Answer: V has the units of liters, and T has the units of kelvin.
Ideal gas law: p·V = n·R·T.
atm · L = mol · L·atm/mol·K · K; both side of equatation have same values.
R = 0,08206 L·atm/mol·K; universal gas constant.
p is pressure of the gas, unit is standard atmosphere (atm).
V is volume of the gas, unit is liters (L).
n is amount of substance of the gas; unit is mole (mol).
T is temperature of the gas, unit is Kelvin (K
Hope this helps!!
Explanation:
Answer:
ok this might not be right but i think its 1,575??
Explanation:
(125)(28)(0.450)
i think you just multiply the grams, change in temp, then the cp by each other not completely sure tho
Answer:
![\boxed{\text{18.2 kJ}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctext%7B18.2%20kJ%7D%7D)
Explanation:
The formula for the heat involved is
![q = m\Delta_{\text{f}}\text{H}](https://tex.z-dn.net/?f=q%20%3D%20m%5CDelta_%7B%5Ctext%7Bf%7D%7D%5Ctext%7BH%7D)
Data:
m = 115 g
![\Delta_{\text{f}}\text{H} = \text{158.3 J/g}](https://tex.z-dn.net/?f=%5CDelta_%7B%5Ctext%7Bf%7D%7D%5Ctext%7BH%7D%20%3D%20%5Ctext%7B158.3%20J%2Fg%7D)
Calculation:
![q = \text{115 g} \times \dfrac{\text{158.3 J}}{\text{1 g}}\\\\q = \text{18 200 J} = \textbf{18.2 kJ}}\\\\\text{It takes }\boxed{\textbf{18.2 kJ}} \text{ to melt the p-xylene}](https://tex.z-dn.net/?f=q%20%3D%20%5Ctext%7B115%20g%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B158.3%20J%7D%7D%7B%5Ctext%7B1%20g%7D%7D%5C%5C%5C%5Cq%20%3D%20%5Ctext%7B18%20200%20J%7D%20%3D%20%5Ctextbf%7B18.2%20kJ%7D%7D%5C%5C%5C%5C%5Ctext%7BIt%20takes%20%7D%5Cboxed%7B%5Ctextbf%7B18.2%20kJ%7D%7D%20%5Ctext%7B%20to%20melt%20the%20p-xylene%7D)
Answer:
x, Z, M, H, A, K, E, I, L, R, O, W