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Bond [772]
4 years ago
11

During the lab, you measured the pH of common household items. Write net Brønsted equations that show the acidic or basic nature

of the following substances, given the pH. For polyprotic acids, only show one proton transfer. Remember that spectator ions are not included. (Use the lowest possible coefficients. Omit states-of-matter in your answer.) (a) native lime (calcium oxide) with a pH of 10.50
Chemistry
1 answer:
tamaranim1 [39]4 years ago
8 0

Answer:

CaO + H₂O ⟶ Ca²⁺ + 2OH⁻

Explanation:

CaO + H₂O  ⟶     Ca²⁺   +   2OH⁻

base    acid       conj. acid  conj. base

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the god ran down the street

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3 years ago
3. What are the energy levels of electrons?
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Energy levels inside an atom are the specific energies that electrons can have when occupying specific orbitals.
7 0
3 years ago
How many atoms of Oxygen are there in 3 C1207?
natka813 [3]

Answer:

9 atoms

Explanation:

Explanation: In 1 formula unit of Al(NO3)3 , there are (clearly!) 9 atoms of oxygen, 3 nitrogen atoms, and 1 aluminum atom. I have gone on before that the mole ( NA , Avogadro's number) is simply a much larger number, i.e. NA = 6.022×1023 .

7 0
3 years ago
Read 2 more answers
White phosphorus (P4) is used to produce phosphorus oxides, such as P4O10. How many kilograms of P4O10 would be produced from 41
eduard

Answer:

0.93956kg

Explanation:

Equation for the reaction is given as;

P4 + 5O2 --> P4O10

From the equation, 1 mole of P4 produces 1 mole of P4O10.

Mass = Number of moles * molar mass

Mass of P4 = 1 mol * 123.88 g/mol = 123.88 grams

Mass of P410 = 1 mol * 283.886 g/mol = 283.886 g

This means 123.88 grams of P4 produced 283.886 g of P4O10.

How many kilograms of P4O10 would be produced from 410. g of pure P4 ?

123.88 = 283.886

410 = x

x = (410 * 283.886 ) / 123.88

x = 939.56g

Upon converting to Kilogram by dividing by 1000

x = 0.93956kg

8 0
4 years ago
What is the solubility in moles/liter for chromium(III) iodate at 25 oC given a Ksp value of 5.0 x 10-6. Write using scientific
Radda [10]

<u>Answer:</u> The solubility of chromium (III) iodate is 2.07\times 10^{-2}M

<u>Explanation:</u>

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of chromium (III) iodate follows:

Cr(IO_3)_3(aq.)\rightleftharpoons Cr^{3+}(aq.)+3IO_3^-(aq.)  

                                s                3s

The expression of K_{sp} for above equation follows:

K_{sp}=s\times (3s)^3

We are given:  

K_{sp}=5.0\times 10^{-6}

Putting values in above expression, we get:

5.0\times 10^{-6}=s\times (3s)^3\\\\s=2.07\times 10^{-2}M

Hence, the solubility of chromium (III) iodate is 2.07\times 10^{-2}M

6 0
3 years ago
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