Answer:
1. First one is true : as per periodic table down the group , the elements has increasing order of shell & with that the London dispersion forces brings the inter-molecules close together and bromine converted into liquid .
2. second one is False because carbon-carbon bonds are not weak bonds they form mutual covalent bonds which are stronger bonds and cannot be easily disrupted .
3. A single carbon atom has the valency of 4 so it can be bonded with four hydrogen atom at the same time .
Explanation:
The pH of the solution is 2.54.
Explanation:
pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.
The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.
So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the
![K_{a}=\frac{[H^{+}][HB] }{[reactant]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BHB%5D%20%7D%7B%5Breactant%5D%7D)
[HB] is the concentration of base.


Then
![pH = - log [x] = - log [ 0.283 * 10^{-2}]\\ \\pH = 2 + 0.548 = 2.54](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5Bx%5D%20%3D%20-%20log%20%5B%200.283%20%2A%2010%5E%7B-2%7D%5D%5C%5C%20%5C%5CpH%20%3D%202%20%2B%200.548%20%3D%202.54)
So the pH of the solution is 2.54.
Answer is D organs because cell forms tissues, tissues form organs and organs forms the organ system
Answer:
3.15 × 10⁻⁶ mol H₂/L.s
1.05 × 10⁻⁶ mol N₂/L.s
Explanation:
Step 1: Write the balanced equation
2 NH₃ ⇒ 3 H₂ + N₂
Step 2: Calculate the rate of production of H₂
The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s
Step 3: Calculate the rate of production of N₂
The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s