Answer:
Explanation:
Hello!
In this case, as we know the mass of the total sample, we can first compute the mass of oxygen:
Next, we compute the moles of each element:
Now, we divide the moles by 0.184 moles, the fewest ones, to obtain:
Therefore, the empirical formula is:
Regards!
Sulfur dioxide is an example of air pollution.
It is the main component of acid rain. It is released into the air as a result of the burning of fossil fuels.
CH₃-CH=CH₂ propene
-----------------------------------
Answer:
The answer to the question is;
The equilibrium constant,K, at 25 °C for the reaction between Ni²⁺(aq) and Zn(s), which form Ni(s) and Zn²⁺ is 2.04×10¹⁷.
Explanation:
The half reactions are as follows
Ni²⁺ (aq) + 2e⁻ -> Ni (s)
Zn (s) -> Zn²⁺ (aq) + 2e⁻
For the Ni²⁺/Ni system we have the potential given as -0.23V (Reduction)
For the Zn²⁺/Zn sytem, the potential is -0.76. Here however, we should note that the zinc is being oxidized and therefore the potential is positive, that is;
Zn/Zn²⁺ = 0.76
Therefore the voltage for the sum of the reactions on both sides of the process is
-0.23 V + 0.76 V = 0.53 V
We then call upon the Nernst equation to calculate the equilibrium constant as follows
E⁰
= 
Where:
E⁰ = Standard cell potential = 0.53 V
n = Number of moles of electrons = 2 moles of e⁻
K = Equilibrium constant
Therefore we have
0.53 V = 
Therefore log K = 17.905
and K = 
= 2.04×10¹⁷.
Answer:
final volume = 10.5 Liters N₂(g) at 21°C and 823Torr*
Explanation:
*Note=>No specified mass value of N₂(g) is defined in the problem. Therefore for a starting point, the gas sample is assumed to be 1.00 mole N₂(g) at STP conditions 22.4L
Determine volume of N₂(g) at 21°C(=294K) and 823 Torr (= 2.286 Atm).
Start with Volume of N₂(g) at 0°C and 1 Atm pressure => 22.4L and adjust to final volume of N₂(g) based upon 21°C(=294K) and 823 Torr (= 2.286 Atm).
V(final) = 22.4L(294K/273K)(360 Torr/823 Torr) = 22.4L(294/273)(360/823) = 10.55 Liters final volume.
Note: The volume of 1 mole (assumed) of any gas at STP (0°C/1 Atm) is 22.4 Liters. To convert to non-STP conditions, convert temperature and pressure factors (changes) that reflect what happens when the gas is expanded or decreased; but, these adjustments are taken independently for each variable of interest. The following notes explain.
For the increase in temperature from 0°C(=273K) to 21°C(=294K) one must apply a temperature ratio that will increase volume. That is, the change in volume due to the temperature change is 294K/273K. If a 273K/294K ratio were used the volume would have decreased. Not so for heating a sample of gas.
For the increase in pressure one should expect a decrease in volume. Therefore apply a pressure ration that will effectively decrease the volume of the gas. That is, to decrease a 22.4L sample at STP multiply the standard volume by a ratio of pressures that will decrease 22.4L to a smaller volume. That is, V(final by pressure effects) multiply by 360Torr/823Torr to decrease the STP VOLUME (22.4L) to the new non-standard volume. If 823Torr/360Torr were used, the final volume would not be smaller, but larger. Such is the physical effect of an increasing pressure change.
