Answer:
80 ml
Explanation:
Moles to start = .04 L * .6 M/L = .024 moles
this is how much must be in the final dilution
.2 moles / liter * x liter = .024 moles
x = .024 / .2 = .12 liter final volume
start with . 040 then add .080 to get .12 liter
.080 liter = 80 ml
Answer:
V₂ = 167 cm₃
Explanation:
Because you are only dealing with volume and temperature, you can use Charles' Law to find the missing value. This formula looks like this:
V₁ / T₁ = V₂ / T₂
In this formula, "V₁" and "T₁" represent the initial volume and temperature. "V₂" and "T₂" represent the final volume and temperature. You have been given values for all of the variables except for final volume. Therefore, by plugging these values into the formula, you can simplify to find your answer.
V₁ = 100 cm³ T₁ = 150 °C
V₂ = ? T₂ = 250 °C
V₁ / T₁ = V₂ / T₂ <---- Charles' Law formula
(100 cm³) / (150 °C) = V₂ / (250 °C) <---- Insert values
(0.667) = V₂ / (250 °C) <---- Simplify left side
(0.667) x (250 °C) = V₂ <---- Rearrange
167 = V₂ <---- Simplify left side
Answer:
0.536 mol HNO₃
Explanation:
Step 1: Convert 37.0 g of NO₂ into mol NO₂:
37.0 g NO₂ x 1 mol NO₂/46.0 g NO₂ = 0.8043 mol NO₂
Step 2: Convert 0.8043 mol NO₂ into mol HNO₃:
0.8043 mol NO₂ x 2 mol HNO₃/3 mol NO₂ = 0.5362 mol HNO₃
Step 3: Round to 3 significant figures:
0.5362 mol HNO₃ -> 0.536 mol HNO₃
The balanced equation for the above reaction is as follows;
2Ca + O₂ ---> 2CaO
stoichiometry of Ca to O₂ is 2:1
we first need to find the limiting reactant
number of Ca moles - 6.84 mol
number of O₂ moles - 4.00 mol
if Ca is the limiting reactant
if 2 mol of Ca reacts with 1 mol of O₂
then 6.84 mol of Ca reacts with - 6.84 / 2 = 3.42 mol of O₂
this means that Ca is the limiting reactant and O₂ is in excess
therefore amount of CaO produced depends on amount of limiting reactant present
stoichiometry of Ca to CaO is 2:2
number of moles of Ca reacted = number of CaO moles formed
number of moles of CaO formed - 6.84 mol
answer is 6.84 mol