Depends on the grade of the test. But for the most part if it’s below a 78, it’ll drop, If it’s above it’ll raise. It shouldn’t drop or raise it more than 3 point/percent.
Answer:
0.2
Explanation:
Given parameters:
Mass of helium = 0.628g
Mass of neon = 11.491g
Mass of argon = 7.613g
Unknown:
Mole fraction of neon = ?
Solution:
The mole fraction of an element is the number of moles of that element to the total number of moles in the gas mixture.
We need to calculate the number of moles of each element first;
Number of moles = ![\frac{mass}{molar mass}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D)
Molar mass of Helium = 4g/mol
Molar mass of Neon = 20g/mol
Molar mass of Argon = 40g/mol
Number of moles of He =
= 0.16moles
Number of moles of Ne =
= 0.58moles
Number of moles of Ar =
= 0.19moles
Total number of moles = 0.16moles + 0.58moles + 0.19moles = 0.93moles
Mole fraction Neon =
= 0.2
Answer:
![\boxed {\boxed {\sf 0.078 \ L }}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%200.078%20%5C%20L%20%7D%7D)
Explanation:
We are asked to find the volume of a solution given the moles of solute and molarity.
Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:
![molarity= \frac{moles \ of \ solute}{liters \ of \ solution}](https://tex.z-dn.net/?f=molarity%3D%20%5Cfrac%7Bmoles%20%5C%20of%20%5C%20solute%7D%7Bliters%20%5C%20of%20%5C%20solution%7D)
We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.
- moles of solute = 0.14 mol KCl
- molarity= 1.8 mol KCl/ L
- liters of solution=x
Substitute these values/variables into the formula.
![1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}](https://tex.z-dn.net/?f=1.8%20%5C%20mol%20%5C%20KCl%2F%20L%20%3D%20%5Cfrac%20%7B%200.14%20%5C%20mol%20%5C%20KCl%7D%7Bx%7D)
We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.
![\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}](https://tex.z-dn.net/?f=%5Cfrac%20%7B1.8%20%5C%20mol%20%5C%20KCl%2FL%7D%7B1%7D%20%3D%20%5Cfrac%7B0.14%20%5C%20mol%20%5C%20KCl%7D%7Bx%7D)
![1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl](https://tex.z-dn.net/?f=1.8%20%5C%20mol%20%5C%20KCl%2F%20L%20%2Ax%20%3D%201%2A0.14%20%5C%20mol%20%5C%20KCl)
![1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl](https://tex.z-dn.net/?f=1.8%20%5C%20mol%20%5C%20KCl%2F%20L%20%2Ax%20%3D%200.14%20%5C%20mol%20%5C%20KCl)
Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.
![\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}](https://tex.z-dn.net/?f=%5Cfrac%20%7B1.8%20%5C%20mol%20%5C%20KCl%2F%20L%20%2Ax%7D%7B1.8%20%5C%20mol%20%5C%20KCl%2FL%7D%20%3D%20%5Cfrac%7B0.14%20%5C%20mol%20%5C%20KCl%7D%7B1.8%20%5C%20mol%20%5C%20KCl%2FL%7D)
![x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B0.14%20%5C%20mol%20%5C%20KCl%7D%7B1.8%20%5C%20mol%20%5C%20KCl%2FL%7D)
The units of moles of potassium chloride cancel.
![x= \frac{0.14 }{1.8 L}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B0.14%20%7D%7B1.8%20L%7D)
![x=0.07777777778 \ L](https://tex.z-dn.net/?f=x%3D0.07777777778%20%5C%20L)
The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.
![x \approx 0.078 \ L](https://tex.z-dn.net/?f=x%20%5Capprox%200.078%20%5C%20L)
There are approximately <u>0.078 liters of solution.</u>
Answer:
249 L
Explanation:
Step 1: Write the balanced equation
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)
Step 2: Calculate the moles of CO₂ produced from 5.00 moles of C₃H₈
The molar ratio of C₃H₈ to CO₂ is 1:3. The moles of CO₂ produced are 3/1 × 5.00 mol = 15.0 mol
Step 3: Convert "30.0°C" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 30.0°C + 273.15 = 303.2 K
Step 4: Calculate the volume of carbon dioxide
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 15.0 mol × 0.0821 atm.L/mol.K × 303.2 K/1.50 atm
V = 249 L