<span>6.
Because the general formula for alkenes is CnH2n</span>
The equation of state for a hypothetical ideal gas is known as the ideal gas law, sometimes known as the general gas equation. i.e. PV = nRT or P1V1 = P2V2.
- According to the ideal gas law, the sum of the absolute temperature of the gas and the universal gas constant is equal to the product of the pressure and volume of one gram of an ideal gas.
- Robert Boyle, Gay-Lussac, and Amedeo Avogadro's observational work served as the basis for the ideal gas law. The Ideal gas equation, which simultaneously describes every relationship, is obtained by combining all of their observations into a single statement.
- When applying the gas constant R = 0.082 L.atm/K.mol, pressure, volume, and temperature should all be expressed in units of atmospheres (atm), litres (L), and kelvin (K).
- At high pressure and low temperature, the ideal gas law basically fails because molecule size and intermolecular forces are no longer negligible but rather become significant considerations.
Learn more about ideal gas law here:
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Answer:
to do day-to day activities easily and make surrounding comfortable for man
Explanation:
as an example , binary numbers help to work computers ,calculators.they help us lot .
scientific methods also need to write large and small numbers in short ways .so, scientists find various patterns .
Answer:
Boiling point of the solution is 100.78°C
Explanation:
This is about colligative properties.
First of all, we need to calculate molality from the freezing point depression.
ΔT = Kf . m . i
As the solute is nonelectrolyte, i = 1
0°C - (-2.79°C) = 1.86 °C/m . m . 1
2.79°C / 1.86 m/°C = 1.5 m
Now, we go to the boiling point elevation
ΔT = Kb . m . i
Final T° - 100°C = 0.52 °C/m . 1.5m . 1
Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C
Answer:
0.00370 g
Explanation:
From the given information:
To determine the amount of acid remaining using the formula:
where;
v_1 = volume of organic solvent = 20-mL
n = numbers of extractions = 4
v_2 = actual volume of water = 100-mL
k_d = distribution coefficient = 10
∴




Thus, the final amount of acid left in the water = 0.012345 * 0.30
= 0.00370 g