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3 years ago

Will give BRAINLIEST!!! Will give BRAINLIEST

Chemistry

1 answer:

pantera1 [17]3 years ago

6 0

Answer:

The calculated concentration of acid will be higher than the actual concentration of acid

Explanation:

We have information that all enable us to calculate the concentration of KOH in the solution. From the question, we have;

Mass of KOH= 14.555g

Molar mass of KOH= 56.1056 g/mol

Volume of solution= 500 ml

Number of moles of KOH= ???

From;

m/M= CV

m= mass of KOH

M= molar mass of KOH

C= concentration of KOH solution

V= volume of solution

Substituting values;

14.555g/56.1056 g/mol = C× 500/1000

0.259 moles = 0.5C

C= 0.259/0.5

C= 0.518 M

If the acid is HA, the reaction equation is;

KOH(aq) + HA(aq) ----> KA(aq) + H2O(l)

The concentration of the acid is usually determined via titration. This involves delivering a particular volume of acid in a burette into the base and watching out for the volume of acid used at end point. If there are air bubbles in the burette, then more volume of acid is recorded than that actually used and this will make the calculated concentration of the acid to be higher than the actual concentration of acid present.

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A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

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