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Andrei [34K]
3 years ago
6

PLEASE HELPWhat do you know about valence electrons?

Chemistry
2 answers:
Ahat [919]3 years ago
6 0

Answer:

Valence electrons are the electrons in the outermost shell, or energy level, of an atom. For example, oxygen has six valence electrons, two in the 2s subshell and four in the 2p subshell. We can write the configuration of oxygen's valence electrons as 2s²2p⁴

k0ka [10]3 years ago
4 0

Answer:

Valence electrons are the electrons in the outermost shell, or energy level, of an atom

Explanation:

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An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH
Burka [1]

Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
8 0
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All of the halogens, group 17, have seven valence electrons. which of these would represent the oxidation number of the halogens
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but when it react with other element it has an oxidation number of -1
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John was considered to be a pillar in the early church?
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Read 2 more answers
1) For the following reaction, 8.44 grams of carbon (graphite) are allowed to react with 9.63 grams of oxygen gas.C(graphite)(s)
Yuki888 [10]

Answer:

The answers to the question are

(a) 13.24 g

(b) (O₂)

(c) 4.8252 g

(2) 0.662 M/L

Explanation:

(a) To solve the question we write the equation as follows

C + O₂ → CO₂

That is one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide

number of moles of graphite = 8.44/12 = 0.703 moles

number of moles of oxygen = 9.63/32 = 0.3009

However since one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide, therefore, 0.3009 moles of oxygen will react with 0.3009 moles of  carbon to fore 0.3009 moles of  CO₂

The maximum mass of carbon dioxide that  can be formed = mass = moles × molar mass

= 0.3009×44 = 13.24 g

(b) The formula for  the limiting reagent (O₂)

Finding the limting reagent is by checking the mole balance of the reactants available to the moles specified in th stoichiometry of the reaction and selecting the reagent with the list number of moles

(c) The mass of excess reagent = 0.703 moles - 0.3009 moles = 0.4021 moles

mass of excess reagent = 0.4021 × 12 = 4.8252 g

(2) The molarity is given by number of moles per liter of solution, therefore

molar mass of mgnesium iodide = 278.1139 g/mol, number of moles of magnesium iodide in 23 g = 23g/ 278.1139 g/mol= 8.3 × 10⁻² M

Therefore the moles in 125 mL = (8.3 × 10⁻² M)/(125 mL) = (8.3 × 10⁻² M)/(0.125 L) = 0.662 M/L

3 0
3 years ago
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