Answer:
The answer to your question is: density = 0.993 g/ml
Explanation:
Data
mass beaker empty = 29.3 g
volume of liquid = 15 ml
mass beaker + liquid = 44.2 g
Formula
density = mass / volume
Process
mass of liquid = 44.2 - 29.3
= 14.9 g
density = 14.9 / 15
= 0.993 g/ml
molar concentration of AgNO₃ solution = 0.118 mole/L
Explanation:
Because we have the volume of the solution and there is no information about the density of the solution I will asume that you ask for the molar concentration.
number of moles = mass / molecular weight
number of moles of AgNO₃ = 10 / 170 = 0.0588
molar concentration = number of moles / volume (L)
molar concentration of AgNO₃ solution = 0.0588 / 0.5
molar concentration of AgNO₃ solution = 0.118 mole/L
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molar concentration
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Answer:
use the rule of speed
Explanation:
speed =distance over time
Answer: 1, magnesium & mg
2, two energy shells
Explanation:
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane