Answer:
16.2 J
Explanation:
Step 1: Given data
- Specific heat of liquid bromine (c): 0.226 J/g.K
- Volume of bromine (V): 10.0 mL
- Initial temperature: 25.00 °C
- Final temperature: 27.30 °C
- Density of bromine (ρ): 3.12 g/mL
Step 2: Calculate the mass of bromine
The density is equal to the mass divided by the volume.
ρ = m/V
m = ρ × V
m = 3.12 g/mL × 10.0 mL
m = 31.2 g
Step 3: Calculate the change in the temperature (ΔT)
ΔT = 27.30 °C - 25.00 °C = 2.30 °C
The change in the temperature on the Celsius scale is equal to the change in the temperature on the Kelvin scale. Then, 2.30 °C = 2.30 K.
Step 4: Calculate the heat required (Q) to raise the temperature of the liquid bromine
We will use the following expression.
Q = c × m × ΔT
Q = 0.226 J/g.K × 31.2 g × 2.30 K
Q = 16.2 J
<h3><u>Answer;</u></h3>
= 9.45 × 10^23 molecules
<h3><u>Explanation; </u></h3>
The molar mass of Na2SO4 = 142.04 g/mol
Number of moles = mass/molar mass
= 223/142.04
= 1.57 moles
But;
1 mole = 6.02 × 10^23 molecules
Therefore;
1.57 moles = ?
= 1.57 × 6.02 × 10^23 molecules
<u>= 9.45 × 10^23 molecules </u>
Answer: Metals are shiny and lustrous with a high density. They have very high melting and boiling points because metallic bonding is very strong, so the atoms are reluctant to break apart into a liquid or gas.
Explanation:
sorry for not answering in less than 1 minute
A gas can be treated as ideal gas when it is at higher temperatures or low pressures relative to its critical temperature and pressure. Because, at high temperatures and low pressures, the density of gas decreases and at low densities real gases behave as ideal gases.
