We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to 
= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is 
= 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³
Answer:
Explanation:
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In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:
Thus, we insert mass, specific heat and temperatures to obtain:
In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:
Now, we plug in to obtain:
NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.
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Answer: D. Potassium bromide
Answer:
B. 0.2.
Explanation:
<em>n = mass/molar mass</em>
mass of CaCO₃ = 20 g, molar mass of CaCO₃ = 100.0869 g/mol.
<em>∴ n = mass/molar mass = </em>(20 g)/(100.0869 g/mol) <em>= 0.1998 ≅ 0.2 mol.</em>
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<em>So, the right choice is: B. 0.2.</em>
Answer:
In 1851 gold-seekers from around the world began pouring into the colonies, changing the course of Australian history. The gold rushes greatly expanded Australia's population, boosted its economy, and led to the emergence of a new national identity.
Explanation:
