Answer:
0.129 L = 129.0 mL.
Explanation:
- NaOH neutralizes acetic acid (CH₃COOH) according to the balanced reaction:
<em>NaOH + CH₃COOH → CH₃COONa + H₂O. </em>
- According to the balanced equation: 1.0 mole of NaOH will neutralize 1.0 mole of CH₃COOH.
<em>no. of moles of CH₃COOH = mass/molar mass </em>= (2.0 g)/(60 g/mol) = <em>0.033 mole. </em>
<em>
</em>
no. of moles = (0.258 mol/L)(V)
- At neutralization: no. of moles of NaOH = no. of moles of CH₃COOH
∴ (0.258 mol/L)(V) = 0.033 mole
<em>∴ The volume of NaOH</em> = (0.033 mole)/(0.258 mol/L) = <em>0.129 L = 129.0 mL.</em>
From Boyle's law, we can use the equation
P1V1 = P2V2
to find the volume when the pressure is increased to 760 kPa:
V2 = P1V1 / P2
V2 = (210 kPa)(15.0 L) / 760 kPa
V2 = 4.14 Liters
Therefore, the volume is 3.99 Liters after the pressure increased to 760 kPa.
But if the pressure is increased to 790 kPa, the volume of the krypton will decrease to
V2 = P1V1 / P2
V2 = (210 kPa)(15.0 L) / 790 kPa
V2 = 3.99 Liters
Answer:
An example of a metal with an organic radical is: Pentamethylcyclopentadiene, which is the union between Zinc and an organic compound, where in order to obtain it there is a release of dihydrogen.
Another example is lead tetraethyl, it is a compound for industrial use, it is dangerous for the human body, toxic and can be used as fuel.
Explanation:
In radical chemistry a chemical species is called both organic and inorganic, which one of the most used examples is methyl, where it is a chemical structure that has one carbon and 4 hydrogens attached to this central carbon.
These chemical compounds when united with metals, release one of the unions that they have with hydrogens, it is because in reactions they release hydrogens.
Answer:
The reaction will shift to the right
Explanation:
In the reaction:
NO(g) + NO₂(g) ⇌ N₂O₃(g)
Kp of reaction is:
= 0.03
<em>Where P represents the pressures in equilibrium</em>
Replacing in the kP formula the initial pressures:
= 0.001
As the <em>Reaction quotient (Q) </em>is less than kP, <em>the reaction will shift to the right </em>producing more N₂O₃ until Q = kP.
