Answer:
The answer is 6.25g.
Explanation:
First create your balanced equation. This will give you the stoich ratios needed to answer the question:
2C8H18 + 25O2 → 16CO2 + 18H2O
Remember, we need to work in terms of NUMBERS, but the question gives us MASS. Therefore the next step is to convert the mass of O2 into moles of O2 by dividing by the molar mass:
7.72 g / 16 g/mol = 0.482 mol
Now we can use the stoich ratio from the equation to determine how many moles of H2O are produced:
x mol H2O / 0.482 mol O2 = 18 H2O / 25 O2
x = 0.347 mol H2O
The question wants the mass of water, so convert moles back into mass by multiplying by the molar mass of water:
0.347 mol x 18 g/mol = 6.25g
Vas happenin
Hope your day is going well
Atoms are referred to basic unit of matter
Hope this helps *smiles*
Answer:
See explanation
Explanation:
For this question, we have to remember the effect of an atom with high <u>electronegativity</u> as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an <u>inductive effect</u>. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be <u>weaker</u> and the compound will be more acid (because is easier to produce the hydronium ion
).
With this in mind, for A in the last compound, we have <u>2 Br atoms</u> near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have <u>more acidity</u>. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.
In B, the difference between the molecules is the <u>position</u> of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a <u>higher inductive effect</u> and more <u>acidity</u>.
See figure 1
I hope it helps!
Answer:
E₁ ≅ 28.96 kJ/mol
Explanation:
Given that:
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,
Let the activation energy for a catalyzed biochemical reaction = E₁
E₁ = ??? (unknown)
Let the activation energy for an uncatalyzed biochemical reaction = E₂
E₂ = 50.0 kJ/mol
= 50,000 J/mol
Temperature (T) = 37°C
= (37+273.15)K
= 310.15K
Rate constant (R) = 8.314 J/mol/k
Also, let the constant rate for the catalyzed biochemical reaction = K₁
let the constant rate for the uncatalyzed biochemical reaction = K₂
If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:
K₁ = 3.50 × 10³
K₂ = 1
Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;
we can use the formula for Arrhenius equation;

If
&





E₁ ≅ 28.96 kJ/mol
∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol