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:
V₂ = 0.62 L
Explanation:
Given data:
Initial volume = 2.4 L
Initial temperature = 25°C
Final temperature = -196°C
Final volume = ?
Solution:
Initial temperature = 25°C (25+273 = 298 K)
Final temperature = -196°C ( -196+273 = 77 K)
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 2.4 L × 77 K / 298 k
V₂ = 184.8 L.K / 298 K
V₂ = 0.62 L
The heat required to raise the temperature to a specific temperature change of a sample is related to the specific heat capacity of the substance. In this case, the heat can be calculated through mCpΔT = 350 g * 0.39 J/g C *25 C. This is equal to 3412. 5 Joules. Closest answer is C.
Answer:
Mass Number
Explanation:
In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.
Please vote brainliest!
Answer:
Explanation:
mole of O₂ = 
= .25 moles
mole of CO₂
= 
= .1818 moles
moles of SO₂
= .125 moles
Total moles of gas
= .5568 moles.
total volume of gas mixture
= 22.4 x .5568 liter ( volume of one mole of any gas = 22.4 liter)
= 12.47 liter.
gas will exert partial pressure according to their mole fraction
gas having greatest no of moles in the total mole will have greatest mole fraction so
O₂ will have greatest partial pressure.
