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Nataliya [291]
3 years ago
10

Calculate the concentration of OH-in a solution that has a concentration of H+ = 8.1 x 10^−6 M at 25°C. Multiply the answer you

get by 1010 and enter that into the field to 2 decimal places.
Chemistry
1 answer:
Nat2105 [25]3 years ago
7 0

Answer:

The answer is 12.35

Explanation:

From the question we are given that the concentration of H^{+} is 8.1 * 18^{-6}M

 Generally The rate equation is given as

                                                           K_{w} = [H^{+} ][OH^{-} ]

and K_{w} the rate constant has a value 1 * 10^{-14}

     Substituting and making [OH^{-}] the subject we have

                                                 [OH^{-} ] = \frac{1 * 10^{-14}}{[H^{+}]} = \frac{1 * 10^{-14}}{8.1 *10^{-6}} =1.235 * 10^{-9}

                                                  [OH ^ {-}] = 1.235 * 10^{-9}M

                            Multiply the value by 10^{10} as instructed from the question we have  

                       Answer =   1.235 * 10 ^{-9} * 10^{10} = 12.35

Hence the answer in 2 decimal places is 12.35

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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0
3 years ago
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