The mass of ni(NO3)2 that dissolved in 25.0 ml of 0.100m ni(NO3)2 solution is calculated as follows
fin the number of moles = molarity x volume in liters
=25 x0.100/ 1000= 2.5 x10^-3 moles
mass = mass x molar mass
= 2.5 x10^-3 moles x 182.71 g/mol = 0.457 grams
Answer:
25.7 kJ/mol
Explanation:
There are two heats involved.
heat of solution of NH₄NO₃ + heat from water = 0
q₁ + q₂ = 0
n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ × 1 mol NH₄NO₃/80.0 g NH₄NO₃
∴ n = 0.100 mol NH₄NO₃
q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln
m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g
q₂ = mcΔT = 58.0 g × 4.184 J°C⁻¹ g⁻¹ × ((20.39-21)°C) = -2570.19 J
q₁ + q₂ = 0.100 mol ×ΔHsoln – 2570.19 J = 0
ΔHsoln = +2570.19 J /0.100 mol = +25702 J/mol = +25.7 kJ/mol
Answer:
Solution of isopropanol is 10.25 molal
Explanation:
615 g of isopropanol (C3H7OH) per liter
We gave the information that 615 g of solute (isopropanol) are contained in 1L of water. We need to find out the mass of solvent, so we use density.
Density of water 1g/mL → Density = Mass of water / 1000 mL of water
Notice we converted the L to mL
Mass of water = 1000 g (which is the same to say 1kg)
Molality are the moles of solute in 1kg of solvent, so let's convert the moles of isopropanol → 615 g . 1mol / 60g = 10.25 moles
Molality (mol/kg) = 10.25 moles / 1kg = 10.25 m
Answer:
The answer to your question is below:
Explanation:
Having exactly the same data as the previous experiment I think that having the same data as the previous experiment is extremely important but not the most important, for me is the second most important.
Using the same procedure and variables as the previous experiment For me, this is the most importan thing when a scientist is designing an experiment, because if he or she follow exactly the same procedure and variables, then the results will be very close.
Conducting an experiment similar to the previous experiment This characteristic is important but not the most important.
Using the same laboratory that was used in the previous experiment It is not important the laboratory, if the procedure and variables are the same, your experiment must give the same results in whatever laboratory.
