Answer:
Explanation:
Combustion means the process by which the burning of any substance, whether gaseous, liquid or solid, occurs. In this process, the fuel oxidizes and gives off heat, and, frequently, light.
Combustion reactions are reactions where oxygen intervenes as a reagent. Oxygen has the ability to combine with various elements to produce oxides, where then oxidation is the combination of oxygen with another substance. There are oxidations that are extremely slow, but when oxidation is rapid it is called combustion.
When a substance containing carbon and hydrogen (a hydrocarbon) undergoes complete combustion, or burning, oxygen is consumed and carbon dioxide is produced, and water. Incomplete combustion can also occur when part of the fuel does not react completely because oxygen is not enough.
a. What is the pKa of X-281? Express your answer numerically.
At 25∘C, for any conjugate acid-base pair
pKa + pKb = 14.00
b. What is pKb of the conjugate base of X-281? (Assume 25 ∘C.) Express your answer numerically.
Answer:
a. pka = 3,73.
b. pkb = 10,27.
Explanation:
a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:
HX + H₂O ⇄ H₃O⁺ + X⁻
Where ka is defined as:
In equilibrium, molar concentrations are:
[HX] = 0,089M - x
[H₃O⁺] = x
[X⁻] = x
pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:
[H₃O⁺] = 0,004M
Thus:
[X⁻] = 0,004M
And:
[HX] = 0,089M - 0,004M = 0,085M
ka = 1,88x10⁻⁴
And pka = 3,73
b. As pka + pkb = 14,00
pkb = 14,00 - 3,73
pkb = 10,27
I hope it helps!
Answer:
C15H24O
Explanation:
TO GET THE EMPIRICAL FORMULA, WE NEED TO KNOW THE MASSES AND CONSEQUENTLY THE NUMBER OF MOLES OF EACH OF THE INDIVIDUAL CONSTITUENT ELEMENTS.
FIRSTLY, WE CAN GET THE MASS OF THE CARBON FROM THAT OF THE CARBON IV OXIDE. WE NEED TO KNOW THE NUMBER OF MOLES OF CARBON IV OXIDE GIVEN OFF. THIS CAN BE CALCULATED BY DIVIVDING THE MASS BY THE MOLAR MASS OF CARBON IV OXIDE. THE MOLAR MASS OF CARBON IV OXIDE IS 44G/MOL
The combustion of 1.376 g of butylated hydroxytoluene (BHT) produced 4.122 g CO2 and 1.350 g H2O. Calculations yield an empirical formula of CH2O, indicating one carbon, two hydrogen, and one oxygen atom.
To determine the empirical formula of butylated hydroxytoluene (BHT), we can follow these steps:
1. **Find moles of CO2 and H2O produced:**
\[ \text{moles of } CO_2 = \frac{\text{mass of } CO_2}{\text{molar mass of } CO_2} \]
\[ \text{moles of } H_2O = \frac{\text{mass of } H_2O}{\text{molar mass of } H_2O} \]
2. **Find the mole ratio:**
Divide the moles of each element (C, H, and O) in CO2 and H2O by the smallest number of moles.
3. **Write the empirical formula:**
Use the mole ratios to write the empirical formula.
Let's perform the calculations:
\[ \text{Molar mass of } CO_2 = 12.01 \, \text{(C)} + 2 \times 16.00 \, \text{(O)} = 44.01 \, \text{g/mol} \]
\[ \text{Molar mass of } H_2O = 2 \times 1.01 \, \text{(H)} + 16.00 \, \text{(O)} = 18.02 \, \text{g/mol} \]
\[ \text{moles of } CO_2 = \frac{4.122 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.0938 \, \text{mol} \]
\[ \text{moles of } H_2O = \frac{1.350 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.0749 \, \text{mol} \]
Divide by the smallest number of moles (0.0749) to get a ratio close to 1:1:
\[ \text{C} : \text{H} : \text{O} \approx 1.25 : 1 : 1 \]
The ratio is approximately 1:1:1, so the empirical formula is CH2O.
Learn more about empirical formula here:
#SPJ6
The relationship between concentration and rate of reaction is directly proportional. As the concentration of reactants increases, the rate of the reaction also increases.
The rate of a chemical reaction is determined by the collision of reactant particles. When the concentration of reactants is higher, there are more particles per unit volume, increasing the likelihood of successful collisions. More collisions lead to more effective collisions, where the particles have sufficient energy and proper orientation to form products.
As a result, the reaction proceeds at a faster pace. This direct relationship between concentration and rate of reaction is described by the rate law of the specific reaction. The greater the concentration of the reactants, the more rapid the reaction will occur.
Concentration and rate of reaction are directly proportional; an increase in concentration leads to a corresponding increase in the reaction rate.
Read more about concentration here: brainly.com/question/17206790
#SPJ6