Answer:
Decreasing the volume of solvent in the solution of molecule A
Explanation:
We know that one of the factors that affect the rate of reaction is the concentration of the reactants. The greater the concentration of reactants, the faster the rate of reaction (the greater the frequency of collision between reactants).
Hence, when we decrease the volume of solvent in the solution of molecule A, the concentration of the solution increases and consequently more particles of molecule A are available to collide with particles of molecule B resulting in a higher rate of reaction.
The mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.
The given parameters:
- <em>Density of the octane, ρ = 0.703 g/ml</em>
- <em>Volume of octane, v = 3.79 liters</em>
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The mass of the octane burnt is calculated as follows;

The combustion reaction of octane is given as;

From the reaction above:
228.46 g of octane -------------------> 704 g of CO₂ gas
2,664.37 of octane --------------------> ? of CO₂ gas

Thus, the mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.
Learn more about combustion of organic compounds here: brainly.com/question/13272422
Answer:
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
Explanation:
Data Given
M1 = 6.00 M
M2 = 2.5 M
V1 = 250 mL
V2 = ?
Solution:
As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.
Now
first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution
For this Purpose we use the following formula
M1V1=M2V2
Put values from given data in the formula
6 x V1 = 2.5 x 250
Rearrange the equation
V1 = 2.5 x 250 /6
V1 = 104 mL
So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M aqueous solution of ammonium hydroxide
But we have to prepare 250 mL of the solution.
so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.
in this question you have to tell about the amount of water that is 146 mL
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
Answer:
A.) ![K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D)
Explanation:
The general Kb expression is:
![K_b = \frac{[HA][OH^-]}{[A^-]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BHA%5D%5BOH%5E-%5D%7D%7B%5BA%5E-%5D%7D)
In this equation
-----> Kb = equilibrium constant
-----> [HA] = acid
-----> [A⁻] = base
Since liquids are not included in equilibrium expressions, H₂O should not be present. The products are in the numerator while the reactant are in the denominator. In this reaction, CH₃NH₂ is acting as a base and CH₃NH₃⁺ is acting as an acid.
As such, the expression is:
![K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D)
Well you didn’t post any but I’m hoping that either plastic, lead, wood, glass, or paper, are an option.