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1 year ago

Without using Appendix B, predict the sign of ΔS° for(b) KBr(s)→ KBr(aq)

Chemistry

1 answer:

Volgvan1 year ago

8 0

Since the sign is positive, the entropy increased by 88.48 J/K.

Examine the phases of the species present to determine whether a physical or chemical process will cause an increase or decrease in entropy. Keep in mind "Silly Little Goats" to aid you in telling.

[1 Sf K+1 + 1 Sf Br-1 (aq)] ([1Sf(KBr (s))])

[1(102.5) + 1(82.42)] - [1(96.44)] = 88.48 J/K

If the entropy has grown, we say that Delta S is positive, and if it has dropped, we say that Delta S is negative. Due to its ionic nature, KBr is soluble in water and causes the 'K(+)' ions to hydrate.

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CONCLUSIONS: Suppose a student titrated a sample of monoprotic acid of unknown concentration using a previously standardized sol

Zolol [24]

Concentration of unknown acid is 0.061 M

Given:

Concentration of NaOH = 0.125 M

Volume of NaOH = 24.68 mL

Volume of acid solution = 50.00 mL

To Find:

Concentration of the unknown acid

Solution: Concentration is the abundance of a constituent divided by the total volume of a mixture. The concentration of the solution tells you how much solute has been dissolved in the solvent

Here we will use the formula for concentration:

M1V1 = M2V2

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2 years ago

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How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

meriva

Answer:

$\large \boxed{\text{77.4 mL}}$

Explanation:

Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

V/mL: 249

c/mol·L⁻¹: 0.0443 0.285

1. Calculate the moles of Ba(OH)₂

$\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}$

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

$\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}$

3. Calculate the volume of HCl

$V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}$

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