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:
pH = 4.34
Explanation:
pH= -1/2(logKa) -1/2(log C)
= -1/2( log 5.98*10^-8) -1/2(log 0.0353)
=-1/2(-7.22)-1/2(-1.45)
=3.61+0.725= 4.34
Hydrogen-1, Carbon-13, Nitrogen-15, Fluorine-19, and Phosphorus-31 are the most useful. Out of these, Hydrogen-1 and Carbon-13 in NMR are the most useful nuclei because the these atoms are the most commonly present in organic molecules.
last one? don't take my word though
Explanation:
the suns heat is related to nuclear fusion
Answer:
The disruption of the bonds or attractions occurs during protein hydrolysis which results in the loss for the primacy structure. The peptide bonds is the bond affected in this scenario.
The disruption of the bonds however only exist in the process of denaturation and this results in a change in the confirmation which could be secondary, tertiary, and quaternary structural related. And example of the bonds affected include salt bridges, disulfide bridges, hydrogen bonds etc.
