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Rama09 [41]
3 years ago
11

An acid and a base solution were mixed together in a crucible and all the water in the mixture was evaporated off. What is left

in the crucible?
Chemistry
1 answer:
Ratling [72]3 years ago
5 0

Answer:

Base on the question , an acid and  a base solution was mixed together in a crucible and all the water in the mixture was evaporated off. The only substance left in the crucible is the salt solution.

Explanation:

Acids are substances that dissolves in water to produce hydrogen ion(H⁺) while base are substances that dissolves in water to produce hydroxide ion (OH⁻).

The chemical reaction between an acid and a base solution is known as a neutralization reaction. The reaction between these 2 compounds will produce salt and water.  An acid solution will react with a base solution to form salt and water.  An example of this kind of reaction is between HCl (Hydrochloric acid) and NaOH(sodium hydroxide).

HCl + NaOH → NaCl + H₂O . The product form here is sodium chloride(salt) and water

Base on the question , an acid and  a base solution was mixed together in a crucible and all the water in the mixture was evaporated off. The only substance left in the crucible is the salt solution. Note that the other product which is water has been evaporated and we are left with only salt solution.

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2 years ago
A 25.00 mL sample of 0.320 M KOH is titrated with 0.750 M HBr at 25 °C.
Fudgin [204]

Answer:

a. pH = 13.50

b. pH = 13.15

Explanation:

Hello!

In this case, since the undergoing chemical reaction between KOH and HBr is:

HBr+KOH\rightarrow KBr+H_2O

As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:

pOH=-log([OH^-])=-log(0.320)=0.50

Thus, the pH is:

pH+pOH=14\\pH=14-pOH=14-0.50\\pH=13.50

Which is the same answer for a and b as they ask the same.

Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:

n_{KOH}=0.02500mL*0.320mol/L=0.00800mol\\\\n_{HBr}=0.005L*0.750mol/L=0.00375mol

It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:

n_{KOH}^{remaining}=0.00425mol

Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):

[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M

So the pOH and the pH turn out:

pOH=-log(0.142)=0.849\\pH+pOH=14\\pH=14-pOH=14-0.849\\pH=13.15

Best regards!

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<u>Activation Energy = Threshold Energy - Average Kinetic Energy</u>

<u>This means Activation energy decreases on increasing kinetic energy</u>

On increasing Temperature average kinetic energy of the molecule increases which reduces the activation energy and the reaction occur faster in that case.

Catalyst also reduces the Activation energy.

<u>Er = Threshshold energy for reaction at 30 degree</u>

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