There are 1000 mg in 1 g
and there are 1000 g in 1 kg
Start by converting 1.34 mg to grams by dividing 1.34 mg by 1000 g = 0.00134 g
Then convert 0.00134 g to kg by dividing 0.00134 g by 1000 kg = 1.34×10^-6 kg OR 0.00000134 kg
Given:
n = 12 moles of oxygen
T = 273 K, temperature
p = 75 kPa, pressure
Use the ideal gas law, given by

where
V = volume
R = 8.3145 J/(mol-K), the gas constant
Therefore,

Answer: 0.363 m³
You add the protons and the nuetrons together i think
Answer:
In the final solution, the concentration of sucrose is 0.126 M
Explanation:
Hi there!
The number of moles of solute in the volume taken from the more concentrated solution will be equal to the number of moles of solute in the diluted solution. Then, the concentration of the first solution can be calculated using the following equation:
Ci · Vi = Cf · Vf
Where:
Ci = concentration of the original solution
Vi = volume of the solution taken to prepare the more diluted solution.
Cf = concentration of the more diluted solution.
Vf = volume of the more diluted solution.
For the first dillution:
26.6 ml · 2.50 M = 50.0 ml · Cf
Cf = 26.6 ml · 2.50 M / 50.0 ml
Cf = 1.33 M
For the second dilution:
16.0 ml · 1.33 M = 45.0 ml · Cf
Cf = 16.0 ml · 1.33 M / 45.0 ml
Cf = 0.473 M
For the third dilution:
20.0 ml · 0.473 M = 75.0 ml · Cf
Cf = 20.0 ml · 0.473 M / 75.0 ml
Cf = 0.126 M
In the final solution, the concentration of sucrose is 0.126 M
Because they don't care and think the ozone layer is fine.