flaccid resulting in the wilting of the plant.
- The wilted plant on the left has lost its turgor as opposed to the plant on the right that has turgid cells.
Answer:
SbcI3
Explanation:
The symbol of antimony is 'Sb'.
The symbol of chlorine is 'Cl'
First write down the symbol of the first element.
Use the prefix to determine the atoms of first element. If there is no prefix on element then there is only 1 atom.
Now write down the symbol of the second element.
Use the prefix to determine the atoms of second element.
Use prefix as 'mono' for '1', 'di' for '2', 'tri' for '3' and so on.
Answer: well the receptacle connect the stalk to the flower and to support the flower and keeps the flower in an elevated position so as to attract the insects
Explanation: I don’t know if this helped
Lewis Structure is drawn in following steps,
1) Calculate Number of Valence Electrons: # of Valence electrons in Mg = 2
# of Valence electrons in I = 7
# of Valence electrons in I = 7
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Total Valence electrons = 16
2) Draw Mg as a central atom surround it by two atoms of Iodine.3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.
Total Valence electrons 16
- Four electrons - 4
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12
4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.
The final lewis structure formed is as follow,