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padilas [110]
3 years ago
8

Debbie was doing a science lab, and she was instructed to decide whether a piece of plastic would float or sink without actually

putting the plastic in the water. how could she do this?
Chemistry
2 answers:
miv72 [106K]3 years ago
7 0

''Find the density of the plastic and compare it to the density of water''

Aleksandr [31]3 years ago
3 0

Answer:

Debbie can take the following step

1. Find the mass of the substance in kilograms by measuring on a chemical balance and record

2. Fine the volume of the plastic(m^3). if it is a cylindrical plastic, volume will be area multiplied by height(\pi r^{2} h)

3. Density of a substance is the ratio of the mass of a substance to its volume.Find its density and record

4. Compare the density of the substance to that of water which is 1000kg/m^3.

5. If it is greater than that of water then it will sink, if it is less than that of water it will float

Explanation:

Debbie can take the following step

1. Find the mass of the substance in kilograms by measuring on a chemical balance and record

2. Fine the volume of the plastic(m^3). if it is a cylindrical plastic, volume will be area multiplied by height(\pi r^{2} h)

3. Density of a substance is the ratio of the mass of a substance to its volume.Find its density and record

4. Compare the density of the substance to that of water which is 1000kg/m^3.

5. If it is greater than that of water then it will sink, if it is less than that of water it will float

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If a buffer contains 1.05M B and 0.750M BH+ has the pH of 9.5. What would be the pH after 0.005mol of HCL is added to 0.5L of so
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\displaystyle \text{pK}_b = \text{pOH} -\log{\frac{[\text{Salt}]}{[\text{Base}]}}\\\phantom{\text{pK}_b} = 4.5 - \log{\frac{0.750}{1.05}} \\\phantom{\text{pK}_b} =4.64613.

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After adding the HCl:

  • n(\text{B}) = 0.525 - 0.005 = 0.520\;\text{mol};
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Assume that the volume is still 0.5 L:

  • \displaystyle [\text{B}] = \frac{n}{V} = \frac{0.520}{0.5} = 1.04\;\text{mol}\cdot\text{dm}^{-3}.
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What's will be the pH of the solution?

Apply the Henderson-Hasselbalch equation again:

\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}} = 4.64613 + \log{\frac{0.760}{1.04}} = 4.50991

\text{pH} = \text{pK}_w - \text{pOH}= 14 - 4.50991 = 9.49.

The final pH is slightly smaller than the initial pH. That's expected due to the hydrochloric acid. However, the change is small due to the nature of buffer solutions: adding a small amount of acid or base won't significantly impact the pH of the solution.

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