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Sveta_85 [38]
2 years ago
8

44 mL of H2CrO4 is needed to neutralize 176mL of a 10M solution of Pd(OH)4. What is the molarity of this solution?

Chemistry
1 answer:
balandron [24]2 years ago
3 0

Answer: The molarity of this solution is 2.5 M.

Explanation:

Given : V_1 = 44 mL,           V_2 = 176 mL

M_1 = 10 M

Therefore, the value of molarity of the solution is calculated using following formula.

M_1 \times V_1 = M_2 \times V_2

Substitute the values into above formula as follows.

M_1 \times V_1 = M_2 \times V_2\\10 M \times 44 mL = M_2 \times 176 mL\\M_2 = \frac{10 M \times 44 mL}{176 mL}\\= 2.5 M

Hence, molarity of the solution is 2.5 M.

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A reddish layer of rust can form on old iron nails and bikes. In this chemical reaction, iron reacts slowly with oxygen in the a
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Answer:

Iron slowly reacts with oxygen and forms rust. In this case, the reactants are iron and oxygen. The product is rust, or iron oxide. The chemical equation looks like this:

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The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti
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NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}

Here the base is a benzoate ion, which is a weak base and reacts with water.

C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or [OH^{-}] = 10^{^{-pOH}}

[OH^{-}] = 10^{^{-4.96}}

[OH^{-}] = 1.1\times 10^{-5}

The base dissociation equation kb = \frac{Product}{Reactant}

kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = 1.6\times 10^{-10}

And value of [OH-] we have calculated as 1.1\times 10^{-5} and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}

1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}

[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}

[C6H5COO^{-}] = 0.76 M or 0.76\frac{mol}{L}

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = 0.76(\frac{mol}{L}) \times (0.50L)

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

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