There are 22 bonding parts
Answer:
1400KJ/mol⁻¹
Explanation:
Amount of heat required can be found by:
Q = m × c × ΔT
<em>Where m is the mass, c is the specific heat capacity (4.2KJ for water) and ΔT is the change in temperature.</em>
Q = 24 × 4.2 × (23 - 9)
= 24 × 4.2 × 14
= 1411.2KJ/mol⁻¹
= <u>1400KJ/mol⁻¹</u> (to 2 significant figures)
Answer:
option b
Explanation:
When the energy is released the process is called exothermic reaction. This happens when the bonds are broken in the reactants and the system release energy.
Answer:
a
Explanation:
The formation of ion occurs when an atom that is said to be neutral gains or losses electrons.
At the time it gains electrons, it is regarded that a negative ion (anion) is formed.
When it loses electron, it is regarded that a positive ion (cation) is formed.
Atomic number = No of protons and electrons occurring in a neutral atom.
Given that:
Protons = 14
electron = 18
Net Charge = no of proton - no of electron
= 14 - 18
= -4
Mass number = 14 + 15 = 29
So, the chemical symbol = 
For ion with
27 proton, 32 neutrons and 25 electrons
Net charge = 27 - 25
= +2
Mass number = 27 + 32 = 59
Thus, the chemical symbol = 
Answer:
Rate constant k = 1.57*10⁻⁵ s⁻¹
Explanation:
Given reaction:

Expt [A] M [B] M Rate [M/s]
1 3.40 4.16 1.82*10^-4
2 4.59 4.16 3.32*10^-4
3. 3.40 5.46 1.82*10^-4
![Rate = k[A]^{x}[B]^{y}](https://tex.z-dn.net/?f=Rate%20%3D%20k%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D)
where k = rate constant
x and y are the orders wrt to A and B
To find x:
Divide rate of expt 2 by expt 1
![\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2](https://tex.z-dn.net/?f=%5Cfrac%7B3.32%2A10%5E%7B-4%7D%20%7D%7B1.82%2A10%5E%7B-4%7D%20%7D%20%3D%5Cfrac%7B%5B4.59%5D%5E%7Bx%7D%20%5B4.16%5D%5E%7By%7D%20%7D%7B%5B3.40%5D%5E%7Bx%7D%20%5B4.16%5D%5E%7By%7D%20%7D%5C%5C%5C%5Cx%20%3D2)
To find y:
Divide rate of expt 3 by expt 1
![\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0](https://tex.z-dn.net/?f=%5Cfrac%7B1.82%2A10%5E%7B-4%7D%20%7D%7B1.82%2A10%5E%7B-4%7D%20%7D%20%3D%5Cfrac%7B%5B3.40%5D%5E%7Bx%7D%20%5B5.46%5D%5E%7By%7D%20%7D%7B%5B3.40%5D%5E%7Bx%7D%20%5B4.16%5D%5E%7By%7D%20%7D%5C%5C%5C%5Cy%20%3D0)
Therefore: x = 2, y = 0
![Rate = k[A]^{2}[B]^{0}](https://tex.z-dn.net/?f=Rate%20%3D%20k%5BA%5D%5E%7B2%7D%5BB%5D%5E%7B0%7D)
To find k
Use rate for expt 1:
![k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7BRate1%7D%7B%5BA%5D%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B1.82%2A10%5E%7B-4%7DM%2Fs%20%7D%7B%5B3.40%5D%5E%7B2%7D%20%7D%20%3D1.57%2A10%5E%7B-5%7D%20s-1)