In this question, you are given the NaOH volume but asked for concentration.
Don't forget that for every 1 mol of NaOH there will be 1 mol OH- ion, but for every 1 mol of H2SO4 there will be 2 mol of H- ion.
To neutralize you need the same amount of OH- and H+, so the equation should be:
OH-= H+
<span>35.50cm3 * x*1= 25cm3* 0.2mol/dm3 *2
</span>x= 10/35.5 mol/dm3= 0.2816/dm3
Answer:-A. It is less than 890 kJ/mol because the amount of energy required to break bonds is less than the amount of energy released in forming bonds.
Explanation: Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and
for the reaction comes out to be positive.
Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and
for the reaction comes out to be negative.
In the formation of new bonds more energy is released than is required to break the existing bonds, heat is released.
In the formation of bonds less energy is released than is required to break the existing bonds, heat is absorbed.
Answer:
The velocity of the particle is 2 m/s,
Explanation:
Kinetic energy is defined as energy of the body due to its motion. It is given by :

Where :
m = mass of the object
v = velocity of the object
We have , particle with mass m and its kinetic energy is twice its mass.




And unit of velocity are m/s , so the velocity of the particle is 2 m/s.
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a <u>sp³</u> hybridization because they are involved only in single bonds.