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hram777 [196]
2 years ago
14

How many grams of P2Cl5 are produced if 2.00 moles of P reacts with an excess of Cl2?

Chemistry
1 answer:
Roman55 [17]2 years ago
5 0

Answer:

m_{P_2Cl_5 }=239.2125gP_2Cl_5

Explanation:

Hello there!

In this case, according to the reaction:

4P+5Cl_2\rightarrow 2P_2Cl_5

Thus, since there is a 4:2 mole ratio of P to P2Cl5 and the molar mass of the later is 239.2125 g/mol, we obtain the following mass as the produced one:

m_{P_2Cl_5 }=2.00molP*\frac{2molP_2Cl_5}{4molP} *\frac{239.2125gP_2Cl_5 }{1molP_2Cl_5} \\\\m_{P_2Cl_5 }=239.2125gP_2Cl_5

Regards!

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Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of
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The first thing we need to do here is to recognize the unit of molarity and the units of the given percentage of nitric acid.

Molarity is mol HNO3 / L of solution. This is our aim

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What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

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