Molarity = number of moles of solute/liters of solution
number of moles of solute = molarity x liters of solution
Part (a): <span>30.00 ml of 0.100m Cacl2
number of moles of CaCl2 = 0.1 x 0.03 = 3x10^-3 moles
1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine
Part (b): </span><span>10.0 ml of 0.500m bacl2
number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles
1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine
Part (c): </span><span>4.00 ml of 1.000m nacl
number of moles of NaCl = 1 x 0.004 = 0.004 moles
1 mole of NaCl contains 1 mole of chlorine, therefore 4x10^-3 moles of NaCl contains 4x10^-3 moles of chlorine
Part (d): </span><span>7.50 ml of 0.500m fecl3
number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles
1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine
Based on the above calculations, the correct answer is (d)</span>
Answer:
Molarity of NaOH = 1.8 M.
Explanation:
From the question given above, the following data were obtained:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Volume = 500 mL
Molarity of NaOH =?
Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Mole of NaOH =?
Mole = mass / molar mass
Mole of NaOH = 36 / 40
Mole of NaOH = 0.9 mole
Next, we shall convert 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Finally, we shall determine the molarity of NaOH. This can be obtained as follow:
Mole of NaOH = 0.9 mole
Volume = 0.5 L
Molarity of NaOH =?
Molarity = mole / Volume
Molarity of NaOH = 0.9 / 0.5
Molarity of NaOH = 1.8 M
Answer:
ΔU=-369.2 kJ/mol.
Explanation:
We start from the equation:
Δ(H)=ΔU+Δ(PV), which is an extension of the well known relation: H=U+PV.
If Δ(PV) were calculated by ideal gas law,
PV=nRT
Δ(PV)=RTΔn.
Where Δn is the change of moles due to the reaction; but, this reaction does not give a moles change (Four moles of HCl produced from 4 moles of reactants), so Δ(PV)=0.
So, for this case, ΔH=ΔU.
The enthalpy of reaction given is for one mole of reactant, so the enthalpy of reaction for the reaction of interest must be multiplied by two:
ΔU=-369.2 kJ/mol.
Answer:
Empirical formula is Li₂CO₃.
Explanation:
Percentage of oxygen= 65.0%
Percentage of lithium = 18.7%
Percentage of carbon= 16.3%
Empirical formula = ?
Solution:
Number of gram atoms of C = 16.3/12 = 1.4
Number of gram atoms of Li = 18.7/6.94 = 2.7
Number of gram atoms of O = 65.0/ 16 = 4.1
Atomic ratio:
Li : C : O
2.7/1.4 : 1.4/1.4 : 4.1/1.4
2 : 1 : 3
Li : C : O = 2 : 1 : 3
Empirical formula is Li₂CO₃.
Answer:
20% of phosphorus
Explanation:
A fertilizer is used to improve the fertility of soils. Most fertilizers contains the element nitrogen, phosphorus and potassium.
They are often designated NPK fertilizers.
Now we know that the numbers 10-20-20 depicts the nitrogen-phosphorus and potassium content of the fertilizer.
From the designation,
The actual percentage is 20% of phosphorus.
10% of nitrogen
20% of potassium
