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:D molecular attraction increases as temp decreases
Explanation:A,B would require data not available. C is missing. The question is a mess.
Answer: 400K
Explanation:
Given that,
Original volume of balloon V1 = 3.0L
Original temperature of balloonT1 = 27°C
Convert the temperature in Celsius to Kelvin
(27°C + 273 = 300K)
New volume of balloon V2 = 4.0L
New temperature of balloon T2 = ?
Since volume and temperature are given while pressure is constant, apply the formula for Charle's law
V1/T1 = V2/T2
3.0L/300K = 4.0L/T2
To get the value of T2, cross multiply
3.0L x T2 = 4.0L x 300K
3.0LT2 = 1200LK
Divide both sides by 3.0L
3.0LT2/3.0L = 1200LK/3.0L
T2 = 400K
Thus, at a temperature of 400 Kelvin, the balloon would have a volume of 4.0L.
Answer: The nuclide symbol of X is
Explanation:
The given nuclear reaction is a type of alpha decay process. In this process, the nucleus decays by releasing an alpha particle. The mass number of the nucleus is reduced by 4 units and atomic number is also decreased by 2 units. The particle released is a helium nucleus.
The general equation representing alpha decay process is:
For the given equation :
As the atomic number and mass number must be equal on both sides of the nuclear equation:
Thus the nuclide symbol of X is