There is a key piece of information that we are missing.
we need the following:
Kb of water= 0.512
the change in boiling point (ΔTb) can be calculated using the following formula:
ΔTb= Kb x m
we already have Kb, but we need to determine the molality (m).
1) let's convert the grams of glucose to moles using the molar mass of it. The molecule formula of glucose is C₆H₁₂O₆.
molar mass C₆H₁₂O₆= (6 x 12.0) + (12 x 1.01) + (6 x 16.0)= 180 g/mol
2) let's determine the Kilograms of water.
info:
density of water= 1.0 g/ mL or 1 grams = 1 mL
1000 grams= 1 kilogram
3) let's plug in the values to solve for molality
finally, we can solve for change in boiling point.
ΔTb= Kb x m
ΔTb= (0.512) (0.545m)=
0.279°C
Answer:
The answer to your question is Cm = 2.093 J/kg°C
Explanation:
Data
mass metal = mass of water
Δ water = 5°
Δ metal = 10°
Cm = ?
Cw = 4.186 J/Kg°C
Formula
for metal for water
mCmΔTm = mCwΔTw
Cancel mass of metal
Cm ΔTm = CwΔTw
solve for Cm
Cm = (CwΔTw) / ΔTm
Substitution
Cm = (4.186 x 5) / 10
Simplification
Cm = 20.93 / 10
Result
Cm = 2.093 J/kg°C
Answer:
The reagents -
Explanation:
is an electron donating group which directs the electrophile to meta position reduction of nitro group with produces amine.
Acylation of amine with produces required product.
The complete reaction is as follows.
Answer:
pKa = 3.72
Explanation:
Let's consider the dissociation of a generic monoprotic weak acid.
HA(aq) ⇄ H⁺(aq) + A⁻(aq)
For a weak acid, we can find the value of the acid dissociation constant (Ka) using the following expression:
where,
[H⁺] is the molar concentration of H⁺
Ca is the initial concentration of the acid
First, we need to find [H⁺] from pH.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -2.40 = 3.98 × 10⁻³ M
Then,
Finally,
pKa = -log Ka = -log 1.89 × 10⁻⁴ = 3.72
Answer:
3 × 10⁴ kJ
Explanation:
Step 1: Write the balanced thermochemical equation
C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(g) ΔH = -2220 kJ
Step 2: Calculate the moles corresponding to 865.9 g of H₂O
The molar mass of H₂O is 18.02 g/mol.
865.9 g × 1 mol/18.02 g = 48.05 mol
Step 3: Calculate the heat produced when 48.05 moles of H₂O are produced
According to the thermochemical equation, 2220 kJ of heat are evolved when 4 moles of H₂O are produced.
48.05 mol × 2220 kJ/4 mol = 2.667 × 10⁴ kJ ≈ 3 × 10⁴ kJ