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Ulleksa [173]
3 years ago
13

1. In this experiment, what property of NaCl is used to separate it from the other two components? Is this a chemical or physica

l property? NaCl is the only compound in this mixture that is soluble in water, so its solubility in water is the property used to separate it. This is a physical property, because you are doing nothing to change the actual chemical composition. It's just solvated by water. 2. In this experiment, what property of CaCO3 is used to separate it from SiO2? Is this a chemical or physical property? The property of CaCO3 used to seperate it from SiO2 is its reactivity with acetic acid. By adding the acetic acid, you develop carbon dioxide (which bubbles out of solution), and CaCl2, which is soluble in water. You are undergoing a chemical reaction, so this is a chemical property. 3. An unknown sample of a mixture of salt and sand weighs 7.52 g before washing with water and 3.45 g after. What is the percent of sand in the sample? (Show all work.) 7.52g salt and sand, 3.45g sand, 4.07g salt (the difference in before and after water shows how much salt you had) 3.45 g sand / 7.52g total * 100 = 45.9% sand To check that this is correct, find the percentage of salt as well. 4.07g salt / 7.52g * 100 = 54.1% salt 45.9% + 54.1% = 100%. Everything is accounted for, so this must be the composition of the mixture.
Chemistry
1 answer:
irinina [24]3 years ago
4 0

Answer 1) In the mixture of sand, NaCl and CaCO_{3} first separation was done by dissolving the mixture in water. In the first step the NaCl will get dissolved whereas, CaCO_{3} will be undissolved and sand will settle at the bottom. Here, the physical property of solubility of NaCl in water is taken into consideration. The collected water is then filtered off and evaporated to get the NaCl back from the mixture with some loss. No chemical change occurs in case of NaCl extraction from water.


Answer 2) When CaCO_{3} was to be removed from the undissolved part. The chemical property of CaCO_{3} was used. Where it gets dissolved in acidic medium. And this way we can extract CaCO_{3} and remove sand from it. We change the chemical composition of CaCO_{3} by adding HCl to the mixture and dissolve CaCO_{3} to form CaCl_{2} which gets dissolved into the solution of HCl. Here, first decantation occurs and then extraction is done. Second, extraction is done using potassium carbonate in it which separates CaCl_{2} from sand.


Answer 3) Here, Unknown mixture of salt and sand weighed =7.52 g (before washing);


Unknown mixture of salt and sand weighed =3.45 g (after washing);


To calculate the amount of salt in it, we can simply subtract the values of before and after washing change in weights.


The property of NaCl being soluble in water it will go away with washing leaving behind the sand only, after washing.


So, Weight of salt was = 7.52g - 3.45 g = 4.07g


To find the percentage of sand that was mixed with salt =

(3.45 g sand / 7.52g of mixture) X 100 = 45.9% sand was mixed with salt.


To verify whether the correct percentage of salt and sand was calculated we can recheck the value for salt as well.

(4.07 g of salt / 7.52g of mixture) X 100 = 54.1% salt


On adding we get, 45.9% + 54.1% = 100%.


Which confirms that the calculations are correct.

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Calculate the mass of NaCO3 used in experiment. SHOW WORK — 15 points!!
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The mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g

<h3>Calculating mass </h3>

From the question we are to calculate the mass of NaHCO₃ (sodium bicarbonate) used in the experiment

From the given information

Mass of empty evaporating dish = 46.233g

Mass of evaporating dish + Sodium bicarbonate = 48.230g

∴ Mass of sodium bicarbonate (NaHCO₃) = [Mass of evaporating dish + Sodium bicarbonate] - [Mass of empty evaporating dish]

Mass of sodium bicarbonate (NaHCO₃) = 48.230g - 46.233g

Mass of sodium bicarbonate (NaHCO₃) = 1.997 g

Hence, the mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g

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When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit
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25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

How many moles of this process?

Relative atomic mass from a modern periodic table:

  • K: 39.098;
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Molar mass of \text{KNO}_3:

M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of the process = Number of moles of \text{KNO}_3 dissolved:

\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}.

What's the enthalpy change of this process?

Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ} for 0.212162\;\text{mol}. By convention, the enthalpy change \Delta H measures the energy change for each mole of a process.

\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}.

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

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