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>
Answer:
C) the energy transferred between objects t different temperatures
Answer:
A common characteristic of most Alkali Metals is their ability to displace H2(g) from water. This is represented by their large, negative electrode potentials. In this event, the Group 1 metal is oxidized to its metal ion and water is reduced to form hydrogen gas and hydroxide ions. The general reaction of an alkali metal (M) with H2O (l) is given in the following equation:2M(s)+2H2O(l)⟶2M+(aq)+2OH−(aq)+H2(g)(1)
Explanation:
Answer:
whats the question so i can answer?
Explanation:
The given molecule is
H3C - C ≡ C - CH3
The numbering order is shown below:
H3C - C ≡ C - CH3
1 2 3 4
So, the alkyne group is in the second position.
The carbon chain has four carbons.
Hence, the IUPAC name of the given compound is:
2-butyne.
