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aleksklad [387]
2 years ago
5

Phosphine, an extremely poisonous and highly reactive gas, will react with oxygen to form tetraphosphorus decaoxide and water. P

H3(g) + O2(g) → P4O10(s) + H2O(g) [unbalanced]Calculate the mass of P 4O 10( s) formed when 225 g of PH 3 reacts with excess oxygen
Chemistry
1 answer:
pantera1 [17]2 years ago
3 0

Answer:

The mass of P₄O₁₀ (s) formed is 471.03 g

Explanation:

4PH₃(g) + 8O₂(g) →  P₄O₁₀ (s)  + 6H₂O (g)

This is the ballanced equation.

Every 4 moles of phosphine, 1 mol of tetraphosphorus decaoxide is generated.

Moles of PH₃ = Mass PH₃ / Molar mass PH₃

Moles PH₃ = 225 g / 33.9 g/m

Moles PH₃ = 6.63 moles

So the rule of three, will be:

If 4 moles of PH₃ generate 1 mol  P₄O₁₀

Then, 6.63 moles of  PH₃ generate (6.63  .1)/4 = 1.66 moles

If we want to know the mass:

Moles . Molar mass = mass

Molar mass  P₄O₁₀  = 283.88 g/m

1.66m  . 283.88 g/m = 471.03 g

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3 years ago
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A mixture is made of 40 ml of salt water to 200 ml of solution. What percent of the solution is salt water?
Serggg [28]

Answer:

16.7%

Explanation:

40 ml of salt water + 200 ml of solution = 240 ml

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2 years ago
Which statement is correct about molarity and volume?
Oksana_A [137]

Answer:

Explanation:

If the choices are:

A. Molarity is defined as moles of solute per liter of solvent.

B. % by mass is defined as grams of solute per 100 g of solvent.

C. % by volume is defined as grams of solute per 100 L of solution.

D. Molarity is defined as moles of solute per liter of solution.

E. All of the above.

then the ans is E

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If greenhouse gases in the atmosphere act like a blanket around the Earth, how is this "blanket” being changed by human activiti
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!!!Need answer for chem homework ASAP PLS !!!!
stealth61 [152]

Answer:

The answer to your question is: kc = 6.48

Explanation:

Data

             Given                 Molecular weight

CaO =    44.6 g                    56 g

CO₂ =     26 g                       44 g

CaCO₃ = 42.3 g                  100 g

Find moles

        CaO                 56 g ----------------  1 mol

                                 44.6 g --------------   x

                               x = (44.6 x 1) / 56 = 0.8 mol

        CO₂                 44 g -----------------  1 mol

                                26 g ----------------    x

                               x = (26 x 1 ) / 44 = 0.6  moles

        CaCO₃            100 g ---------------   1 mol

                                42.3g --------------    x

                              x = (42.3 x 1) / 100 = 0.423 moles

       

Concentrations

 

        CaO     =    0.8 / 6.5  = 0.12 M

         CO₂    =     0.6 / 6.5 = 0.09 M

        CaCO₃  =   0.423 / 6.5 = 0.07 M

Equilibrium constant =   \frac{[products]}{[reactants]}

Kc = [0.07] / [[0.12][0.09]

Kc = 0.07 / 0.0108

kc = 6.48

7 0
3 years ago
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