Here you go! There are 0.9307 moles in 123.0 g of the compound. I solved this by using a fence post method. I calculated the number of grams in one mol of (NH4)2 SO4 and got 132.16.

I did this by finding the atomic mass of each element on the periodic table (my work is in the color blue for this step)

After that, i divided the given mass by the mass of one mol of the compound.

The answer is 0.9307 moles!! I hope this helped you! :))

7 0

2 years ago

How many liters of space will 7.80 moles of methane gas (CH4) occupy at STP

Alchen [17]

Answer:

V CH4(g) = 190.6 L

Explanation:

assuming ideal gas:

PV = RTn

∴ STP: T =298 K and P = 1 atm

∴ R = 0.082 atm.L/K.mol

∴ moles (n) = 7.80 mol CH4(g)

∴ Volume CH4(g) = ?

⇒ V = RTn/P

⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)

⇒ V CH4(g) = 190.6 L

4 0

3 years ago

5.2 X 1026 molecules of CH4

olya-2409 [2.1K]

<h3 /><h3>Answer above </h3>

I cant send this with no text 

Have a nice day

8 0

2 years ago

How many moles of ammonia are produced when 5.0 moles of hydrogen react with excess nitrogen? Use this balanced equation for the

wel

3 mol H₂ → 2 mol NH₃

5 mol H₂ → x mol NH₃

x=2*5.0/3=3.3

n(NH₃)=3.3 mol

7 0

3 years ago