Answer:
Formula is NA2SO4•10H2O
Sodium sulfate decahydrate.
Explanation:
Sodium sulfate is soluble in water. Soduim sulphare form various hydrates, so if the solution is open to the atmosphere for a week at least a lot of the water will have evaporated leaving behind a solid hydrate of Soduim sulfate.
Mass of anhydrous Na2SO4 = molar mass * number of moles
Molar mass = (23*2) + (32*1) + (16*4)
= 146 g/mol
Mass = 146*1
= 146 g of NA2SO4
NA2SO4•nH2O --> Na2SO4 + nH2O
Molar mass of hydrate, NA2SO4.nH2O
= (142 + 18n) g/mol
Mass of NA2SO4.nH2O = (142 + 18n)*1
= (142 + 18n) g
Mass of the residue = 322.2 g
Therefore, 142 + 18n = 322.2
18n = 180.2
n = 10
Formula is NA2SO4•10H2O
Sodium sulfate decahydrate.
0.0588 mol MgSO4 / 0.0588 = 1 mol MgSO4
0.412 mol H2O / 0.0588 = 7 mol H2O
Formula is MgSO4•7H2O
10. The first one is red, on the spectrum longest to shortest is red, orange, yellow, green, blue and violet
11. Judging by the diagram I believe its radar, I'm not sure though.
12. Gamma Rays
13. Ultraviolet
14. Decreases
15. Sound waves
That's all I can answer for now since I have to go but I hope this helped.
I’m pretty sure its b or a
Answer:
29.75 Kg of NH₃
Solution:
In order to calculate the theoretical yield, first we will identify the limiting reactant.
According to equation,
6 g (3 moles) H₂ requires = 28 g (1 mole) N₂
So,
5250 g H₂ will require = X g of N₂
Solving for X,
X = (5250 g × 28 g) ÷ 6 g
X = 25433 g of N₂
Hence, it is found that H₂ is the limiting reactant because N₂ is provided in excess (32700 g). Therefore,
As,
6 g (3 mole) H₂ produced = 34 g ( 2 moles) of NH₃
So,
5250 g H₂ will produce = X g of NH₃
Solving for X,
X = (5250 g × 34 g) ÷ 6 g
X = 29750 g of NH₃
Or,
X = 29.75 Kg of NH₃
