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Kryger [21]
3 years ago
11

Determine the [OH−] of a solution that is 0.150 M in CO2−3. Kb(CO32−)=1.8×10−4

Chemistry
1 answer:
Novay_Z [31]3 years ago
5 0

The [OH−] of a solution : 0.0052 M

<h3>Further explanation</h3>

Kb is the base ionization constant  

LOH (aq) ---> L⁺ (aq) + OH⁻ (aq)  

\rm Kb=\dfrac{[L][OH^-]}{[LOH]}

Dissociation of weak base of CO₃²⁻ :

Reaction

CO₃²⁻ + H₂O⇒ HCO₃⁻ + OH⁻

use ICE method :

I : 0.15               0             0

C : x                   x             x

E : 0.15-x            x            x

\tt Kb=\dfrac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}\\\\1.8\times 10^{-4}=\dfrac{x^2}{0.15-x\approx 0.15}\\\\x^2=1.8\times 10^{-4}\times 0.15\\\\x=0.0052

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Answer:

light is the result of electrons moving between defined energy levels in an atom called shells.

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3 0
3 years ago
The combustion of 135 mg of a hydrocarbon produces 440 mg of CO2 and 135 mg H2O. The molar mass of the hydrocarbon is 270 g/mol.
NeX [460]

Answer:

Molecular formula = C20H30

Explanation:

NB 440mg = 0.44g, 135mg= 0.135g

From the question, moles of CO2= 0.44/44= 0.01mol

Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01

Also from the question, moles of H2O = 0.135/18= 0.0075mole

Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H

To get the empirical formula, divide by smallest number of mole

Mol of C = 0.01/0.01=1

Mol of H = 0.015/0.01= 1.5

Multiply both by 2 to obtain a whole number

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Mol of H= 1.5×2 = 3

Empirical formula= C2H3

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5 0
3 years ago
Why is the answer C for this problem?
pshichka [43]

Answer:

\boxed{\text{(C) X}$_{3}$P$_{2}}

Explanation:

Step 1. Identify the Group that contains X

We look at the consecutive ionization energies and hunt for a big jump between them

\begin{array}{crc}n & IE_{n} & IE_{n} - IE_{n-1}\\1 & 730 & \\2 & 1450 & 720\\3 & 7700 & 6250\\4 & 10500 & 2800\\\end{array}

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Step 2. Identify the Compound

X can lose two valence electrons to reach a stable octet, and P can do the same by gaining three electrons.

We must have 3 X atoms for every 2 P atoms.

The formula of the compound is \boxed{\text{X}$_{3}$P$_{2}}$}.

4 0
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