Answer:
NO would form 65.7 g.
H₂O would form 59.13 g.
Explanation:
Given data:
Moles of NH₃ = 2.19
Moles of O₂ = 4.93
Mass of NO produced = ?
Mass of produced H₂O = ?
Solution:
First of all we will write the balance chemical equation,
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:
NH₃ : NO NH₃ : H₂O
4 : 4 4 : 6
2.19 : 2.19 2.19 : 6/4 × 2.19 = 3.285 mol
Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:
O₂ : NO O₂ : H₂O
5 : 4 5 : 6
4.93 : 4/5×4.93 = 3.944 mol 4.93 : 6/5 × 4.93 = 5.916 mol
we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.
Mass of water = number of moles × molar mass
Mass of water = 3.285 mol × 18 g/mol
Mass of water = 59.13 g
Mass of nitrogen monoxide = number of moles × molar mass
Mass of nitrogen monoxide = 2.19 mol × 30 g/mol
Mass of nitrogen monoxide = 65.7 g
17, to be neutral it has to have equal number of protons and electrons
<u>Answer:</u> The molar solubility of
is ![1.25\times 10^{-3}mol/L](https://tex.z-dn.net/?f=1.25%5Ctimes%2010%5E%7B-3%7Dmol%2FL)
<u>Explanation:</u>
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The balanced equilibrium reaction for the ionization of calcium fluoride follows:
![PbI_2\rightleftharpoons Pb^{2+}+2I^-](https://tex.z-dn.net/?f=PbI_2%5Crightleftharpoons%20Pb%5E%7B2%2B%7D%2B2I%5E-)
s 2s
The expression for solubility constant for this reaction will be:
![K_{sp}=[Pb^{2+}][I^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BPb%5E%7B2%2B%7D%5D%5BI%5E-%5D%5E2)
We are given:
![K_{sp}=7.9\times 10^{-9}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D7.9%5Ctimes%2010%5E%7B-9%7D)
Putting values in above equation, we get:
![7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L](https://tex.z-dn.net/?f=7.9%5Ctimes%2010%5E%7B-9%7D%3D%28s%29%5Ctimes%20%282s%29%5E2%5C%5C%5C%5C7.9%5Ctimes%2010%5E%7B-9%7D%3D4s%5E3%5C%5C%5C%5Cs%3D1.25%5Ctimes%2010%5E%7B-3%7Dmol%2FL)
Hence, the molar solubility of
is ![1.25\times 10^{-3}mol/L](https://tex.z-dn.net/?f=1.25%5Ctimes%2010%5E%7B-3%7Dmol%2FL)
Correct question
The density of liquid mercury is 13.6 g/mL. What is its density in units of lb/in3? (2.5 cm = 1 in., 2.205 lbs= 1 kg., 1000 g =1 kg, 1 mL = 1 cm³)
Answer:
![\rho0.4916\ lb/in^3](https://tex.z-dn.net/?f=%5Crho0.4916%5C%20lb%2Fin%5E3)
Explanation:
Given that;-
The density = 13.6 g/mL
Also, 1 kg = 2.205 lb
1 kg = 1000 g
So, 1000 g = 2.205 lb
1 g = 0.002205 lb
Also,
1 in = 2.54 cm
1 in³ = 16.39 cm³
1 cm³ = 1 mL
So, 1 in³ = 16.39 mL
1 mL = 0.061 in³
The expression for the calculation of density is shown below as:-
![\rho=\frac{m}{V}](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%7Bm%7D%7BV%7D)
Thus,
![\rho=\frac{13.6\ g}{1\ mL}=\frac{13.6\times 0.002205\ lb}{0.061\ in^3}=0.4916\ lb/in^3](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%7B13.6%5C%20g%7D%7B1%5C%20mL%7D%3D%5Cfrac%7B13.6%5Ctimes%200.002205%5C%20lb%7D%7B0.061%5C%20in%5E3%7D%3D0.4916%5C%20lb%2Fin%5E3)