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Andreas93 [3]
3 years ago
12

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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Explanation:

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We are given the amount of carbon dioxide to produce from the reaction. This will be our starting point.

 

560 L CH4 ( 1 mol CH4/ 22.4 L CH4 ) (2 mol O2/ 1 mol CH4 ) ( 22.4 L O2 / 1 mol <span>O2</span><span>) = 1120 L O2</span>

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A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g
Lynna [10]

Answer:

\%NaHCO_3=61.2\%

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

NaHCO_3+HA\rightarrow NaA+H_2O+CO_2

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%

Best regards.

5 0
3 years ago
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