B. 3.0 mol·L⁻¹ NaCl
Explanation:
Freezing point is a colligative property: it depends only on the number of particles in solution.
The for freezing point depression ΔT_f is
ΔT_f = iK_fb
where
i = the number of moles of particles available from one mole of solute
K_f = the molal freezing point depression constant
b = the molal concentration of the solute
All your solutions are aqueous NaCl. They differ only in their concentrations.
Thus, the most concentrated solution will have the greatest freezing point depression and the lowest freezing point.
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♥ In fact, the similarity of the numbers 85,19,17,4&2 is <span>that they are all divisors of 6460.
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The graph would not have plummeted if various species were planted due to improvement of the potato crops.
<h3>What is variety?</h3>
The term variety has to do with the planting of different species of plants on the same farm. We know that the planting of different species of the same crop helps to improve the resistance to diseases of a plant.
Hence, the graph would not have plummeted if various species were planted due to improvement of the potato crops.
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63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction
The balanced reaction equation is;
2 Na + 3 N2--> 2 NaN3
Number of moles of Na = 50.0 g/23 g/mol = 2.17 moles of Na
Number of moles of Nitrogen = 40.5 g/28 g/mol = 1.45 moles of N2
We have to obtain the limiting reactant, this is the reactant that yields the least number of moles of product.
For Na
2 moles of Na yields 2 moles of NaN3
2.17 moles of Na yields 2.17 moles of NaN3 (reaction is 1:1).
For N2
3 moles of N2 yields 2 moles of NaN3
1.45 moles of N2 yields 1.45 * 2/3 = 0.97 moles of NaN3
So, N2 is the limiting reactant. Mass of product formed depends on the limiting reactant.
Mass of NaN3 = 0.97 moles of NaN3 * 65 g/mol = 63.1 g of NaN3
Therefore, 63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction
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