Answer:
32.4 mol
Explanation:
Given data:
Number of moles of C atom present = ?
Number of moles of glucose = 5.4 mol
Solution:
Glucose formula = C₆H₁₂O₆
There are 6 moles of C atoms are present in one mole of glucose.
In 5.4 moles of glucose:
5.4 mol × 6 = 32.4 mol
Answer:
1.602 L (or) 1602 mL
Explanation:
Molarity is the amount of solute dissolved per unit volume of solution. It is expressed as,
Molarity = Moles / Volume of Solution ----- (1)
Rearranging above equation for volume,
Volume of solution = Moles / Molarity -------(2)
Data Given;
Molarity = 0.00813 mol.L⁻¹
Mass = 1.55 g
First calculate Moles for given mass as,
Moles = Mass / M.mass
Moles = 1.55 g / 119.002 g.mol⁻¹
Moles = 0.0130 mol
Now, putting value of Moles and Molarity in eq. 2,
Volume of solution = 0.0130 mol / 0.00813 mol.L⁻¹
Volume of solution = 1.60 L
or,
Volume of solution = 1602 mL
Explanation:
The given data is as follows.
= 100 mm Hg or 
= 0.13157 atm
= 
= (1080 + 273) K = 1357 K
= 
= (1220 + 273) K = 1493 K
= 600 mm Hg or 
= 0.7895 atm
R = 8.314 J/K mol
According to Clasius-Clapeyron equation,
![log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]](https://tex.z-dn.net/?f=log%28%5Cfrac%7B0.7895%7D%7B0.13157%7D%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B2.303%20%5Ctimes%208.314%20J%2Fmol%20K%7D%5B%5Cfrac%7B1%7D%7B1357%20K%7D%20-%20%5Cfrac%7B1%7D%7B1493%20K%7D%5D)
![log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]](https://tex.z-dn.net/?f=log%20%286%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B19.147%7D%5B%5Cfrac%7B%281493%20-%201357%29%20K%7D%7B1493%20K%20%5Ctimes%201357%20K%7D%5D)
0.77815 = 
= 
J/mol
= 
= 221.9 kJ/mol
Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.
Answer:
I don't speck or understand spanish
Explanation:
I can understand Si that means yes but that all
I cant entirely tell for now but an article on rodioactivity should solve the problem
