The first answer is B and the second answer is B
we can do soo by noticing any colour change, or by the formation of an insoluble precipitate or by the evolution of a gas.
Answer:
The molecular weight for the compound is 60.1 g/mol
Explanation:
We need to determine the molality of solute to find out the molar mass of it.
We apply the colligative property of freezing point depression:
ΔT = Kf . m . i
If the compound was also found to be nonvolatile and a non-electrolyte,
i = 1.
Freezing T° of pure solvent - Freezing T° of solution = Kf . m
0°C - (-2.05°C) = 1.86°C/m . m
2.05°C / 1.86m/°C = m → 1.10 mol/kg
To determine the moles of solute we used, we can multiply molality by the mass of solvent in kg → 202.1 g . 1kg/1000g = 0.2021 kg
1.10 mol/kg . 0.2021kg = 0.223 moles
Molar mass→ g/mol → 13.39 g / 0.223 mol = 60.1 g/mol
Answer:
the carbon monoxide will release blue fumes
Answer:
The nucleus consists of 9 protons and 10 neutrons. Nine electrons occupy available electron shells.
