Given that 1 mole contains 6.02x10^23 molecules, 3.0x10^23 is just around half a mole. Then we check the number of moles for each choice:
A. This is approximately half a mole, since the molar mass of Br2 is 159.8 g/mol.
B. He has a molar mass around 4 g/mol, so this is 1 mole.
C. H2 has a molar mass of 2.02 g/mol, so this is 2 moles.
D. Li has a molar mass of around 6.97 g/mol, so this is around 2 moles.
Therefore the only choice that fits is A. 80 g of Br2.
<h3><u> Answer</u>;</h3>
= 4.0 L
<h3><u>Explanation;</u></h3>
Boyle's law states that the volume of a fixed mass of a gas is inversely proportional to pressure at a constant temperature.
Therefore; <em>Volume α 1/pressure</em>
<em>Mathematically; V α 1/P</em>
<em>V = kP, where k is a constant;</em>
<em>P1V1 = P2V2</em>
<em>V1 = 0.5 l, P1 =203 kPa, P2 = 25.4 kPa</em>
<em>V2 = (0.5 × 203 )/25.4 </em>
<em> = 3.996 </em>
<em> ≈ </em><em><u>4.0 L</u></em>
Answer:
The answer should be 1000 kg / m3
<h3><u>Answer;</u></h3>
<em>-49 °C</em>
<h3><u>Explanation and solution;</u></h3>
- Considering the fact that, the specific heat capacity of aluminum is 0.903 J/g x C, and the heat of vaporization of water at 25 C is 44.0 KJ/mol.
Moles water = 0.48 g / 18.02 g/mol
=0.0266 moles
<em>Heat lost by water</em> = 0.0266 mol x 44.0 kJ/mol
=1.17 kJ => 1170 J
<em>But heat lost =heat gained</em>
<em>Therefore;</em> Heat gained by aluminium = 1170 J
1170 = 55 x 0.903 ( T - 25) = 49.7 T - 1242
1170 + 1242 = 49.7 T
T = 48.5 °C ( 49 °C <em>at two significant figures)</em>
<em>Hence</em>, final temperature = 49 °C
