Answer: A) NO increases the rate at which
molecules are formed.
Explanation:
A catalyst is a substance which increases the rate of a reaction by taking the reaction through a different path which involves lower activation energy and thus more reactant molecules can cross the energy barrier by undergoing collisions and convert to products.

Thus NO will increase the rate of reaction by lowering the activation energy and thus the colllisions among
and
molecules will incraese which in turn will lead to formatioon of more
molecules.
It's about .0934 Liters at STP.
The hydrogen deficiency index( HDI) of strigol is = 10
<h3>How to calculate HDI:</h3>
The hydrogen deficiency index is used to measure the number of degree of unsaturation of an organic compound.
Strigol is an example of an organic compound because it contains carbons and hydrogen.
To calculate the HDI using the molecular formula given (C19H20O6) the formula for HDI is used which is:

where C = number of carbon atoms = 19
n= number of nitrogen atoms = 0
h= number of hydrogen atoms = 20
X = number of halogen atoms = 0
Note that oxygen was not considered because it forms two bonds and has no impact.
There for HDI =

HDI=

HDI =

HDI = 10
Therefore, the hydrogen deficiency index of strigol is = 10
Learn more about unsaturated compounds here:
brainly.com/question/490531
Answer:
Do you need 3 ways or just one?
1. Temperature.
2. Pressure.
3. Polarity.
Explanation:
Eh hope these help, Idr understand the question but those are 3 ways to increase the solubility of a solid in water.
Answer:
Option B. 4 moles of the gaseous product
Explanation:
Data obtained from the question include:
Initial volume (V1) = V
Initial number of mole (n1) = 2 moles
Final volume (V2) = 2V
Final number of mole (n2) =..?
Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:
V1/n1 = V2/n2
V/2 = 2V/n2
Cross multiply
V x n2 = 2 x 2V
Divide both side by V
n2 = (2 x 2V)/V
n2 = 2 x 2
n2 = 4 moles
Therefore, 4 moles of the gaseous product were produced.