[Co(NH₃)₅Br]²⁺
Ligands and charges on them,
5 × NH₃ = 5 × 0 = 0
1 × Br⁻¹ = 1 × -1 = -1
Charge on sphere = +2
So, putting values in equation,
Co + (0)₅ - 1 = +2
Co + 0 - 1 = +2
Co - 1 = +2
Co = +2 + 1
Co = +3
Result:
Oxidation state of Co in [Co(NH₃)₅Br]²⁺ is +3.
Answer:
378mL
Explanation:
The following data were obtained from the question:
Pressure (P) = 99.19 kPa
Temperature (T) = 28°C
Number of mole (n) = 0.015 mole
Volume (V) =...?
Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:
For Pressure:
101.325 KPa = 1 atm
Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm
For Temperature:
T(K) = T(°C) + 273
T(°C) = 28°C
T(K) = 28°C + 273 = 301K.
Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:
PV = nRT
Pressure (P) = 0.98 atm
Temperature (T) = 301K
Number of mole (n) = 0.015 mole
Gas constant (R) = 0.0821atm.L/Kmol.
Volume (V) =...?
0.98 x V = 0.015 x 0.0821 x 301
Divide both side by 0.98
V = (0.015 x 0.0821 x 301) /0.98
V = 0.378 L
Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:
1L = 1000mL
Therefore, 0.378L = 0.378 x 1000 = 378mL
Therefore, the volume of N2 collected is 378mL
Answer:
ΔG = -6.5kJ/mol at 500K
Explanation:
We can find ΔG of a reaction using ΔH, ΔS and absolute temperature with the equation:
ΔG = ΔH - TΔS
Computing the values in the problem:
ΔG = ?
ΔH = 2kJ/mol
T = 500K
And ΔS = 0.017kJ/(K•mol)
Replacing:
ΔG = 2kJ/mol - 500K*0.017kJ/(K•mol)
ΔG = 2kJ/mol - 8.5kJ/mol
<h3>ΔG = -6.5kJ/mol at 500K</h3>
- Separated by an actual physical barrier
<span>- Geographic isolation can be overcome in some circumstances but the majority of the population will be isolated and therefore will diverge into different species
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