We can use the heat equation,
Q = mcΔT
where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 11.2 kJ = 11200 J
m = <span>145 g
</span>c = ?
ΔT = (67 - 22) °C = 45 °C
By applying the formula,
11200 J = 145 g x c x 45 °C
c = 1.72 J g⁻¹ °C⁻¹
Hence, specific heat of benzene is 1.72 J g⁻¹ °C⁻¹.
Sharing of two electrons make a <u>Covalent </u>bond.
<u>Explanation: </u>
Attractions among the atoms bring them together. So the electrons from each of the atoms are attracted towards the nucleus of those two atoms, that “share” the electrons produces a covalent bond.
It is also named as molecular bond, a bond that entails the sharing of a pair of electrons among the atoms. When the atoms share the electrons among themselves, it produces a molecule, which is more stable than the atom.
If the attractions between the atoms are strong enough and if every atom has enough space for the electrons in its outermost energy level then there occurs covalent bonding. So electrons are very important in the covalent bond formation.
Answer:
anion
Explanation:
anion undergoes reduction.
cation undergoes oxidation.
Answer:
5.56 × 10⁻⁸
Explanation:
Step 1: Given data
- Concentration of the weak acid (Ca): 0.187 M
Step 2: Calculate the concentration of H⁺
We will use the following expression.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
![Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%20%7D%7BCa%7D%20%3D%20%5Cfrac%7B%281.02%20%5Ctimes%2010%5E%7B-4%7D%29%5E%7B2%7D%20%7D%7B0.187%7D%20%3D%205.56%20%5Ctimes%2010%5E%7B-8%7D)
You have to use everything that is given since you have to know which is the limiting reactant. We find the limiting reactant by calculating the number of moles of each reactant and compare the number of moles. The limiting reactant would be the one that is consumed fully by the reaction.
