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3 years ago

Chemical equation for SO + H2O

Chemistry

1 answer:

Zanzabum3 years ago

3 0

Do you mean SO2? if so its <span>SO2 + H2O =====> H2SO3 </span>

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2 years ago

You are on an alien planet where the names for substances and the units of measures are very unfamiliar. Nonetheless, you obtain

Margarita [4]

Answer: 12

Explanation.

This problem was recently posted and deals with the conversion of units,so I give the same answer.

You need to convert the 9 quibs of skvarnick to sleps, to determine how many sleps you have.

The conversion of units is easily undertaken once you build your conversion factors.

Conversion factors are ratios (fractions) equivalent to one, which permits use the identity property of the multiplication to change one unit to a different equivalent one.

Remember that the identity property of multiplication states that any amount multiplied by 1 remains unchanged, i.e.

$a.1=a$

Then, build your conversion factor from the definitions given:

12 sleps is equal to 9 quibs.

12 sleps = 9 quibs

in virtue of the division property of equality, you can divide both sides by 12 sleps and get:

1 = 12 sleps / 9 quibs ← conversion factor

Now you can multiply the amount 9 quibs by the conversion factor, which is equal to 1, to find the equivalent amount of sleps:

identity property: 9 quibs = 9 quibs × 1

substitution property: 9 quibs = 9 quibs × 12 sleps/ 9 quibs

You can see that the unit quibs appear in the numerator and the denominator so it cancels out, and the result will be in sleps.

So, all you have to do now is the operations with the numbers:

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3 years ago

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3 years ago

Ethylenediamine (en) forms an octahedral complex with Ni2+(aq) with the formula [Ni(en)3]2+. Ni2+(aq) + 3 en ⇌ [Ni(en)3]2+(aq) K

yawa3891 [41]

Answer:

$3.125\times 10^{-19} mol/L$ is the concentration of $Ni^{2+}(aq)$ in the solution.

Explanation:

$Ni^{2+}(aq) + 3 en\rightleftharpoons [Ni(en)_3]^{2+}(aq)$

Concentration of nickel ion = $[Ni^{2+}]=x$

Concentration of nickel complex= $[[Ni(en)_3]^{2+}]=\frac{0.16 mol}{2 L}=0.08 mol/L$

Concentration of ethylenediamine = $[en]=\frac{0.80 mol}{2 L}=0.40 mol/L$

The formation constant of the complex = $K_f=4.0\times 10^{18}$

The expression of formation constant is given as:

$K_f=\frac{[[Ni(en)_3]^{2+}]}{[Ni^{2+}][en]^3}$

$4.0\times 10^{18}=\frac{0.08 mol/L}{x\times (0.40 mol/L)^3}$

$x=\frac{0.08 mol/L}{4.0\times 10^{18}\times (0.40 mol/L)^3}$

$x=3.125\times 10^{-19} mol/L$

$3.125\times 10^{-19} mol/L$ is the concentration of $Ni^{2+}(aq)$ in the solution.

6 0

3 years ago