Answer:
Cl⁻, Na⁺, OH⁻
Explanation:
The titration is:
CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)
In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.
Thus, ions present after the equivalence point are:<em> Cl⁻, Na⁺</em> (Don't react, spectator ions), and <em>OH⁻</em>.
Answer: 0.07868 mol H₂O
Explanation:
1) Chemical equation:
Cu₂O +H₂ → 2Cu + H₂O
2) mole ratios:
1 mol Cu₂O : 1 mol H₂ : 2 mol Cu : 1 mol H₂O
3) Convert 10.00 g of Cu to grams, using the atomic mass:
Atomic mass of Cu: 63.546 g/mol
number of moles = mass in grams / atomic mass = 10.00g / 63.546 g/mol
number of moles = 0.1574 mol
4) Use proportions
2mol Cu 0.1574 mol Cu
--------------- = ---------------------
1 mol H₂O x
⇒ x = 0.1574 mol Cu × 1 mol H₂O / 2mol Cu = 0.07868 mol H₂O
That is the answer
Answer:
40.4 kJ
Explanation:
Step 1: Given data
- Heat of sublimation of CO₂ (ΔH°sub): 32.3 kJ/mol
Step 2: Calculate the moles corresponding to 55.0 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
n = 55.0 g × 1 mol/44.01 g = 1.25 mol
Step 3: Calculate the heat (Q) required to sublimate 1.25 moles of CO₂
We will use the following expression.
Q = n × ΔH°sub
Q = 1.25 mol × 32.3 kJ/mol = 40.4 kJ
Answer:
Q1) Distillation relies on evaporation to purify water
Explanation:
contaminated water is heated to form steam. inorganic compounds and large non-volatile organic molecules do not evaporate with the water and are left behind. The steam then cools and condenses to form purified water
Answer:
168°C is the melting point of your impure sample.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample = = ?
Depression in freezing point =
Depression in freezing point is also given by formula:
= The freezing point depression constant
m = molality of the sample = 0.275 mol/kg
i = van't Hoff factor
We have: = 40°C kg/mol
i = 1 ( non electrolyte)
168°C is the melting point of your impure sample.