Answer:
<h3>The answer is 0.75 g/mL</h3>Explanation:
The density of a substance can be found by using the formula
From the question
mass = 90 g
volume = 120 mL
We have
We have the final answer as
<h3>0.75 g/mL</h3>Hope this helps you
What splits into carbon and oxygen atoms
1 answer:
You might be interested in
The standard ambient temperature and pressure are
Temperature =298 K
Pressure = 1atm
The density of gas is 1.5328 g/L
density = mass of gas per unit volume
the ideal gas equation is
PV = nRT
P = pressure = 1 atm
V = volume
n = moles
R= gas constant = 0.0821 Latm/mol K
T = 298 K
moles = mass / molar mass
so we can write
n/V = density / molar mass
Putting values
Thus molar mass of gas is 37.50g/mol
Answer : The excess reactant in the combustion of methane in opem atmosphere is molecule.
Solution : Given,
Mass of methane = 23 g
Molar mass of methane = 16.04 g/mole
The Net balanced chemical reaction for combustion of methane is,
First we have to calculate the moles of methane.
=
= 1.434 moles
From the above chemical reaction, we conclude that
1 mole of methane react with the 2 moles of oxygen
and 1.434 moles of methane react to give moles of oxygen
The Moles of oxygen = 2.868 moles
Now we conclude that the moles of oxygen are more than the moles of methane.
Therefore, the excess reactant in the combustion of methane in open atmosphere is molecule.