15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.
<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>
Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2
<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO
</em>
Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>
Answer:
A
Explanation:
I looked up aromatic hydrocarbon and this one looks like a replica of benzene
Answer:
<em>In equipment reliability, the likelihood that a given event will occur</em>
Explanation:
Answer:
8.354 nanometers
Explanation:
To treat a diffusive process in function of time and distance we need to solve 2nd Ficks Law. This a partial differential equation, with certain condition the solution looks like this:

Where Cs is the concentration in the surface of the solid
Cx is the concentration at certain deep X
Co is the initial concentration of solute in the solid
and erf is the error function
Then we solve right side,

And we need to look up the inverse error function of 0.001964 resulting in: 0.00174055
Then we solve for x:

<u>Answer:</u> The amount of water required to prepare given amount of salt is 398.4 mL
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

We are given:
Molarity of solution = 0.16 M
Given mass of manganese (II) nitrate tetrahydrate = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Putting values in above equation, we get:

Volume of water = Volume of solution = 398.4 mL
Hence, the amount of water required to prepare given amount of salt is 398.4 mL