Answer:
Rate constant = 0.0237 M-1 s-1, Order = Second order
Explanation:
In this problem, it can be observed that as the concentration decreases, the half life increases. This means the concentration of the reactant is inversely proportional to the half life.
The order of reaction that exhibit this relationship is the second order of reaction.
In the second order of reaction, the relationship between rate constant and half life is given as;
t1/2 = 1 / k[A]o
Where;
k = rate constant
[A]o = Initial concentration
k = 1 / t1/2 [A]
Uisng the following values;
k = ?
t1/2 = 113
[A]o = 0.372M
k = 1 / (113)(0.372)
k = 1 / 42.036 = 0.0237 M-1 s-1
Answer:
Magnesium
Explanation:
Using the KLMN styled electronic configuration, the electronic configuration of sulphur with atomic number 16 is 2, 8,6
What this means is that it needs extra 2 electrons to fill into its M shell to attain the octet configuration.
Now let’s look at Magnesium, with atomic number 12, the electronic configuration it has is 2,8,2.
This means it has 2 extra electrons to give away so as to attain its own stability.
The sulphur atom will gladly accept the two electrons which the magnesium atom wants to give away. This makes it the perfect element to be reacted with sulphur to make it attain its octet configuration
Answer:
The jewelry is 2896.54_Kg/m^3 less dense than pure silver
Explanation:
Density of jewellery = (mass of jewellery) ÷ (volume of jewellery)
=3.25g ÷ 0.428mL = 0.00325Kg÷0.000000428m^3 = 7583.46Kg/m^3
The density of silver is 10490_Kg/m^3 which is (10490 - 7583.46) 2896.54_Kg/m^3 more dense than the jewellery
The density of Silver [Ag]
The weight of Silver per cubic centimeter is 10.49 grams or the weight of silver per cubic meter is 10490 kilograms, that is the density of silver is 10490 kg/m³; at 20°C (68°F or 293.15K) at a pressure of one atmospheres.
Answer:
4.504g of acetic acid
Explanation:
The acetic acid in reaction with NaOH produce acetate ion, thus:
CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺
<em>That means the moles of acetate buffer comes, in the first, from the acetic acid</em>
As you need 500mL (0,500L) of a 0.150M acetate buffer, moles are:
0.500L × (0.150mol / 1L) = <em>0.075 moles of acetate</em>. That is:
0.075mol = [CH₃COO⁻] + [CH₃COOH]
Thus, grams of acetic acid you need to prepare the buffer are:
0.075 moles acetic acid × (60.05g / 1mol) = <em>4.504g of acetic acid</em>
Answer:
Sulfur — S
calcium — Ca
carbon — C
silver — Ag
gold — Au
helium — He
hydrogen — H
should become easier for you to memorise them as you go on in life (meaning going up a grade… grade 7 to 8 to 9 to 10 until you reach college and whatnot)
