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nataly862011 [7]
3 years ago
7

Where would you expect to find an element with the greatest electronegativity?

Chemistry
1 answer:
beks73 [17]3 years ago
4 0
As you move from left to right of the periodic table, the element becomes more negative.
You might be interested in
Sodium chloride reacts with copper sulfate to produce sodium sulfate and copper chloride. 2 upper N a upper C l (a q) plus upper
V125BC [204]

Aqueous sodium chloride,

NaCl

, will not react with aqueous copper(II) sulfate,

CuSO

4

, because the two potential products are soluble in aqueous solution.

The chemical equation given to you is actually incorrect because copper(II) chloride,

CuCl

2

, is not insoluble in aqueous solution. In fact, it is quite soluble.

This means that the reaction does not produce an insoluble solid that precipitates out of solution.

Sodium chloride and copper(II) sulfate are both soluble ionic compounds that dissociate completely in aqueous solution

Explanation:

8 0
3 years ago
Read 2 more answers
In a 74.0-g aqueous solution of methanol, CH4O, the mole fraction of methanol is 0.140. What is the mass of each component?
Salsk061 [2.6K]

Answer:

The correct answer is 16.61 grams methanol and 57.38 grams water.

Explanation:

The mole fraction (X) of methanol can be determined by using the formula,  

X₁ = mole number of methanol (n₁) / Total mole number (n₁ + n₂)

X₁ = n₁/n₁ + n₂ = 0.14

n₁ / n₁ + n₂ = 0.14 ---------(i)

n₁ mole CH₃OH = n₁ mol × 32.042 gram/mol (The molecular mass of CH₃OH is 32.042 grams per mole)

n₁ mole CH₃OH = 32.042 n₁ g

n₂ mole H2O = n₂ mole × 18.015 g/mol  

n₂ mole H2O = 18.015 n₂ g

Thus, total mole number is,  

32.042 n₁ + 18.015 n₂ = 74 ------------(ii)

From equation (i)

n₁/n₁ + n₂ = 0.14

n₁ = 0.14 n₁ + 0.14 n₂

n₁ - 0.14 n₁ = 0.14 n₂

n₁ = 0.14 n₂ / 1-0.14

n₁ = 0.14 n₂/0.86 ----------(iii)

From eq (ii) and (iii) we get,  

32.042 × 0.14/0.86 n₂ + 18.015 n₂ = 74

n₂ (32.042 × 0.14/0.86 + 18.015) = 74

n₂ = 74 / (32.042 × 0.14/0.86 + 18.0.15)

n₂ = 3.1854 mol

From equation (iii),  

n₁ = 0.14/0.86 n₂

n₁ = 0.14/0.86 × 3.1854

n₁ = 0.5185 mol

Now, presence of water in the mixture is,  

= 3.1854 mole × 18.015 gram per mole  

= 57.38 grams

Methanol present in the mixture is,  

= 0.5185 mol × 32.042 gram per mole

= 16.61 grams

8 0
3 years ago
Name the following ionic compounds that contain transition metals. Use the periodic table provided
Goryan [66]

The correct name of the compound CuCl is copper I chloride.

An ionic compound is formed two atoms when one loose electrons and the other atoms gains those electrons. Ionic bonds are actually an ion pair, a pair of oppositely charged ions held together by electrostatic interaction. Transition metals are elements with incompletely filled d orbitals. Copper is a transition element.

The compound CuCl is named copper I chloride. We can see that the copper is in the +1 oxidation state in this compound hence the name.

Learn more: brainly.com/question/1527403

4 0
2 years ago
Estimate the ph of the resulting solution prepared by mixing 1.0 mole of solid disodium phosphate (na2hpo4) and 1.25 mole of hyd
GenaCL600 [577]

The HCl added = 1.25 moles

and the moles of Na2HPO4 = 1 mole

Now when acid is added in the given solution of Na2HPO4

One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4

Na2HPO4  + H+  ---> NaH2PO4

Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of  H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution

So this will result into formation of a buffer of phosphoric acid and NaH2PO4

NaH2PO4  + H+  ---> H3PO4

pKa of H3PO4 = 2.1

so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58

so the pH will be in between 2.1 to 7.2

7 0
3 years ago
An analytical chemist is titrating 118.3 mL of a 0.3500 M solution of butanoic acid (HC3H7CO2) with a 0.400 M solution of KOH. T
TEA [102]

Answer:

pH = 12.33

Explanation:

Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.

The titration reaction is

HA + KOH ---------------------------- A⁻ + H₂O + K⁺

number of moles of HA :   118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA

number of  moles of OH  : 115.4 mL/1000ml/L x 0.400 mol/L  = 0.046 mol A⁻

therefore the weak acid will be completely consumed and what we have is  the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.

n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH

pOH = - log (KOH)

M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M

pOH = - log (0.0021) = 1.66

pH = 14 - 1.96 = 12.33

Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.

4 0
3 years ago
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