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Svetllana [295]
3 years ago
13

What are some examples of mixtures?

Chemistry
1 answer:
Flura [38]3 years ago
8 0

Here are 5 examples of suspensions mixtures:

<span>1-mercury shaken in oil</span>

<span>2-oil shaken in water</span>

<span>3-powdered chalk in water </span>

<span>4-dust in air</span>

<span><span>5-soot in air</span></span>

<span><span>hope this helps
</span></span>

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A buffer is prepared by combining 25 ml of 0.5O MNH(aq) with 25 mL of 0.20 M HCI. What is the pH of the buffer? (KA (NH4) = 5.6
V125BC [204]

Answer:  e) 9.43

Explanation: A buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid and the pH of the buffer is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

There are 25 mL of 0.50 M NH_3 which is a weak base. 25 mL of 0.20 M HCl, a strong acid are added. Ammonia reacts with HCl to form its conjugate acid, ammonium ion. The net ionic equation will be:

NH_3(aq)+H^+(aq)\rightarrow NH_4^+(aq)

From above reaction, ammonia and HCl react in 1:1 mol ratio. Let's calculate the moles of each we have before the reaction. There is also, 1:1 mol ratio between HCl and ammonium ion.

moles of ammonia = 25mL(\frac{1L}{1000mL})(\frac{0.50mol}{1L})

= 0.0125 mol

moles of HCl = 25mL(\frac{1L}{1000mL})(\frac{0.20mol}{1L})

= 0.005 mol

Excess moles of ammonia = 0.0125 - 0.005 = 0.0075 mol

moles of ammonium ion formed = 0.005 mol

Total volume of the solution = 0.025 mL + 0.025 mL = 0.050 L

concentration of ammonia in buffer = \frac{0.0075mol}{0.050L}

= 0.15 M

concentration of ammonium ion in buffer = \frac{0.005mol}{0.050L}

= 0.10 M

pKa is calculated from given Ka as:

pKa=-logKa

pKa=-log5.6*10^-^1^0

pKa = 9.25

Plug in the values in Handerson equation:

pH=9.25+log(\frac{0.15}{0.10})

pH = 9.25 + 0.18

pH = 9.43

So, the correct choice is e) 9.43 .

7 0
3 years ago
The Ksp for CaCO3 is 3.8 x 10-9. Determine the concentration in moles/liter of calcium ion in a saturated solution of calcium ca
balu736 [363]

Answer: c. 6.2\times 10^{-5}moles/L

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as K_{sp}

The equation for the ionization of the calcium carbonate is given as:

CaCO_3\leftrightharpoons Ca^{2+}+CO_3^{2-}

We are given:

K_{sp}=3.8\times 10^{-9}

By stoichiometry of the reaction:

1 mole of CaCO_3 gives 1 mole of Ca^{2+} and 1 mole of CO_3^{2-}.

Expression for the equilibrium constant of CaCO_3 will be:

K_{sp}=[Ca^{2+}][CO_3^{2-}]

3.8\times 10^{-9}=[s][s]

3.8\times 10^{-9}=s^2

s=6.2\times 10^{-5}moles/L

Hence, the concentration in moles/liter of calcium ion in a saturated solution of calcium carbonate is 6.2\times 10^{-5}moles/L

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4 years ago
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Explanation:

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