Answer:
I think Sammy is right because I'm sure the compound contains ionic bonds
Answer:
The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous
Explanation:
<u>Step 1:</u> Data given
ΔH∘=20.1 kJ/mol
ΔS is 45.9 J/K
<u>Step 2:</u> When is the reaction spontaneous
Consider temperature and pressure = constant.
The conditions for spontaneous reactions are:
ΔH <0
ΔS > 0
ΔG <0 The reaction is spontaneous at all temperatures
ΔH <0
ΔS <0
ΔG <0 The reaction is spontaneous at low temperatures ( ΔH - T*ΔS <0)
ΔH >0
ΔS >0
ΔG <0 The reaction is spontaneous at high temperatures ( ΔH - T*ΔS <0)
<u>Step 3:</u> Calculate the temperature
ΔG <0 = ΔH - T*ΔS
T*ΔS > ΔH
T > ΔH/ΔS
In this situation:
T > (20100 J)/(45.9 J/K)
T > 437.9 K
T > 164.75 °C
The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous
Explanation:
The given data is as follows.
Current (I) = 3.50 amp, Mass deposited = 100.0 g
Molar mass of Cr = 52 g
It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.
Therefore, 100 g of Cr will be deposited by "z" grams of electricity.

z = 
= 185576.9 C
As we know that, Q = I × t
Hence, putting the given values into the above equation as follows.
Q = I × t
185576.9 C =
t = 53021.9 sec
Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.
15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.
<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>
Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2
<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO
</em>
Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>