I need help on this.

Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below.

zavuch27 [327]

15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.

Step 1. Convert grams of O_2 to moles of O_2

Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2

Step 2. Use the molar ratio of HgO:O_2 to convert moles of O_2 to moles of HgO

Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = 15.63 mol HgO

7 0

3 years ago

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Please help me with this question please!!!

Deffense [45]

Answer:

A

Explanation:

I looked up aromatic hydrocarbon and this one looks like a replica of benzene

3 0

3 years ago

8. What is probability?

nata0808 [166]

Answer:

In equipment reliability, the likelihood that a given event will occur

Explanation:

6 0

3 years ago

If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperature of 1100˚C for 8 hours, wh

RideAnS [48]

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:

$\frac{C_{s}-C{x}}{C_{s}-C_{o}}=erf(x/2\sqrt{D*t})$

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,

$\frac{C_{s}-C{x}}{C_{s}-C_{o}}=\frac{1018atoms/cm3-1016atoms/cm3}{1018atoms/cm3}=0.001964$

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

Then we solve for x:

$x=0.00174055*2*\sqrt{D*t} =0.00174055*2*\sqrt{2*10^{-12}cm^{2}/s*8h*3600s/h}=8.35464*10^{-7}cm$

6 0

3 years ago

If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount

Schach [20]

Answer: The amount of water required to prepare given amount of salt is 398.4 mL

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

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:

$0.16M=\frac{16\times 1000}{251\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{16\times 1000}{251\times 0.16}=398.4mL$

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

4 0

3 years ago