Answer:
The ΔH of the reaction is + 12.45 KJ/mol
Explanation:
Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)
heat capacity of water = 4.18 Jk-1 Mol-1
Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)
Molar mass of NaHCO3 = 84 g/mol
Mole of NaHCO3 = 14.5 / 84 = 0.173 mol
Step 1 : Calculate the heat energy (Q) lost by the water.
Q = M x C x ΔT
Q = -100 x 4.18 x (-5.14)
Q = 2148.5 joules
Q = 2.1485 K J
Step 2: Calculating the ΔH of the reaction?
ΔH = Q / number of moles of NaHCO3
ΔH = 2.1485 / 0.173
ΔH = 12.42 KJ/mol
The concept of resonance is required for certain molecules because the localized electron model assumes electrons are located between a given pair of atoms in a molecule.
It produces nitrogen gas and water
4NH3+3O2+heat 2N2+6H 2O
Tin metal reacts with hydrogen fluoride to produce tin(II) fluoride and hydrogen gas according to the following balanced equation.
Sn(s)+2HF(g)→SnF2(s)+H2(g)
Sn(s)+2HF(g)→
SnF
2
(s)+
H
2
(g)
How many moles of hydrogen fluoride are required to react completely with 75.0 g of tin?
Step 1: List the known quantities and plan the problem.
Known
given: 75.0 g Sn
molar mass of Sn = 118.69 g/mol
1 mol Sn = 2 mol HF (mole ratio)
Unknown
mol HF
Use the molar mass of Sn to convert the grams of Sn to moles. Then use the mole ratio to convert from mol Sn to mol HF. This will be done in a single two-step calculation.
g Sn → mol Sn → mol HF
Step 2: Solve.
75.0 g Sn×1 mol Sn118.69 g Sn×2 mol HF1 mol Sn=1.26 mol HF
75.0 g Sn×
1
mol Sn
118.69
g Sn
×
2
mol HF
1
mol Sn
=1.26 mol HF
Step 3: Think about your result.
The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of HF is required for the reaction. The answer has three significant figures because the given mass has three significant figures.
Answer: 0.0508mL
Explanation: Using the basic formula that states: C acid * V acid = C base * V base. we have:0.568 * 17.88 = 20 * C base.
therefore concentration of the base is 1.0156/20 = 0.0508 mL
