Answer:
75.9mL of 2.75M LiOH solution
Explanation:
Molarity is an unit of concentration defined as the ratio between moles and liters. And the molar mass of LiOH is 23.95g/mol. With this information, it is possible to know the volume of solution you should add to supply 5.0g, thus:
5.0g LiOH × (1mol / 23,95g) × (1L / 2.75mol) × (1000mL / 1L) = <em>75.9mL of 2.75M LiOH solution</em>
<em />
I hope it helps!
<span>Answer: option B. 3.07 g
Explanation:
1) given reaction:
S(s) + O₂ (g) → SO(g)
2) Balanced chemical equation:
</span><span>2S(s) + O₂ (g) → 2SO(g)
3) Theoretical mole ratios:
2 mol S : 1 mol O₂ : 2 mol SO
3) number of moles of 4.5 liter SO₂ at</span><span> 300°C and 101 kPa
use the ideal gas equation:
pV = nRT
with V = 4.5 liter
p = 101 kPa
T = 300 + 273.15 K = 573.15 K
R = 8.314 liter×kPa / (mol×K)
=> n = pV / (RT) =
n = [101 kPa × 4.5 liter] / [8.314 (liter×kPa) / (mol×K) × 573.15 K ]
n = 0.0954 mol SO
4) proportion with the theoretical ratio S / SO
2 mol S x
-------------- = ----------------------
2 mol SO 0.0954 mol SO
=> x = 0.0954 mol S.
5) Convert mol of S to grams by using atomic mass of S = 32.065 g/mol
mass = number of moles × atomic mass
mass = 0.0954 mol × 32.065 g/mol = 3.059 g of S
6) Therefore the answer is the option B. 3.07 g
</span>
The purpose of the experiment is to see what the different areas effect the plant.
I believe your answer is the second option.
40g(1mol NH3/17.04g per mol), N2's molar mass=28.02g
3CuO+2NH3=>3Cu+N2+3H2O
(2.3473mol NH3)(1 mol N2/2 mol NH3)(28.02g/1mol)=32.8873g N2
15.5/32.8873=47.1306% yield
