Yes this is true otherwise it would not be a solution
Answer:
29.42 Litres
Explanation:
The general/ideal gas equation is used to solve this question as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K
According to the information provided in this question;
mass of nitrogen gas (N2) = 25g
Pressure = 0.785 atm
Temperature = 315K
Volume = ?
To calculate the number of moles (n) of N2, we use:
mole = mass/molar mass
Molar mass of N2 = 14(2) = 28g/mol
mole = 25/28
mole = 0.893mol
Using PV = nRT
V = nRT/P
V = (0.893 × 0.0821 × 315) ÷ 0.785
V = 23.09 ÷ 0.785
V = 29.42 Litres
Answer:
![\large \boxed{\text{-92 $^{\circ}$C}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B-92%20%24%5E%7B%5Ccirc%7D%24C%7D%7D)
Explanation:
We can use the Ideal Gas Law and solve for T.
pV = nRT
Data
p = 1.25 atm
V = 25.0 L
n = 2.10 mol
R = 0.082 06 L·atm·K⁻¹mol⁻¹
Calculations
1. Temperature in kelvins
![\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%20%7Brcl%7DpV%20%26%20%3D%20%26%20nRT%5C%5C%5Ctext%7B1.25%20atm%7D%20%5Ctimes%20%5Ctext%7B25.0%20L%7D%20%26%20%3D%20%26%20%5Crm%5Ctext%7B2.10%20mol%7D%20%5Ctimes%200.08206%20%5Ctext%7B%20L%7D%5Ccdot%5Ctext%7Batm%7D%5Ccdot%5Ctext%7BK%7D%5E%7B-1%7D%5Ctext%7Bmol%7D%5E%7B-1%7D%20%5Ctimes%20T%5C%5C31.25%26%3D%260.09847T%5Ctext%7B%20K%7D%5E%7B-1%7D%5C%5CT%26%20%3D%20%26%5Cdfrac%7B31.25%7D%7B%5Ctext%7B0.098%2047%20K%7D%5E%7B-1%7D%7D%5C%5C%5C%5C%26%20%3D%20%26%5Ctext%7B181%20K%7D%5Cend%7Barray%7D)
2. Temperature in degrees Celsius
![\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%20%7Brcl%7DT%20%26%20%3D%20%26%20%28181%20-%20273.15%29%20%5C%2C%20%5E%7B%5Ccirc%7D%5Ctext%7BC%7D%5C%5C%26%20%3D%20%26%20-92%20%5C%2C%20%5E%7B%5Ccirc%7D%5Ctext%7BC%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20temperature%20of%20the%20gas%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B-92%20%5C%2C%20%5E%7B%5Ccirc%7D%7D%5Ctextbf%7BC%7D%7D%24%7D)
Molality= mol/ Kg
if we assume that we have 1 kg of water, we have 3.19 moles of solute.
the formula for mole fraction --> mole fraction= mol of solule/ mol of solution
1) if we have 1 kg of water which is same as 1000 grams of water.
2) we need to convert grams to moles using the molar mass of water
molar mass of H₂O= (2 x 1.01) + 16.0 = 18.02 g/mol
1000 g (1 mol/ 18.02 grams)= 55.5 mol
3) mole of solution= 55.5 moles + 3.19 moles= 58.7 moles of solution
4) mole fraction= 3.19 / 58.7= 0.0543
Answer:
carbon
Explanation:
cabonis not a conductor of heat