Answer:
0.643 mol.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 4.0 atm).
V is the volume of the gas in L (V = 4000 mL = 4.0 L).
n is the no. of moles of the gas in mol (n = ??? mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 30ºC + 273 = 303 K).
<em>∴ n = PV/RT </em>= (4.0 atm)(4.0 L)/(0.0821 L.atm/mol.K)(303 K) = <em>0.643 mol.</em>
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
Answer:
The concentration of an anthracene solution is ![c = 3.560 *10^{-5}M](https://tex.z-dn.net/?f=c%20%3D%203.560%20%2A10%5E%7B-5%7DM)
Explanation:
From the question we are told that
The incident beam
is ![=1532](https://tex.z-dn.net/?f=%3D1532)
The fluorescence intensity is ![I = 775](https://tex.z-dn.net/?f=I%20%3D%20775)
The length of the medium is b = 0.875 cm
The molar extinction coefficient is ![\epsilon = 9.5 *10^3 M^{-1} cm^{-1}](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%209.5%20%2A10%5E3%20M%5E%7B-1%7D%20cm%5E%7B-1%7D)
The proportionality constant k = 0.30
According to Lambert law the Absorbance of the anthracene solution is mathematically represented as
![A = log (I_O/I)](https://tex.z-dn.net/?f=A%20%3D%20log%20%28I_O%2FI%29)
Where ![I_o =P_o](https://tex.z-dn.net/?f=I_o%20%3DP_o)
and A is the Absorbance
Substituting value
![A = log( (1532)/(775))](https://tex.z-dn.net/?f=A%20%3D%20log%28%20%281532%29%2F%28775%29%29)
![=0.2960](https://tex.z-dn.net/?f=%3D0.2960)
Generally beers law can be represented mathematically as
![A = \epsilon c l](https://tex.z-dn.net/?f=A%20%3D%20%5Cepsilon%20c%20l)
where c is the concentration of an anthracene solution
Making c the subject of the formula
![c = \frac{A}{cl}](https://tex.z-dn.net/?f=c%20%3D%20%5Cfrac%7BA%7D%7Bcl%7D)
Substituting 0.875 cm for length = b ,
We have
![c = \frac{0.2960}{9.5*10^{3} * 0.875}](https://tex.z-dn.net/?f=c%20%3D%20%5Cfrac%7B0.2960%7D%7B9.5%2A10%5E%7B3%7D%20%2A%200.875%7D)
![c = 3.560 *10^{-5}M](https://tex.z-dn.net/?f=c%20%3D%203.560%20%2A10%5E%7B-5%7DM)
B. steong acid strong base
In the balanced equation, we see that the ratio of
is 1:2 respectively. That being said, we can then set up a proportion to solve for the number of moles of water produced from 13.35 moles of oxygen:
![\frac{1mol_{O_{2}}}{2mol_{H_{2}O}} =\frac{13.35mol_{O_{2}}}{xmol_{H_{2}O}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1mol_%7BO_%7B2%7D%7D%7D%7B2mol_%7BH_%7B2%7DO%7D%7D%20%3D%5Cfrac%7B13.35mol_%7BO_%7B2%7D%7D%7D%7Bxmol_%7BH_%7B2%7DO%7D%7D%20%20)
Then we cross multiply and solve for x:
![x=26.7mol_{H_{2}O}](https://tex.z-dn.net/?f=%20x%3D26.7mol_%7BH_%7B2%7DO%7D%20)
Therefore the answer is B) 26.70 mol.