<h3>
Answer:</h3>
The structure of Electrophile (2-methylpropan-2-ylium) is shown in attached picture highlighted by green color.
<h3>Explanation:</h3>
Electrophiles are those species which are electron deficient and loves electron. These species often carry positive charge or contain atoms which have incomplete octet.
Also, we know that Benzene undergo Electrophilic Substitution reactions in which hydrogen atom is replaced by other electrophile. In given statement the electrophile to be reacted with benzene to form tert-Butylbenzene is generated from 2-methylpropene. 2-methylpropene when treated with phosphoric acid adds one proton across the double bond (electrophilic addition reaction) and generate a positively charged tertiary carbocation (stable) (highlighted green) which acts as an electrophile. Further on reaction with benzene this electrophile is substituted with proton and results in the formation of tert-Butylbenzene.
Answer:
The answer is Nitrogen has a larger radius and one less proton than oxygen.
Explanation:
Nitrogen is a gas and gas expands to fill up an area so oxygen would be pushed out and the Nitrogen would take over.
Answer:
(a) -0.00017 M/s;
(b) 0.00034 M/s
Explanation:
(a) Rate of a reaction is defined as change in molarity in a unit time, that is:
Given the following reaction:
We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:
Reaction rate is also equal to the rate of formation of products divided by their coefficients:
Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:
(b) Using the relationship derived previously, we know that:
Rate of appearance of nitrogen dioxide is given by:
Which is obtained from the equation:
If we multiply both sides by 4, that is:
This yields:
[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]
The molarity of the aqueous acid solution : 0.0101 M
<h3>Further explanation </h3>
Titration is a procedure for determining the concentration of a solution by reacting with another solution that is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range
Acid-base titration formula
Ma. Va. na = Mb. Vb. nb
Ma, Mb = acid base concentration
Va, Vb = acid base volume
na, nb = acid base valence (amount of H⁺, OH⁻)
Va=12.5 ml
na = 2 (H₂SO₄⇒2H⁺+SO₄²⁻⇒2 ion H⁺)
Mb=0.015
Vb=16.8 ml
nb = 1(LiOH⇒Li⁺+OH⁻⇒1 ion OH⁻)
The molarity of the acid(Ma) :