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3 years ago

52. Classify each compound as ionic or molecular. If it is ionic,

Chemistry

1 answer:

Darina [25.2K]3 years ago

8 0

Answer:

Explanation:

1...............

CoCl₂ is an ionic compound

CF₄ is molecular

BaSO₄ is an ionic compound

NO is molecular

An ionic compound is a compound which forms between a metal and non-metal. It involves the formation of a bond between two ions of significant electronegative difference. It results from the electrostatic force of attraction between two atoms in which one has lost an electron(metal) and the other that has gained an electron(non-metal).

CoCl₂ is an ionic compound

BaSO₄ is also ionic

A molecular compound is formed as a result of covalent bonds between atoms. This is due to the fact that the two atoms have similar electronegativities and would both pull the shared electrons closely. Most molecules are usually formed this way.

CF₄ and NO are both molecular compounds

2...... The ionic compounds are the ones containing metals:

In CoCl₂ the metal is Co, cobalt

BaSO₄ has Ba

Co is a transition metal. Transition elements are known to have various oxidation states. This implies that they can form more than one type of ion. The oxidation number of Co are -1, 0, +1, +2, +3 and +4

Ba will only form one type of ion as an alkaline earth metal. It has an oxidation state of +2 only. It will only form one type of ion.

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A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

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2 years ago

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Nitrogen dioxide gas is formed by reaction of nitrogen monoxide gas and oxygen gas

NeTakaya

Answer:

For the formation of nitrogen dioxide, the reactants used are nitrogen monoxide and oxygen gas. 2 moles of nitrogen monoxide gas react with 1 mole of oxygen gas to produce 2 moles of nitrogen dioxide gas.

Explanation:

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2 years ago

A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is

Charra [1.4K]

Answer:

$[H^{+}] = 0.761 \frac{mol}{L}$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

$pH = 0.119$

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

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