Answer:
Following are the responses to the given choices:
Explanation:
- Air woods is a smoking process as it releases heat.
- Incasereaction produces reaction energy, the response is then exothermic, while absorbs react energy, therefore the response is exothermic.
- Heat is essential for melting ice. Correspondingly, ice melts into liquid water as well as other reactions stop.
- Feel cold due to fuel absorption. That ice pack thus feels cold and brings another micro reaction to the stop.
- Unless the reaction's heat is positive, that process is endothermal.
- The reaction is exothermic unless the heat from the reaction is bad.
- Feel hot due to its loosening energy. A test tubefeelhot is, thus, an exothermic reaction.
<span>1) The electron configuration of Re is [Xe] 4f14 5d5 6s2
</span>
<span>The ion Re (3+) has lost 3 electrons, so its configuration is </span><span>[Xe] 4f14 5d4
</span>
Then, the number outer-shell d electrons is 4.
<span>2) Sc: [Ar] 3d1 4s2
</span>
Sc (3+) has lost 3 electrons -> <span>[Ar] which has not d electrons.
</span>
<span>Then, the number of d electrons is 0</span>
<span>3) Ru: [Kr] 4d7 5s1
</span>
Ru (4+) has lost 4 electrons -> <span>[Kr] 4d4
</span>
Then, the number of outer-shell d electrons is 4
<span>4) Hg: [Xe] 4f14 5d10 6s2
</span>
Hg (2+) has lost 2 electrons -> <span>[Xe] 4f14 5d10
</span>
<span>Then, the number of outer-shell d electrons is 10</span>
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
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:
Scientific Notation: 3.45 x 10^5
E Notation: 3.45e5
Explanation: