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
Answer:
Density stays the same
Explanation:
The density remains the same because cutting the object in half will divide the mass & volume by the same amount. Also, the density of a substance remains the same no matter what size it is.
Less than that of nitrogen and oxygen remember cnof "see noff" . this is the order of electronegativity carbon is less than nitrogen which is less than oxygen which is less than fluorine.
Hello!
The precipitate of the reaction that occurs <span>when a silver nitrate solution is mixed with a sodium chloride solution is silver chloride (AgCl).
The chemical equation for this precipitation reaction is the following:
AgNO</span>₃(aq) + NaCl(aq) → AgCl↓(s) + NaNO₃(aq)
Silver Chloride is a chemical compound that is poorly soluble in water because the chemical bond between Ag⁺ and Cl⁻ is very strong as the two ions are nearly the same size and have similar electronic densities. If the chemical bond is strong, it is more difficult for water molecules to break this bond to surround the individual ions, and thus solubility will decrease.
Have a nice day!
Answer:
7.43 × 10²⁴ m⁻³
Explanation:
Data provided in the question:
Conductivity of a semiconductor specimen, σ = 2.8 × 10⁴ (Ω-m)⁻¹
Electron concentration, n = 2.9 × 10²² m⁻³
Electron mobility, 
= 0.14 m²/V-s
Hole mobility, 
= 0.023 m²/V-s
Now,
σ = 
or
σ = 
here,
q is the charge on electron = 1.6 × 10⁻¹⁹ C
p is the hole density
thus,
2.8 × 10⁴ = 1.6 × 10⁻¹⁹( 2.9 × 10²² × 0.14 + p × 0.023 )
or
1.75 × 10²³ = 0.406 × 10²² + 0.023p
or
17.094 × 10²² = 0.023p
or
p = 743.217 × 10²²
or
p = 7.43 × 10²⁴ m⁻³
