In the context of chemistry, yes. Energy input is always equal to the energy output.
A mass of <span>58.44 g hope this helps and plz rate below and thank me on my profile plz</span>
Answer:
7.46 g.
Explanation:
- Firstly, we need to calculate the amount of heat needed to warm 5.64 kg of water from 21.0°C to 70.0°C using the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by water (Q = ??? J).
m is the mass of water <u><em>(m: we will determine).</em></u>
c is the specific heat capacity of water (c = 4.186 J/g.°C).
ΔT is the temperature difference (final T - initial T) (ΔT = 70.0 °C - 21.0 °C = 49.0 °C).
- To determine the mass of 1.76 L of water we can use the relation:
mass = density x volume.
density of water = 1000 g/L & V = 1.76 L.
∴ mass = density x volume = (1000 g/L)(1.76 L) = 1760.0 g.
∵ Q = m.c.ΔT
<em>∴ Q = m.c.ΔT </em>= (1760.0 g)(4.186 J/g.°C)(49.0 °C) = 360483.2 J ≅ 360.4832 kJ.
- As mentioned in the problem the molar heat of combustion of hexane is - 4163.0 kJ/mol.
<em>Using cross multiplication we can get the no. of moles of hexane that are needed to be burned to release 360.4832 kJ:</em>
Combustion of 1.0 mole of methane releases → - 4163.0 kJ.
Combustion of ??? mole of methane releases → - 360.4832 kJ.
∴ The no. of moles of hexane that are needed to be burned to release 360.4832 kJ = (- 360.4832 kJ)(1.0 mol)/(- 4163.0 kJ) = 0.0866 mol.
- Now, we can get the mass of hexane that must be burned to warm 1.76 L of water from 21.0°C to 70.0°C:
<em>∴ mass = (no. of moles needed)(molar mass of hexane)</em> = (0.0866 mol)(86.18 g/mol) = <em>7.46 g.</em>
To answer this, we use Raoult's Law where the partial pressure is equal to the product of the fraction of the gas in the mixture and its total pressure. Also, we use Dalton's Law of Partial Pressure where the total pressure is equal to the sum of the partial pressure of the gases in the mixture.
Ptotal = 73.44 + 128.52 + 2.04 = 204 atm
201.96 = x (204)
x = 0.99
Explanation:
Independent variable and dependent variable
