<span>The concentration would remain at 2.0m. The problem states that 0 ml from the 2.0 m solution is diluted, therefore implying that none of it was diluted. Therefore, the level of concentration of the new solution would be the same as before.</span>
To answer this question, we will use the following equation:
<span>ln(P2/P1) = (∆Hvap/R)*((1/T1) - (1/T2))
</span>
Now we examine the givens of the problem and transform to standard units if required:
<span>∆Hvap = 30.5 kJ/mol
</span>R is a constant = <span>8.314 x 10^-3 kJ K^-1 mol^-1
T1 </span><span>= 91 celcius = 91 + 273= 364 Kelvin
</span>T2 = 20 celcius = 20 + 273 = 293 k3lvin
P1 is the standard atmospheric pressure = 760 mmHg
P2 is the value to be calculated
Substitute with these values in the equation:
ln(P2/760) = (30.5 / 8.314 x 10^-3) x ((1 / 364) - (1 / 293))
ln(P2/760) = - 2.4662 (Take the exponential both sides to eliminate the ln)
P2 / 760 = e^(-2.4462) = 0.0866
P2 = 0.0866 x 760 = 65.816 mmHg
Answer:
I think carbon and hydrogen
Answer:
1837.89 Lt
Explanation:
The chemical reaction for this situation is:
NaHCO₃ + HCl → NaCL + H₂O + CO₂ ₍g₎
Where the mola mass we need are:
M NaHCO₃ = 84 g/mol
M CO₂ = 44 g/mol
As we have 6.00 Kg of sodium bicarbonate, then:
6 Kg NaHCO₃ = 71.43 moles of NaHCO₃
Due the stoichiometry of this chemaicl reaction:
1 mol NaHCO₃ = 1 mol CO₂
71.43 moles NaHCO₃ = 71.43 moles CO₂
And considering that CO₂ is an ideal gas, we can use the following formula:
PV=nRT
V = (nRT)/P
n = 71.43 mol
R = 0.083 Ltxatm(molxK)
T = 37°C = 310 K
P = 1 atm
So: V = (71.43x0.083x310)/1
V CO₂ = 1837.89 Lt
D. Leaves because leaves in a plant absorb the most sunlight
