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Contact [7]
3 years ago
15

If the hydrogen ion concentration, [H+], in a solution is 4.59 x 10-6 M, what is [OH-]?

Chemistry
1 answer:
atroni [7]3 years ago
6 0

Answer:

2.18x10^-9 M

Explanation:

From the question given,

Hydrogen ion concentration, [H+] = 4.59x10^-6 M

Hydroxide ion, [OH-] =?

The hydroxide ion concentration, [OH-] in the solution can be obtained as follow:

[H+] x [OH-] = 1x10^-14

4.59x10^-6 x [OH-] = 1x10^-14

Divide both side by 4.59x10^-6

[OH-] = 1x10^-14 / 4.59x10^-6

[OH-] = 2.18x10^-9 M

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If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3
ss7ja [257]

Answer:- 27.7 grams of PbI_2 are produced.

Solution:- The balanced equation is:

Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of Pb(NO_3)_2 = 207.2+2(14.01)+6(16)  = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2 = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})

= 39.0gPbI_2

18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})

= 27.7gPbI_2

From above calculations, NaI gives least amount of PbI_2, so the answer is, 27.7 g of PbI_2 are produced.

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Why do isotope decay
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Isotopes decays because they are unstable. Stable isotopes do not decay.

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