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nadya68 [22]
3 years ago
9

How many moles of H2O can be formed when 4.5 moles of NH3 react with 3.2 moles of O2?

Chemistry
1 answer:
dimaraw [331]3 years ago
3 0
Hi Sentpain! Thanks for asking a question here on Brainly. 

Solving: 

<span>4 NH3 +5 O2 = 4 NO + 6 H2O 
</span>
☛ Find the ratio for <span>NH3 and O2. That's 4:3 
</span>☛ <span>4.5 x 5 /4 = 5.6 
</span>☛ Find the ratio for <span>O2 and H2O. That's 5:6
</span>☛ <span>3.2 x 6 / 5 = 3.8
</span>
Answer: <span>3.8 mol H2O - choice A
</span>
<span>Hope that helps! ★ <span>If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia </span></span>
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Sea water contains roughly 158.0g of Nacl per liter. What is the molarity of sodium chloride in sea water?
IrinaK [193]
Molar mass NaCl = 58.5 g/mol

C = 158.0 g/L

Molarity =  C / molar mass

M = 158.0 / 58.5

M = 2.7000 M

hope this helps!
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What is the most soluble salt of the following set? What is the most soluble salt of the following set?
DiKsa [7]

Answer:

(c) AgCN\ (K_{sp} =6.0\times 10^{-17})

Explanation:

The solubility product of a solid is the amount of solid dissociates into its respective ions in the solution. Thus more the value of the Ksp, the more is the salt soluble in the solvent.

So,  Given that:-

Sn(OH)_2\ (K_{sp} =1.6\times 10^{-19})

Al(OH)_3\ (K_{sp} =1.9\times 10^{-33})

AgCN\ (K_{sp} =6.0\times 10^{-17})

Fe(OH)_3\ (K_{sp} =2.6\times 10^{-39})

The salt having highest value of Ksp is AgCN. So, it is most soluble.

4 0
3 years ago
The density of liquid mercury is 13.6 g/mL. What is its density in units of ? (2.54 cm = 1 in., 2.205 lb = 1 kg)
nalin [4]

Correct question

The density of liquid mercury is 13.6 g/mL. What is its density in units of lb/in​3​? (2.5 cm = 1 in., 2.205 lbs= 1 kg., 1000 g =1 kg, 1 mL = 1 cm³)

Answer:

\rho0.4916\ lb/in^3

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}

Thus,

\rho=\frac{13.6\ g}{1\ mL}=\frac{13.6\times 0.002205\ lb}{0.061\ in^3}=0.4916\ lb/in^3

7 0
3 years ago
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