When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.

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A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq )

Alex73 [517]

Answer:

0.00369 moles of HCl react with carbonate.

Explanation:

Number of moles of HCl present initially = $\frac{0.1200}{1000}\times 50.0$ moles = 0.00600 moles

Neutralization reaction (back titration): $NaOH+HCl\rightarrow NaCl+H_{2}O$

According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.

So, excess number of moles of HCl present = number of NaOH added for back titration = $\frac{0.0980}{1000}\times 23.60$ moles = 0.00231 moles

So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles

Hence, 0.00369 moles of HCl react with carbonate.

3 0

3 years ago